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Section P.9 Notes Page 1
P.9 Linear Inequalities and Absolute Value Inequalities
Sometimes the answer to certain math problems is not just a single answer. Sometimes a range of answers
might be the answer. In this section we will discuss the different ways to write the answers to such problems.
The table below is from the text and explains the different types of intervals. Besides writing out the intervals,
these problems will require you to represent your answer on a number line.
Parenthesis indicate that the endpoints are NOT included in an interval. Square brackets indicate the that
endpoints ARE included on the interval. Whenever the interval ends with or , parenthesis are always
used. That is because or are not exact numbers.
Section P.9 Notes Page 2
The table below from the text lists the nine possible types of intervals used to represent different types of
answers. You will notice that there are two types of ways to represent the answer. The first one is called setbuilder notation. This answer is in the form of a set, hence the { and } notation. Your answer always begins
with a { x | and then you write the statement and close it with a }. Then there is interval notation. This is
where you use the brackets and parenthesis as discussed on the previous page. The smallest number always
comes first in interval notation.
In the following problems, express each interval in set-builder notation and graph the interval on a number line.
EXAMPLE: [5, )
For this problem, we will use the table above. We look at the interval notation that was given, and match it to
the form on the table. We see that -5 would be the a. This states our set-builder notation would be:
{x | x 5} . Using the table, we can also express this answer on a number line. The table above states that our
graph will start with -5 and have an arrow pointing to the right. Therefore our correct number line graph is:
5
EXAMPLE: (, 2)
Using the table again, it says that the following is the correct set-builder notation: {x | x 2} . Using the table,
we can also express this answer on a number line. The table above states that our graph will start with 2 and
have an arrow pointing to the left. Therefore our correct number line graph is:
2
Section P.9 Notes Page 3
EXAMPLE: [4, 3)
Using the table again, it says that the following is the correct set-builder notation: {x | 4 x 3} . Using the
table, we can also express this answer on a number line. The table above states that our graph will be between
4 and 3. There is a bracket on 4 and a parenthesis on the 3:
4
3
In a previous section we covered solving linear equations. Now we will solve linear inequalities. To solve
these, just pretend like the inequality symbol is an equals sign, and isolate the variable like was done previously.
The difference is now we will represent our answer using interval notation and a number line.
EXAMPLE: 18 x 45 12 x 9
18x + 45 12x – 9
–45
–45
18x 12x – 54
_
–12x –12x
6x –54
6
6
x 9
After we isolate x, now we need to write this in interval notation using the table.
this would be written as: [, 9] .
The number line would look like this:
–9
EXAMPLE: 4( x 2) 3x 20
4 x 8 3 x 20
+8
+8
–4x 3x + 28
–3x –3x _
–7x 28
–7 –7
x 4
EXAMPLE:
Notice that to solve for x, I needed to divide by –7 in order to get a positive x.
What you also noticed was the inequality sign changed directions. This is a rule.
Whenever you multiply or divide an inequality by a negative, the inequality
sign will ALWAYS flip. It will not flip with addition or subtraction.
Interval notation: (4, ) . Number line:
4
x 8 10 x x 1
10
15
6
x 8 10 x x 1
30
15
6
10
For fraction problems like this, multiply both sides by the LCD.
30 x 8 30 10 x 30 x 1
1 10 1 15
1 6
We write the LCD next to each fraction
3( x 8) 2(10 x) 5( x 1)
3 x 24 20 2 x 5 x 1
5x – 44 5x – 1
–5x
–5x _
–44 –1
Here we reduced and the fractions were eliminated.
Add like terms on the left side of the equation.
Notice that this is not a true statement, so the answer is
NO SOLUTION.
Section P.9 Notes Page 4
Solving Compound Inequalities
These are different than the linear inequalities because now you have three sides of an equation. For these, you
will do the same operations as in linear inequalities, however you will do operations to all three sides of the
equation.
Solve the following inequalities and express the answer in interval notation and on a number line.
EXAMPLE: 8 3 x 5 17
8 3 x 5 17
–5
–5 –5
3 3x 12
3
3
3
Here I am subtracting 5 from all three sides of the equation to isolate x.
I am dividing all three sides by 3 to get a single x.
1 x 4
Now the x has been solve for. I now need to write this in interval notation and a number line. I will use the
same table I used previously to write the answers. The interval notation is: (1, 4). I want you to notice that this
is NOT a coordinate like (x, y). This is an interval that does not include the 1 or the 4. Now I will express the
answer using a number line:
1
EXAMPLE: 6 4
1
2 6 4 x 3
2
–12 8 – x –6
–8 –8
–8
–20 – x –14
–1 –1 –1
4
1
x 3
2
I will multiply everything by 2 to cancel out the fraction.
Everything was multiplied by 2, so now the fraction is gone since 2/2 = 1.
I need to divide everything by –1 so that the x is positive. Doing this will flip the
inequality sign. Remember whenever you multiply or divide by a negative it flips.
20 x 14
All I did here was flip the inequality so that the smaller number comes first.
Notice that the inequality still opens up towards the 20 and the x.
14 x 20
Now we are ready to write the interval notation and the number line using the same
table I used previously.
The interval notation is: [14, 20). Now I will do the number line:
14
20
Section P.9 Notes Page 5
Solving Absolute Value Inequalities
Assume that u is any algebraic expression and c is a positive number.
Depending on what kind of absolute value inequality you have, you will set up the problem based on:
1.) If you have u c then you want to solve the following inequality: c u c .
2.) If you have u c then you want to solve the following inequality: c u c .
3.) If you have u c then you want to solve the following inequality: u c or u c .
4.) If you have u c then you want to solve the following inequality: u c or u c .
EXAMPLE: x 2 11
In this case we have u c . Therefore we will solve the inequality u c or u c .
x 2 11 or x 2 11
+2 +2
+2
+2
x > 13 or
x < -9
Here we set up the inequality. Now we need to solve each one for x.
Now we need to write this with interval notation. Each inequality will be turned into its own interval:
(,9) (13, ) The U in the middle is to indicate an ‘or’. Now we need to express our answer as a number
line. We will express each statement on the same number line. It will look like:
-9
13
Remember that numbers that are less than x go to the left. Numbers greater than x go to the right.
EXAMPLE: 3 x 4 1 7 .
If the absolute value is not isolated, you FIRST must isolate it. We need to get it into the proper form so that we
know what equation to solve. Add 1 to both sides. You will get 3 x 4 8
In this case we have u c . Therefore we will solve the inequality c u c .
8 3x 4 8
+4
+4 +4
–4 3x 12
3
3
3
4
x4
3
Here we set up the inequality. Now we need to solve this for x.
Add 4 to all sides of the equation.
Now divide all sides by 3.
Here is our final inequality. Now we need to write this in interval notation.
4
Interval notatation: ,4 . Number line answer:
3
4
3
4
Section P.9 Notes Page 6
EXAMPLE: 2 3 x 4 8 .
If the absolute value is not isolated, you FIRST must isolate it. We need to get it into the proper form so that we
know what equation to solve. Subtract 4 from both sides. You will get 2 3 x 4 . Now divide both sides by
2. You will get: 3 x 2
In this case we have u c . Therefore we will solve the inequality c u c .
2 3 x 2
–3 –3
–3
–5 –x –1
–1 –1 –1
Here we set up the inequality. Now we need to solve this for x.
Subtract 3 from all sides of the equation.
Now divide all sides by –1 since we want to get x by itself.
5 x 1
Notice here that when we divided by a negative number, the sign switched
directions. This will always happen whenever you multiply or divide an
inequality by a negative.
1 x 5
All I did here was flip the inequality so that the smaller number comes first.
Notice that the inequality still opens up towards the 5 and the x.
Interval notatation: [1, 5]. Number line answer:
1
5
EXAMPLE: Solve: 3 x 4 0
Be careful with this one. Remember that an absolute value will ALWAYS return a positive value. Since you
will always get a number that is either zero or greater than zero, then this problem has infinite solutions
, . Any number will work. The graph would be a number line with everything shaded.
EXAMPLE: Solve: 3 x 4 0 .
For this one the only solution is when it equals zero, so you would only solve the equation 3 x 4 0 , so x is
4/3.
EXAMPLE: Solve: 3 x 4 0
Since zero is not included this would have no solution. Since an absolute value always returns a number 0 .
EXAMPLE: Solve: 3 x 4 0
In this case we have u c . Therefore we will solve the inequality 3 x 4 0 or 3 x 4 0 . Solving this
4
4
4
or x . This is saying the same thing as x . Therefore every
3
3
3
number is included as the answer EXCEPT four-thirds.
would give you the following: x
Section 1.1 Notes Page 1
1.1 Graphs and Graphing Utilities
Cartesian Coordinate System – This is the standard graphing system we will use in this course for graphing all
kinds of equations. The graph is made up of four quadrants as shown below:
The vertical axis is called the y axis and the horizontal axis is called the
x-axis. The place where the two axes come together is called the origin
and the coordinates are (0, 0). We write points in the form (x, y). A point
is also referred to as an ordered pair. The first number x represents the
horizontal change, and the second number y represents the vertical change.
You move to the right for positive x values and to the left for negative x
values. You move up for positive y values and down for negative y
values.
Plotting points To plot a point, start at (0, 0). Then move left or right depending on the x value and up or
down depending on the y value.
EXAMPLE: Plot (-1, 4).
First we start at (0, 0). Since the x value is negative we will first move
1 place to the left. Then from this spot we will move up 4 places since
the y-value is positive.
EXAMPLE: Plot (-4, -2).
First we start at (0, 0). Since the x value is negative we will first move
4 places to the left. Then from this spot we will move down 2 places since
the y-value is negative.
EXAMPLE: Plot (3, -2).
First we start at (0, 0). Since the x value is positive we will first move
3 places to the right. Then from this spot we will move down 2 places since
the y-value is negative.
EXAMPLE: Plot (0, -3).
First we start at (0, 0). Since the x value is zero we will not move in either
direction. We will stay on the y-axis. Then from this spot we will move
down 3 places since the y-value is negative.
Section 1.1 Notes Page 2
Graphing Equations by Making Tables and Plotting Points
The equations we are graphing in this section will be done by making a table of values and then plotting the
resulting points. Usually the question will tell you which values to use for x; otherwise use whichever values
for x you want. You should plot negatives, positives, and the zero. Later in this course we will learn how to
graph these without making tables.
EXAMPLE: Graph the equation y 2 x 1 . Let x = -3, -2, -1, 0, 1, 2, 3. Then indicate the intercepts.
It doesn’t matter if you’ve never graphed an absolute value before. We plot this one the same way we plot
lines. First we need to set up a table like this:
x
y 2 x 1
(x, y)
-3
-2
-1
0
1
2
3
Now what we do is plug in the x value into the equation y 2 x 1 . Then we write our answer as an
ordered pair as show below. Remember that absolute values always return a positive value. For example
5 5.
x
y 2 x 1
(x, y)
-3
y 2 (3) 1 2 4 2 4 2
(-3, -2)
-2
y 2 ( 2) 1 2 3 2 3 1
(-2, -1)
-1
y 2 (1) 1 2 2 2 2 0
(-1, 0)
0
y 2 ( 0) 1 2 1 2 1 1
(0, 1)
1
y 2 (1) 1 2 0 2 0 2
(1, 2)
2
y 2 ( 2) 1 2 1 2 1 1
(2, 1)
3
y 2 (3) 1 2 2 2 2 0
(3, 0)
To get the graph, just plot all these points. You will get this:
Let’s talk about the term intercepts. This is where the graph
crosses either the vertical or horizontal axis. Where ever the graph
crosses the vertical axis is called the y-intercept. Where ever
the graph crosses the horizontal axis is called the x-intercept. The
problem asked us to identify the intercepts. This means we need
to read them off our graph. For the y-intercept, this point would be
(0, 1). There are two places where it crosses the horizontal axis, and
this occurs at (-1, 0) and (3, 0).
Section 1.1 Notes Page 3
EXAMPLE: Graph the equation y x 4 . Let x = -3, -2, -1, 0, 1, 2, 3. Then indicate the intercepts.
2
It doesn’t matter if you’ve never graphed a quadratic equation before. We plot this one the same way we plot
lines. First we need to set up a table like this:
x
y x2 4
(x, y)
-3
-2
-1
0
1
2
3
Now what we do is plug in the x value into the equation y x 2 4 . Then we write our answer as an ordered
pair as show below. Remember that a negative number raised to an even power is positive. For example
(5) 2 (5)(5) 25 .
x
y x2 4
(x, y)
-3
y (3) 2 4 9 4 5
(-3, 5)
-2
y (2) 2 4 4 4 0
(-2, 0)
-1
y (1) 2 4 1 4 3
(-1, -3)
0
y (0) 2 4 0 4 4
(0, -4)
1
y (1) 2 4 1 4 3
(1, -3)
2
y (2) 2 4 4 4 0
(2, 0)
3
y (3) 2 4 9 4 5
(3, 5)
To get the graph, just plot all these points. You will get this:
Now let’s identify the intercepts the same way we did it for the
previous problem. The y-intercept is where the graph crosses the
vertical axis. This point would be at (0, -4). There are two places
where it crosses the horizontal axis, and this occurs at (-2, 0) and (2, 0).
You can also write this as: 2, 0 .
Graphing Utilities
This book does some exercises where it wants you to set up viewing windows. Since I do not require graphing
calculators in this class, I am not requiring you to do these types of exercises. If you already have a graphing
calculator, you may use it for this class. Please ask me after class or in office hours if you have a specific
question on how to use your particular graphing calculator.
Section 1.2 Notes Page 1
1.2 Basics of Functions and Their Graphs
Domain: (input) all the x-values that make the equation defined
Defined: There is no division by zero or square roots of negative numbers
Range: (output) all y-values that a graph uses.
EXAMPLE: Find the domain and range of the following graph (assume graph ends at edge of graph and the
bottom edge of the graph is the x-axis, and the left edge of the graph is the y-axis)
Domain: [0, 4]
Range: [0, 2]
Function Definition: For each input (x) there can only be one output (y).
EXAMPLE: For each relation below, determine whether it is a function. Then give the domain and range for
each relation.
{(1, 2), (3, 7), (2, 9), (8, 11)}
This is a function. Every x goes to only one y.
The domain of this is all the x-values. The answer is {1, 3, 2, 8}. You may leave it this way or order them.
The range of this is all the y-values. The answer is {2, 7, 9, 11}.
{(-3, 4), (5, 6), (7, 4), (-2, 3)}
This is a function. Even though the x-values -3 and 7 both go to 4, each x value goes to only one y-value.
Domain: {-3, 5, 7, -2}
Range: {4, 6, 3} Notice that 4 is repeated, but you only need to write it once.
{(-2, 4), (-1, 6), (0, 3), (-2, 8)}
NOT a function because when x is -2 it goes to both 4 and 8. There are two different y values for one x.
Domain: {-2, -1, 0} Notice again that even though -2 is repeated, it only needs to be written once.
Range: {4, 6, 3, 8}
{(5, 3), (-2, 1), (5, 3), (9, 10)}
This is a function. The same point repeats, but still goes to only one y.
Domain: {5, -2, 9}
Range: {3, 1, 10}
Section 1.2 Notes Page 2
Vertical line test. If you pass an imaginary vertical line through the graph and it only intersects the graph once
then it is a function. Which graphs below are functions?
Function
NOT a function.
Function
EXAMPLE: Is x 2 3 y 2 7 a function?
We don’t have a graph drawn for us or a set of points. We need to see if it is a function algebraically. First
think we need to do is solve for y. We will isolate it and then take the square root of both sides. Don’t forget
that you will get a plus and minus whenever you take the even root of something.
3y 2 7 x2
y2
7 x2
3
7 x2
Notice that for each x we will get two different y values because of the . Therefore we know
3
this is not a function.
y
EXAMPLE: Is x 2 y 5 1 a function?
Again we will solve for y. When we do we will take the odd root of both sides. There will be no plus and
minus here since it was an odd root.
y5 1 x2
y 5 1 x2
To get rid of the fifth power, I took the fifth root. For each x we put in we will only get one
y-value for each x we put in so it IS a function.
So what is the general rule here based on our previous two examples?
Any equation that has a y raised to an even power is NOT a function.
Any equation that has a y raised to an odd power IS a function.
Section 1.2 Notes Page 3
Function notation: f (x) which means “f of x”. This does not mean f times x. It means that we have a
function called f which contains the variable x.
EXAMPLE: Given the function f ( x) 2 x 5 , find the following:
a.) f (3)
Whatever is inside the parenthesis goes in place of x in the original expression. This is really asking us for the y
value when x is 3.
f (3) 2(3) 5
f (3) 1
b.) f ( x 3)
Now we need to replace x in the original equation with x + 3. Then simplify.
f ( x 3) 2( x 3) 5
f ( x 3) 2 x 6 5
f ( x 3) 2 x 1 This is as far as we can go on this one.
c.) f ( x) f (3)
For this one we can replace the f (x) with 2x – 5. We also know f (3) .
f ( x) f (3) 2 x 5 1
f ( x) f (3) 2 x 4 Notice this is not the same as part b, so the f is not distributed to the x and 3.
d.) f ( x h)
For this one just replace the x with the expression x + h.
f ( x h) 2( x h) 5
f ( x h) 2 x 2h 5 This is as far as we can go.
EXAMPLE: Let f ( x)
x4
. Find the following:
2x 3
a.) f (5)
54
2(5) 3
1
f (5)
7
f (5)
We are replacing x with 5.
Section 1.2 Notes Page 4
b.) f ( x h)
( x h) 4
We are replacing x with the quantity (x + h).
2( x h) 3
xh4
f ( x h)
This is as far as we can go.
2 x 2h 3
f ( x h)
c.) f ( x) f (5)
x4 1
We are replacing f(x) with our original function and f(5) we found in part a.
2x 3 7
7 x 4 2x 3 1
Generally if you have two fractions, then combine after common denominators.
7 2x 3 2x 3 7
7( x 4) (2 x 3)
Now add the fractions together now that we have common denominators.
7(2 x 3)
7 x 28 2 x 3
Distribute and simplify.
7(2 x 3)
9 x 31
This is our final answer.
7(2 x 3)
EXAMPLE: Given f ( x) x 2 3x 3 , find f ( x) .
f ( x) ( x) 2 3( x) 3
f ( x) x 2 3 x 3
Replace x with –x and simplify.
We have looked at function notation for equations, but now we will see the relationship between the function
notation and graphs. This next exercise shows how to read values off a graph, but first let’s talk about
symmetry.
Symmetry:
x-axis symmetry
(x-axis is a fold line)
y-axis symmetry
(y-axis is a fold line)
Origin Symmetry
(Graph is in opposite quadrants)
Section 1.2 Notes Page 5
EXAMPLE: Use the graph below to answer the following:
a.) Find the domain
Since we don’t include the endpoints we have (-2, 2) (x values)
b.) Find the range
The answer is (-1, 1] (y-values)
c.) Indicate the intercepts
x-int: (-1, 0) (1, 0) y-int: (0, 1)
d.) Indicate any symmetry this graph has.
You can fold this in half over the y-axis, so it has y-axis symmetry.
EXAMPLE: Use the graph below to answer the following:
a.) Find f (2) :
This is asking you for the y value when x is -2.
The answer is f (2) = 1.
b.) Find all x such that f ( x) 3
This is asking you to find all x that give a y value of 3.
This happens at the point (5, 3), so x = 5.
c.) Is f (3) positive or negative?
This is asking you if the y value at x = 3 is above or
below the x axis. To find this go over to x = 3. We
notice the graph is below the x-axis, so answer is neg.
d.) What is the domain?
This is asking you for all the x values the graph uses.
This would be [-4, 6]. (lowest x to highest x)
e.) What is the range?
This is asking you for all the y values the graph uses.
The answer is [-2, 3]. (lowest y to highest y).
f.) For which values is f ( x) 0 ?
This is asking you which part of the graph has positive
y values. In other words, what part of the graph is
above the x-axis, but not on the x-axis. We have two
places this occurs. [-4, 0) or (4, 6) Notice the values I
gave in the interval notation are x values. We include
the -4 because it is not on the x-axis.
Section 1.2 Notes Page 6
EXAMPLE: Use the graph below to answer the following:
a.) Find f (1) :
This is asking you for the y value when x is -1.
The answer is f (1) = 2.
It does not matter if the x-value has a dot or not.
b.) Find all x such that f ( x) 0
This is asking you to find all x that give a y value of 0.
This happens at x = –2, 3, and 5 .
3
c.) Is f positive or negative?
2
This fraction is the same as -1.5. When you go to this x
value the graph is above the x-axis here, so positive.
d.) What is the domain?
The domain is referring to the x-values the graph uses.
Since there is an open circle at –3 , this x-value is not
included. So the domain is: (–3, 6].
e.) What is the range?
The range is referring to the y-values the graph uses.
Again since there is an open circle at –3 , this y-value
is not included. So the range is (–4, 4].
f.) Indicate the x and y intercepts.
y-int: (0, 3) x-int: (-2, 0), (3, 0), (5, 0).
g.) Indicate what kind of symmetry, if any, this graph has.
This graph does not have any symmetry.
Section 1.3 Notes Page 1
1.3 More on Functions and Their Graphs
Even and Odd functions
If f ( x) f ( x) then the function is even, and symmetric to the y-axis.
If f ( x) f ( x) then the function is odd, and symmetric to the origin.
EXAMPLE: Determine whether the following are even, odd, or neither.
a.) f ( x) x 4 7
We want to put a –x in for x first. You will get: f ( x) ( x) 4 7 This simplifies to f ( x) x 4 7 . Since
this is the same as the original, this function is even.
b.) f ( x) 6 x 5 x 3
We want to put a –x in for x first. You will get: f ( x) 6( x) 5 ( x) 3 This simplifies to
f ( x) 6 x 5 x 3 . Factoring out a negative: f ( x) (6 x 5 x 3 ) . So we have f ( x) f ( x) so this
function is odd.
c.) f ( x) x 2 x
We want to put a –x in for x first. You will get: f ( x) ( x) 2 ( x) This simplifies to f ( x) x 2 x .
There is nothing more I can do to this one. First we notice this is not the same as the original so it is definitely
not even. If we try to factor out a negative we get f ( x) ( x 2 x) . Since you don’t have f(x) inside of the
parenthesis it is not odd either. We will answer neither.
d.) f ( x)
2x
x
If we put in a –x for x we will get: f ( x)
2( x)
x
which simplifies to f ( x)
2x
x
which means the
function will be odd.
Increasing: as x increases, y increases (graph goes uphill as you move from left to right)
Decreasing: as x increases, y decreases (graph goes downhill as you move from left to right)
Constant: as x increases, y does not change (this part of the graph is horizontal)
Relative maximum: a point at which the graph increases and then decreases (peak)
Relative minimum: a point at which the graph decreases and then increases (valley)
Section 1.3 Notes Page 2
EXAMPLE: Use the graph below to answer the following questions
a.) Indicate the interval(s) of which f is increasing
There are two places this occurs. Between the x values of
-3 and 0 the graph is climbing. This also happens between
2 and 4. We write the answer as 3,0 2,4 . We always
use parenthesis because at the endpoints the graph is not
increasing or decreasing.
b.) Indicate the interval(s) of which f is decreasing
There is one place where the graph is falling as we move from
left to right. This is between the x value of 0 and 2, so we
write our answer as (0, 2).
c.) List the number where f has a relative maximum.
This is asking for the x value at which the graph has a local
maximum. This occurs at x = 0.
d.) What is the relative maximum?
This is asking for the y-value of the local max, which is 3.
e.) What is the relative minimum?
The y-value of the local minimum is 0.
EXAMPLE: Use the graph below to answer the following questions
a.) Indicate the interval(s) of which f is increasing
(2,0) (3,5)
b.) Indicate the interval(s) of which f is decreasing
(3,2) (0,3)
c.) List the number(s) where f has a relative minimum.
This is asking for the x value(s) at which the graph has a local
minimum. This occurs at x = -2 and at x = 3.
d.) What is the relative maximum(s)?
This is asking for the y-value of the local max, which is -2.
e.) What is the relative minimum(s)?
The y-value of the local minimum is -3 and -5.
f.) What is the domain?
[3, 5)
g.) What is the range?
[-5, 0]
Section 1.3 Notes Page 3
Piecewise Functions
These functions are made up of different pieces. Each piece is defined for certain values of x.
x 2 if
EXAMPLE: Use the function f ( x)
x 2 if
and use this to determine the graph’s range.
x 3
3
to find f (4) , f (3) and f . Then graph.
x 3
2
a.) f (4) In order to know which equation we are using, look at the number inside the parenthesis, which is 4.
In our function, we need to find what function includes -4. This would be the first equation since -4 is less than
-3. So we put -4 in for x in the first equation. You will get -4 + 2 = -2. So f (4) 2 .
b.) f (3) The equation that includes -3 would be the second one since it is greater than or equal to 3. So we
will place the -3 in for x in the second equation: -3 – 2 = -5. So f (3) 5 .
3
c.) f If you don’t know if this fraction is less or more than -3 then turn it into a decimal. This is -1.5.
2
7
7
3
3
So this would be greater than -3, so we will use the second equation. 2
. So f .
2
2
2
2
Now we do a graph. Notice that there are two different equations we need to graph. To graph these, we will
make a table of values for each one. The most important thing is that we need to plug in x values that match our
conditions in the problem. For example the first equation says we need to use x values that are less than but not
equal to -3. Now we can plug in -3, and this will end up as an open circle on the graph since it is not included.
So we can use x = -5, -4, -3. Three points are enough. For the second equation we need to pick x values that
are greater than or equal to -3. When we plug in -3 we can plot this point as a closed circle since this point is
included. We will use x = -3, -2, -1. Below are a table of values for each equation. To graph this, we plot
points from our table making sure to indicate a closed or open circle:
x
-5
-4
-3
y x2
y 5 2 3
y 4 2 2
y 3 2 1
(x, y)
(-5, -3)
(-4, -2)
(-3, -1)
x
-3
-2
-1
y x2
y 3 2 5
y 2 2 4
y 1 2 3
(x, y)
(-3, -5)
(-2, -4)
(-1, -3)
Notice the open circle at (-3, -1). Also notice that there are no x values plotted to the right of the open circle
because this graph is only defined for values less than or equal to -3. Notice the open circle at (-3, -5). Notice
that there are not x values plotted to the left of this point since we are only allowed to use values for x that are
greater than or equal to -3.
The range is the y-values the graph is using. This would be ALL y values, so the answer is , .
Section 1.3 Notes Page 4
1 2
x if
EXAMPLE: Use the function f ( x) 2
2 x 1 if
use this to determine the graph’s range.
9
. Then graph. and
to find f (1) , f (1) and f
10
x 1
x 1
a.) f (1) In order to know which equation we are using, look at the number inside the parenthesis, which is -1.
In our function, we need to find what function includes -1. This would be the first equation since -1 is less than
1
1
1
1
1. So we put -1 in for x in the first equation. You will get (1) 2 (1) . So f (1) .
2
2
2
2
b.) f (1) The equation that includes 1 would be the second one since it is greater than or equal to 1. So we will
place the 1 in for x in the second equation: 2(1) + 1 = 2 + 1 = 3. So f (1) 3 .
9
If you don’t know if this is less or more than 1 then turn it into a decimal. This would be 0.95,
c.) f
10
2
9
9
1 9
1 9
9
.
. So, f
which is than 1, so we will use the first equation.
2 10
2 10
20
20
10
Now we do a graph. Notice that there are two different equations we need to graph. To graph these, we will
make a table of values for each one. The most important thing is that we need to plug in x values that match our
conditions in the problem. For example the first equation says we need to use x values that are less than but not
equal to 1. Now we can plug in 1, and this will end up as an open circle on the graph since it is not included.
So we can use x = -1, 0, 1. Three points are enough. For the second equation we need to pick x values that are
greater than or equal to 1. When we plug in 1 we can plot this point as a closed circle since this point is
included. We will use x = 1, 2, 3. Below are a table of values for each equation. To graph this, we plot points
from our table making sure to indicate a closed or open circle:
x
-1
0
1
x
1
2
3
1
y x2
2
1
1
1
y (1) 2 (1)
2
2
2
1
y (0) 2 0
2
1
1
y (1) 2
2
2
y 2x 1
y 2(1) 1 3
y 2(2) 1 5
y 2(3) 1 7
(x, y)
1
1,
2
(0, 0)
1
1,
2
(x, y)
(1, 3)
(2, 5)
(3, 7)
The range is the y-values the graph is using. The graph is not using any y value between 0 and 3, so we write
our answer this way: , 0 3, .
Section 1.3 Notes Page 5
if
0
EXAMPLE: Use the function f ( x) x if
x 2 1 if
x 3
10
, f ( 11) .
3 x 0 to find f (0) , f (1) , f
2
x0
a.) f (0) Zero is only included in the third equation. We put a zero in for x in the third equation: 0 2 1 1 .
So we write f (0) 1
b.) f (1) Negative one is included in the middle equation since it is between -3 and 0. We will put in -1 for x.
We will get: (1) 1 . So we write f (1) 1 .
10
Turning it into a decimal we get 1.58, so we use the third equation this time. So
c.) f
2
2
10
10
5
3
1
1
1
.
2
4
2
2
10
f
2
d.) f ( 11)
Turning this into a decimal we get -3.32, so we use the first equation. The answer is 0.
Now we do a graph by making a table of values for each equation. Notice that there are three different
equations we need to graph. To graph these, we will make a table of values for each one. For the first equation
we use the following x values: 1, 2, 3. For the second equation we can use x = -3, -1, 0. For the third equation
we can use x = 0, 1, 2.
x
1
2
3
x
-3
-1
0
y0
y0
y0
y0
(x, y)
(1, 0)
(2, 0)
(3, 0)
y x
y (3) 3
y (1) 1
y (0) 0
(x, y)
(-3, 3)
(-2, 2)
(0, 0)
x
y x2 1
0
1
y (0) 2 1 1 (0, -1)
(1, 0)
y (1) 2 1 0
2
y (2) 2 1 3
(x, y)
(2, 3)
Notice that our first equation is a horizontal line with an open circle at (-3, 0). Notice an open circle at (0, 0),
and closed circle at (0, -1).
The range is the y-values the graph is using. The graph is not using any y values less than 1, so we write our
answer as: 1, .
Section 1.3 Notes Page 6
Difference Quotient
If we wanted to find the slope of a curved line, the only way we can do
this is by estimating it with a straight line. We will start with one point
and then move over by a small amount h. Now we will use the slope
formula. In the picture we have two points, A and B. The coordinates
for these are: ( x, f ( x)) and ( x h, f ( x h)) .
f ( x h) f ( x )
h
In calculus we will try to minimize h so that it is so small that we end up
at a point, which will be the exact slope of the curved line at x.
The slope, also called the difference quotient is:
Now let’s look at some examples finding the difference quotient.
EXAMPLE: Let f ( x) 2 x 3 . Find the difference quotient.
Let’s first find f ( x h) . Once we have this we can put it into the difference quotient formula. Replace x in the
original equation with x + h.
f ( x h) 2( x h) 3
f ( x h) 2 x 2 h 3
Now simplify.
We are ready to substitute this into difference quotient formula. We have f ( x h) and we also know f (x) ,
which is the original equation.
2 x 2h 3 (2 x 3)
Here we have substituted into the formula. Notice the parenthesis around f (x) .
h
2 x 2h 3 2 x 3
Now we distributed the minus sign and the last thing is to simplify.
h
2h
2 The 2x and the 3 canceled and then the h canceled, leaving us with our answer of 2.
h
Does 2 make sense? Yes, because a difference quotient tells you the slope at any value of x. Since we have a
line the slope of 2 will not change no matter what value of x we use.
EXAMPLE: Let f ( x) 3x 2 x 1 . Find the difference quotient.
We will do this the same way as above. First we will find f ( x h) .
f ( x h) 3( x h) 2 ( x h) 1
What is ( x h) 2 ? If you are thinking x 2 h 2 you are wrong. This
is actually ( x h)( x h) which is a FOIL. It is x 2 2 xh h 2 .
f ( x h) 3( x h)( x h) x h 1
f ( x h) 3( x 2 2 xh h 2 ) x h 1
f ( x h) 3 x 2 6 xh 3h 2 x h 1
Now that we have simplified this as much as possible, we will put it into the difference quotient formula.
Section 1.3 Notes Page 7
3 x 6 xh 3h x h 1 (3 x x 1)
Now we will distribute the minus into f (x) .
h
3 x 2 6 xh 3h 2 x h 1 3x 2 x 1
Now cancel and simplify.
h
6 xh 3h 2 h
Now we can factor out an h from the top.
h
h(6 x 3h 1)
Last thing is we can cancel the h from top and bottom.
h
6 x 3h 1 This is our answer. What can you do with this expression? Now if we were in calculus we would
minimize the h (go to zero). What is left can be used to find the slope of this curve at any point x (derivative).
2
2
EXAMPLE: f ( x)
2
3
. Find the difference quotient.
x 1
3
Nothing more we can do to simplify this. Now put it into the formula.
( x h) 1
3
3
x h 1 x 1 Now to simplify this we need add the fractions on the top. Need common denominators.
h
( x 1)
3
( x h 1) 3
( x 1) x h 1 ( x h 1) x 1
Now that we have common denominators, combine the fractions
h
3( x 1) 3( x h 1)
( x 1)( x h 1)
We can simplify the very top part of this fraction
h
3 x 3 3 x 3h 3
( x 1)( x h 1)
Simplify.
h
3h
( x 1)( x h 1)
Now we will clear the double fractions by multiplying by the reciprocal.
h
1
f ( x h)
3h
h( x 1)( x h 1)
Almost done. Just need to cancel out the h from the top and bottom.
3
( x 1)( x h 1)
Whew! Okay this is our answer.
Section 1.4 Notes Page 1
1.4 Linear Functions and Slope
This section is designed to be a review from Intermediate Algebra.
Slope Formula
The slope formula is used to find the slope between two points x1 , y1 and x2 , y2 .
( x2 , y2 )
( x1 , y1 )
The slope is the vertical change divided by the horizontal change.
From our picture, the horizontal change is y 2 y1 and the horizontal
change is x 2 x1 .
From this we get the formula for slope: m
y 2 y1
.
x 2 x1
Positive slopes will rise as you move from left to right.
Negative slopes will fall as you move from left to right.
A slope of zero is a horizontal line.
An undefined or infinity slope is a vertical line.
EXAMPLE: Find the slope of a line passing through the following points. Indicate whether the line rises, falls,
is horizontal or vertical.
a.) (-1, 3) and (2, 4)
To do this problem we can label our point so we know what to put into the slope formula. It doesn’t matter
which point you call x1 or x 2 . I will label the point as the following: x1 1 , y1 3 , x 2 2 , y 2 4 . Now
43
1
we plug these into the slope formula: m
. Since the slope is positive we know this line rises.
2 (1) 3
b.) (4, -1) and (3, -1)
First we label the point as the following: x1 4 , y1 1 , x 2 3 , y 2 1 . Now we plug these into the slope
1 (1)
0
formula: m
0 . Since the slope is zero we know this line is horizontal.
34
1
c.) (3, -2) and (3, -5)
First we label the point as the following: x1 3 , y1 2 , x 2 3 , y 2 5 . Now we plug these into the slope
5 (2) 3
undefined . Since the slope is undefined we know this line is vertical.
formula: m
33
0
Section 1.4 Notes Page 2
Now I will introduce two formulas you will need to know in this section:
Slope-Intercept Formula– this is the standard from of a line which allows you to easily identify the slope and
y-intercept.
y mx b
Here the slope is m and the y-intercept is (0, b).
Point-Slope Formula – this is used when you want to find the equation of a line when you are give a slope and
another point on the line. This other point does not need to be the y-intercept.
y y1 m( x x1 )
EXAMPLE: Use the information and given conditions to write an equation for each line in slope-intercept form
as well as the point-slope form.
a.) Slope = 8, passing through (4, -1).
For this one, we know that m = 4, x1 4 , and y1 1 . We can plug these into our point-slope formula:
y (1) 8( x 4) . When we simplify we get: y 1 8( x 4) . The equation of this line is now written in
slope-intercept form, which is one of our answers. Now we need to write it in slope-intercept form. To do this,
we just need to solve for y. First we distribute the 8: y 1 8 x 32 . Now subtract 1 from both sides to get
our second answer: y 8 x 33 .
3
b.) Slope = , passing through (10, -4).
5
3
For this one, we know that m = , x1 10 , and y1 4 . We can plug these into our point-slope formula:
5
3
3
y (4) ( x 10) . When we simplify we get: y 4 ( x 10) . The equation of this line is now
5
5
written in slope-intercept form, which is one of our answers. Now we need to write it in slope-intercept form.
3
3
3 10
First we distribute the : y 4 x . To multiply the two fractions on the end, multiply
5
5
5 1
3
across the top and bottom. You will get: y 4 x 6 . Now subtract 4 from both sides to get our second
5
3
answer: y x 2 .
5
Section 1.4 Notes Page 3
c.) Passing through (-3, 6) and (3, -2)
This time we are not given a slope, so we first must use the slope formula. We label our points and put them
26 8
4
into the slope formula: m
. When we use the point-slope formula we can use EITHER
3 (3)
6
3
of our given point as the ( x1 , y1 ) . In this case I will use the first point. So x1 3 , and y1 6 . We can plug
these into our point-slope formula:
4
4
y 6 ( x (3)) . When we simplify we get: y 6 ( x 3) . The equation of this line is now written
3
3
in slope-intercept form, which is one of our answers. Now we need to write it in slope-intercept form. First we
4
4
4
distribute the : y 6 x 4 . Now add 6 to both sides to get our second answer: y x 2 .
3
3
3
d.) x-intercept =
1
, y-intercept = 4
2
This time we are not given a slope, so again we must use the slope formula. We want to put these intercepts in
1
a point form, which would be: , 0 and 0, 4 . We label our points and put them into the slope formula:
2
40
4
m
8 . When we use the point-slope formula we can use EITHER of our given point as the
1 1
0
2 2
( x1 , y1 ) . In this case I will use the second point. So x1 0 , and y1 4 . We can plug these into our pointslope formula:
y 4 8( x 0) . When we simplify we get: y 4 8 x . The equation of this line is now written in slopeintercept form, which is one of our answers. Now we need to write it in slope-intercept form. We just need to
add 4 to both sides to get our second answer: y 8 x 4 .
EXAMPLE: Write the following in slope-intercept form and identify the slope and y-intercept. Use this
information to graph the equation.
a.) f ( x)
3
x3
4
This equation can be written in slope-intercept form by replacing f ( x) with y. You will get: y
3
and the y-intercept is (0, -3).
4
To graph this, first plot the y-intercept. Now we need another point on
our line. The definition of the slope is the change in the vertical distance
divided by the change in the horizontal distance. In our slope the top
number is 3. This is our vertical change. Because it is positive we will
move up 3 units from our y-intercept. The bottom number is 4, so we
will need to move 4 units to our right. So from our y-int we will move
up 3 units and 4 units to the right. This will give us our next point. Plot
this and connect our two points with a line.
From here we can identify that the slope is
3
x 3.
4
Section 1.4 Notes Page 4
b.) 4x + 6y + 12 = 0
We need to solve for y in order to put this into slope-intercept form. First isolate y: 6y = -4x – 12. Now
2
2
divide both sides by 6 to get: y x 2 . Now we can identify that the slope is and the y-intercept is
3
3
2
(0, -2). To graph this, first plot the y-intercept. Now the fraction can be
3
2
2
2
or
. If we think of the slope as
then the
written as either
3
3
3
our vertical change is -2. This means we move DOWN 2 units from our
y-intercept. The bottom number is 3, so we will need to move 3 units to our
right. So from our y-int we will move DOWN 2 units and 3 units to the right.
This will give us our next point. Plot this and connect our two points with a line.
2
then we would move UP two units and to the LEFT 3 units.
If we used
3
Notice we will still get another point on the same line, so we can use either fraction.
EXAMPLE: Find the x and y intercepts and use them to graph the following equation: 6x + 9y – 18 = 0.
To find an x-intercept, put a zero in for y and solve: 6x + 9(0) – 18 = 0.
Simplifying will give us 6x – 18 = 0. Solve for x: 6x = 18, so x = 3.
It is important to write our answer as a point. The x-intercept is (3, 0).
To find an y-intercept, put a zero in for x and solve: 6(0) + 9y – 18 = 0.
Simplifying will give us 9y – 18 = 0. Solve for y: 9y = 18, so y = 2.
It is important to write our answer as a point. The y-intercept is (0, 2).
To graph, just plot each point.
EXAMPLE: Find the x and y intercepts and use them to graph the following equation: 6x – 3y + 15 = 0.
To find an x-intercept, put a zero in for y and solve: 6x – 3(0) + 15 = 0.
5
Simplifying will give us 6x + 15 = 0. Solve for x: 6x = -15, so x = .
2
5
It is important to write our answer as a point. The x-intercept is , 0 .
2
To find an y-intercept, put a zero in for x and solve: 6(0) – 3y + 15 = 0.
Simplifying will give us -3y + 15 = 0. Solve for y: -3y = -15, so y = 5.
It is important to write our answer as a point. The y-intercept is (0, 5).
To graph, just plot each point. For fractions, you can change them into
a decimal. Our x-intercept can be written as: (-2.5, 0).
Section 1.5 Notes Page 1
1.5 More on Slope
Parallel lines have the same slope. These lines do not cross.
Perpendicular lines have opposite reciprocal slopes (opposite sign and one fraction is flipped over).
4
3
1
Ex:
and are opposite reciprocals. Also and 2 are opposite reciprocals. Perpendicular lines will
3
4
2
cross at a right angle (90 degrees).
EXAMPLE: Find the slope of a line that is perpendicular to y = -4x – 3.
The slope of this line is -4, and because it says perpendicular, we need to find the opposite reciprocal. The
4
number -4 can be rewritten as the fraction . Because it is a negative, the opposite sign will be positive. If
1
1
we flip over the fraction we get , which is our answer.
4
EXAMPLE: Use the given conditions to write an equation for each line in point-slope form and slope-intercept
form.
a.) Passing through (-2, -7) and parallel to the line whose equation is y 5 x 4
Since we want a line parallel, then this will have the same slope as our given equation, so we know m = -5. We
are also given a point. From here, we will plug this information into the point-slope formula. You should get:
y (7) 5( x (2)) . Simplifying gives you y 7 5( x 2) . This is our first answer. To get this into the
slope-intercept form, we need to solve for y. First distribute: y 7 5 x 10 . Now subtract 7 from both
sides: y 5 x 17 . This is our answer in slope-intercept form.
b.) Passing through (-4, 2) and perpendicular to the line whose equation is y
1
x7
3
Since we want a line perpendicular, then this will have the opposite reciprocal slope as our given equation, so
we know m = -3. We are also given a point. From here, we will plug this information into the point-slope
formula. You should get: y 2 3( x (4)) . Simplifying gives you y 2 3( x 4) . This is our first
answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y 2 3x 12 .
Now add 2 to both sides: y 3 x 10 . This is our answer in slope-intercept form.
c.) Passing through (5, -9) and perpendicular to the line whose equation is x 7 y 12 0
Since we want a line perpendicular, then this will have the opposite reciprocal slope as our given equation. We
1
12
1
first need to solve our given equation for y: y x . The slope of this line is . A line perpendicular
7
7
7
to this one will have a slope of m = 7. From here, we will our slope and given point into the point-slope
formula. You should get: y (9) 7( x 5) . Simplifying gives you: y 9 7( x 5) . This is our first
answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y 9 7 x 35 .
Now subtract 9 from both sides: y 7 x 44 . This is our answer in slope-intercept form.
Section 1.5 Notes Page 2
General Form: Ax + By + C = 0. In the general form, everything is set equal to zero. The constants, A, B,
and C should be written as integers if possible, and A must be written as a positive number.
EXAMPLE: Given y 9 7 x 35 , write this in general form.
In order to solve this, we must bring everything over to one side of the equation and set it equal to zero. I will
move everything from the right side to the left: y 9 7 x 35 0 . Now simplify and write the x and y terms
first: 7 x y 44 0 . The A constant must be written as a positive number, so we will multiply the whole
equation by -1 to get: 7 x y 44 0 .
Average Rate of Change (A.R.C.)
The A.R.C. is an estimate of the slope between x and c. Basically how much does something change between x
and c. The formula is as follows and is derived from the slope formula, much like the difference quotient.
y f ( x 2 ) f ( x1 )
x
x 2 x1
EXAMPLE: Find the A.R.C. for the f ( x) 3 x 2 3 from x1 0 to x 2 2 .
We can substitute these numbers into our general A.R.C. formula:
f (2) f (0)
Now we will work this out. Find f (2) and f (0)
20
93
6 So the slope is -6 between the x values of 0 and 2.
2
EXAMPLE: Find the A.R.C. for the f ( x) x 3 x 2 from x1 1 to x 2 3 .
We can substitute these numbers into our general A.R.C. formula:
f (3) f (1)
3 1
26 2
12
2
Now we will work this out. Find f (3) and f (1)
So the slope is 12 between the x values of 1 and 3.
EXAMPLE: Find the A.R.C. for the f ( x) x from x1 9 to x 2 16 .
We can substitute these numbers into our general A.R.C. formula:
f (16) f (9)
Now we will work this out. Find f (16) and f (9)
16 9
43 1
1
So the slope is between the x values of 9 and 16.
7
7
7
Section 1.6 Notes Page 1
1.6 Transformations of Functions
We will first look at the major graphs you should know how to sketch:
y x
Square Root Function
Domain: [0, )
Range: [0, )
Increasing: (0, )
Decreasing: None
y x2
Standard Quadratic Function
Domain: (, )
Range: [0, )
Increasing: (0, )
Decreasing: (, 0)
y x
Absolute Value Function
Domain: (, )
Range: [0, )
Increasing: (0, )
Decreasing: (, 0)
y x3
Standard Cube Function
Domain: (, )
Range: (, )
Increasing: (, )
Decreasing: None
yx
Identity Function
Domain: (, )
Range: (, )
Increasing: (, )
Decreasing: None
y3 x
Cube Root Function
Domain: (, )
Range: (, )
Increasing: (, )
Decreasing: None
Section 1.6 Notes Page 2
Transformations and Graph Sketches
When we used to graph a line the usual thing to do was make a table of values and plot the points. This method
works but takes a long time. Transformations allows you move a graph up or down, left or right into a new
position. We start with the basic graphs we learned in the last section and will move it based on the following
criteria.
Suppose y = f(x) is the original function (one we looked at in the previous section)
Y = f(x) + k moves f(x) k units up
Y= f(x) – k moves f(x) k units down
Y = f(x - h) moves f(x) h units to the right
Y= f(x + h) moves f(x) h units to the left
Y = -f(x) flips the graph over a horizontal axis
Y = f(-x) flips the graph over a vertical axis
Y = a(f(x)) If a >1 then there is a vertical stretch. If 0 a 1 , then there is a vertical compression.
Let’s look at some examples. For all of these we are just making a sketch of the function.
EXAMPLE: Sketch y x 1 by using transformations.
First we need to recognize what kind of graph we are using. This is the absolute value graph which is a V
shaped graph centered at the origin. Since there is a +1 inside the absolute value we are looking at rule 4 which
says we will move the graph 1 unit to the left. So the graph will look like:
What is the x-intercept? It crosses the x-axis at (-1, 0).
What is the y-intercept? It crosses the y-axis at (0, 1).
EXAMPLE: Sketch y x 1 by using transformations.
This may look similar to the graph above, but it is different because the 1 is on the outside of the absolute
values. This is rule 1 which says the graph will move 1 unit up:
There are no x-intercepts on this one.
The y-intercept is (0, 1).
Section 1.6 Notes Page 3
EXAMPLE: Sketch y x by using transformations.
This one has a negative on the outside of the absolute value. Since the negative is outside we will use rule 5
which says the graph will flip over the horizontal axis. This will flip the graph upside down and the pivot point
is the origin (0, 0):
x-int: (0, 0)
y-int; (0, 0)
EXAMPLE: Sketch y x 1 1 by using transformations.
Now let’s put it all together. Usually you will be using more than one transformation. This problem does puts
the last three examples together. We start with our absolute value graph at the origin. We will move it one
place to the left, then up one unit and then flip it upside down:
x-intercept: (-2, 0) and (0, 0)
y-intercept: (0, 0)
EXAMPLE: Sketch y 2 x 1 1 by using transformations
You will move the graph in the same directions as the last example, but now
there is a number in front of the absolute value. This doubles all of the regular
y-values, causing the graph to be vertically stretched.
EXAMPLE: Sketch y x 2 3 by using transformations.
Now we will look at a different type of graph. This one is the square root. Our transformations now tell us we
will move the square root graph 2 places to the right and 3 units down. We don’t need to flip the graph because
there is no negative outside of the function, or inside next to an x.
Section 1.6 Notes Page 4
EXAMPLE: Sketch y x 2 1 by using transformations.
This one requires us to move the square root graph 2 places to the left and down one unit. Once this is done we
need to flip the graph over the horizontal axis because the negative is on the outside of the equation.
EXAMPLE: Sketch y 3 x 2 by using transformations.
In order to use the transformation rules the x must come first and there must be a one in front of x. In our
problem above we need to first put the x first and then we will factor out a negative:
y 3 x 2
y x32
Here we put the x term first
y ( x 3) 2
Here we factored out a -1.
Now we are ready to graph. Since we already factored out the negative
We need to move the graph 3 places to the right and then up 2 units.
There is a negative inside the function, so we need to use rule 6 of the
transformations which says we will flip the graph over the vertical axis.
EXAMPLE: Sketch y 3 x 1 2 by using transformations.
Be careful you don’t confuse this with the square root graphs we just did. This one looks different. We will
move the regular cube root graph to the right 1 unit and up 2 units.
EXAMPLE: Sketch y 3 ( x 2) 3 by using transformations.
For this one since the negative is already factored out we will move the graph
2 units to the left and up 3 units. We will also flip the graph over
the vertical axis since the negative is inside the radical.
Section 1.6 Notes Page 5
EXAMPLE: Sketch y ( x 2) 4 by using transformations.
3
Here we have the cube graph. Don’t confuse it with the cube root graph. We
need to move the graph 2 units to the right and down four units. There is a
negative outside so we will flip the graph over the horizontal axis.
EXAMPLE: Sketch y (2 x) 3 4 by using transformations.
For this one we need to once again put the x term first and then factor out the
negative sign.
y (2 x) 3 4
y ( x 2) 3 4
y (( x 2)) 3 4
So we move the graph 2 places to the right and down 4 units. The negative is
inside the function, so we need to flip it over the vertical direction.
Does this graph look familiar? It is the same graph as the previous problem. Why? Notice the negative inside
the function above. If I raise a negative to an odd power then the negative can come outside the parenthesis. If
this happens you will get the same equation as the last example.
EXAMPLE: Sketch y ( x 2) 2 1 by using transformations.
Now we have the squaring function. This will be a parabola. We will move
the parabola 2 places to the right and down 1 unit.
EXAMPLE: Sketch y (( x 2)) 2 1 by using transformations.
This one is shifted in the same directions as in the last example. We have
a negative inside the function which means we will shift it over the vertical
axis. If you flip the graph it will look the same as the original, so the graph
is the same as the previous example. Also if you raise a negative to an even
power the negative goes away and you will get the same equation as in the
previous example so that is why the graph looks the same.
Section 1.7 Notes Page 1
1.7 Combinations of Functions; Composite Functions
The following problems deal with finding the domain without a graph, which can be done algebraically. If a
function has no places where you are dividing by zero and no places where you are taking the square root of a
negative then there are no domain restrictions, so the domain would be all reals.
EXAMPLE: Find the domain: y 2 x 5
There is no place where you can divide by zero or take the square root of a negative number, so the domain
would be all reals, indicated by , .
EXAMPLE: Find the domain: y
1
2x 5
Here it is possible to have a zero in the denominator. The denominator is not allowed to be zero, so solve:
5
2 x 5 0 . Solving this you will get x . This means any number but five halves will work. To write this
2
5 5
in interval notation it would be: , , .
2 2
EXAMPLE: Find the domain: y 2 x 5
For this one you need to make sure you do not take the square root of a negative number. The only numbers
that will work are positive numbers, so solve this equation: 2 x 5 0 . It is okay for our answer to equal 0.
5
5
Solving it you will get x . In interval notation this would look like , .
2
2
1
EXAMPLE: Find the domain: y
2x 5
This has two domains restrictions. First the denominator can’t be zero. Also we are not allowed to have
negative numbers under the square root. We will set it up almost the same as before, but this time we will not
include zero. We want to solve: 2 x 5 0 . We don’t want a zero in the denominator, so we don’t include it
5
5
in our answer. Solving this we get x and the interval notation would be , .
2
2
EXAMPLE: Find the domain: y
1
x3
The bottom root has an odd index (little number next to radical, which is 3). This is not a square root. Since we
have an odd index that means that if we take the odd root of a negative number, we will get something defined.
There for the only domain restriction is if the bottom equals zero, so we solve x 3 0 , so x 3 . If we
wanted to write this in interval notation it would be ,3 3, .
3
Section 1.7 Notes Page 2
1
x 9
Since we have a fraction we need to set the denominator equal to zero. Let’s look at the bottom for a second. Is
it possible for us to get a zero on the bottom? The answer is no. If you try 3 as you would suspect is the answer
it will not work since it will give you 18 since there is a plus sign. Since the bottom will never be zero that
means we have no domain restrictions, so we can use any real number for x. Interval notation: , .
EXAMPLE: Find the domain: y
2
x3
x6
Since we have a fraction we need to set the denominator equal to zero. Solving will give us x 6 . On the top
we have a square root, so we will solve the equation x 3 0 . We will get x 3 . Now we need to put both
these statements together. It tells us that any number greater than and including 3 will work, except for 6. In
interval notation it would be written as: 3, 6 6, .
EXAMPLE: Find the domain: y
Composite Functions – a way of combining two functions
( f g )( x) f ( g ( x)) This is pronounced “f of g of x” DOES NOT MEAN F TIMES G!!!
( g f )( x) g ( f ( x)) This is pronounced “g of f of x” DOES NOT MEAN G TIMES F!!
These are not multiplications. The ( f g )( x) means we place the g function into the f function.
The ( g f )( x) means we place the f function into the g function.
EXAMPLE: Given: f ( x) 5 x 4 and g ( x) 3 x 1 find the following:
( f g )(2) , ( g f )(1) , ( f g )( x) , ( g f )( x)
a.) ( f g )(2) f ( g (2)) First we rewrite this using the definition, replacing x with 2.
Let’s first do g (2) since this is inside the parenthesis. To do this, replace the x with 2 in the g function:
g (2) 3(2) 1 7 Since now we know that g (2) 7 we can now replace g (2) with 7:
By replacing g (2) with 7 now we need to find f (7) . Replace the x in the f function with 7:
f (7) 5(7) 4 31 . So now conclude that ( f g )(2) 31 .
b.) ( g f )(1) g ( f (1)) First we rewrite this using the definition, replacing x with -1.
Let’s first do f (1) since this is inside the parenthesis. To do this, replace the x with -1 in the f function:
f (1) 5(1) 4 9 Since now we know that f (1) 9 we can now replace f (1) with -9:
By replacing f (1) with -9 now we need to find g (9) . Replace the x in the g function with -9:
g 9) 3(9) 1 26 . So now conclude that ( g f )(1) 26 .
c.) ( f g )( x) f ( g ( x)) We use the definition.
Since we don’t have a number for x this time there will not be a number answer for this one. What we can do is
replace the g (x) with the expression 3 x 1 . So now we will do f (3 x 1) . This means where ever there is an
x in the f function we will replace it with 3x + 1:
f (3 x 1) 5(3 x 1) 4 Now simplify:
f (3 x 1) 15 x 5 4 15 x 1 So we write our answer as: ( f g )( x) 15 x 1
Notice what happens if we put a 2 in for x in our answer. We will get 31, which is the same as in part a.)
d.) ( g f )( x) g ( f ( x)) We use the definition.
Section 1.7 Notes Page 3
Since we don’t have a number for x this time there will not be a number answer for this one. What we can do is
replace the f (x) with the expression 5 x 4 . So now we will do g (5 x 4) . This means where ever there is an
x in the g function we will replace it with 5x - 4:
g (5 x 4) 3(5 x 4) 1 Now simplify:
g (5 x 4) 15 x 12 1 15 x 11 So we write our answer as: ( g f )( x) 15 x 11
Notice what happens if we put a -1 in for x in our answer. We will get -26, which is the same as in part b.)
EXAMPLE: Given: f ( x) x 3 and g ( x) 2 x 2 1 find the following:
( f g )(1) , ( g f )(0) , ( f g )( x) , ( g f )( x)
For this one the process is the same as I described about. I will only show the algebraic steps here.
a.) ( f g )(1) f ( g (1))
g (1) 2(1) 2 1
g (1) 1
f ( g (1)) f (1) 1 3 4
( f g )(1) 4
b.) ( g f )(0) g ( f (0))
f (0) 0 3
f (0) 3
g ( f (0)) g (3) 2(3) 2 1 17
( g f )(0) 17
c.) ( f g )( x) f ( g ( x))
f (2 x 2 1) (2 x 2 1) 3
( f g )( x) 2 x 2 2
d.) ( g f )( x) g ( f ( x))
g ( x 3) 2( x 3) 2 1 2( x 2 6 x 9) 1 2 x 2 12 x 18 1
( g f )( x) 2 x 2 12 x 17
1
find the following: ( f g )(0) , ( f g )( x) , ( g f )( x) .
x
Express all as single fractions in factored form if possible.
EXAMPLE: Given: f ( x) 4 x 2 and g ( x)
a.) ( f g )(0) f ( g (0))
g (0) undefined
Since this is undefined, that means that ( f g )(0) is also undefined.
Section 1.7 Notes Page 4
b.) ( f g )( x) f ( g ( x))
1
1
1
f 4 4 2 We now want to express this as a single fraction. Use common denominators.
x
x
x
x2 1
4x 2 1
4x 2 1
Now we just need to factor the numerator.
4 2 2 2 2
x
x
x2
x x
(2 x 1)(2 x 1)
( f g )( x)
x2
2
c.) ( g f )( x) g ( f ( x))
1
Since we already have a single fraction we now just need to factor the denominator.
g (4 x 2 )
4 x2
1
( g f )( x)
(2 x)(2 x)
EXAMPLE: Find functions f and g so that ( f g )( x) H ( x) given that H ( x) 1 x 2
3
This problem is the reverse of what we were just doing. Now we have the finished product and we want to go
back to the beginning. We want to find the two functions f and g so that when we find ( f g )( x) we will get
3
the expression 1 x 2 . First let’s find the g (x) . This is usually an ‘inside’ function, one put into something
else. In H we have the expression 1 x 2 . This is inside a set of parenthesis raised to the third power. So we
will let g (x) 1 x 2 . Now the ‘outside’ function is the f (x) . In order to find this function, replace the
‘inside’ function with x in H(x) and then replace H(x) with f(x). In this case we will replace 1 x 2 with x. You
will get f (x) x 3 . Now let’s check to make sure we did it correctly. If we let f (x) 1 x 2 and g ( x) x 3
3
let’s find ( f g )(x) . To do this, we place the g into the f. We will get 1 x 2 , which is equal to H. So the
answers are: f (x) 1 x 2 and g ( x) x 3 .
EXAMPLE: Find functions f and g so that ( f g )( x) H ( x) given that H ( x) 2 x 1 4 x 2
First let’s factor H: H ( x) 2 x 1 2(2 x 1) . Notice we have the expression 2x – 1 repeating. This is our
‘inside’ function, so g ( x) 2 x 1 . To find f (x) , replace the 2x – 1 with x in the H(x) equation and then
replace H(x) with f(x).. You will get f ( x) x 2 x . So our answers are g ( x) 2 x 1 and f ( x) x 2 x .
Section 1.8 Notes Page 1
1.8 Inverse Functions
One application of composition functions is inverses. First, if it is known that f and g are inverses, then:
( f g )( x) x and ( g f )( x) x
EXAMPLE: Given f ( x) 2 x 1 and g ( x)
1
1
x verify that they are inverses.
2
2
We need to show ( f g )( x) x and ( g f )( x) x . Let’s first find ( f g )( x) .
( f g )( x) f ( g ( x)) Start with the definition.
1
1
1
1
( f g )( x) f x We remove the g(x) and replace it with x .
2
2
2
2
1
1
( f g )( x) 2 x 1 Now simplify.
2
2
( f g )( x) x 1 1
( f g )( x) x
We have shown this is true. Now we need to show ( g f )( x) x .
( g f )( x) g ( f ( x)) First start with the definition.
( g g )( x) g 2 x 1 We remove the f(x) and replace it with 2x – 1.
1
1
( g f )( x) (2 x 1)
Now simplify.
2
2
( f g )( x) x
( f g )( x) x
1 1
2 2
We have shown this is true. So we have verified they are inverses.
What is the significance of the x? Why do we get x when we simplify? I’m glad you asked! Let’s look at the
graph of f and g. I will also graph y = x. Notice that f and g are symmetric to the line y = x. This is always the
case with inverses. Notice also that points on the graph of f(x) are reversed on g(x). For example, on the f(x)
line we see the points (2, 3) and (-1, -3). On the graph of g(x) we get the points (3, 2) and (-3, -1).
Section 1.8 Notes Page 2
It is not a coincidence that the points from f(x) are reversed on g(x). It just so happens that this is the way to
find inverses algebraically.
Notation to write “the inverse of f(x)” is f
means we have the inverse of f(x).
1
( x) . This does not mean f raised to the negative one power. It just
EXAMPLE: Verify the following are inverses: f ( x) x 3 and f
We need to show f ( f
1
( x)) x and f
1
( x) x 2 3
( f ( x)) x .
f(f
1
( x))
f ( x 3)
2
We need to show both sides are true:
1
f
1
( f ( x))
f 1 ( x 3 )
( x 2 3) 3
2
x 3 3
x 33 x
x2 x
Both sides are equal to x, so we have verified they are inverses.
How to find an inverse algebraically:
Step 1:
Step 2:
Step 3:
Step 4:
Replace f (x) with y.
Switch x and y.
Solve for y.
Replace y with f 1 ( x) .
EXAMPLE: Given f ( x) 2 x 5 find f
1
( x) . Then verify your answer is correct.
We will follow our four steps to find the inverse.
Step 1: y 2 x 5
Step 2: x 2 y 5
Step 3: x 5 2 y
x5
y
2
x5
f 1 ( x)
Step 4:
2
Now we need to verify our answer is correct.
We need to show f ( f 1 ( x)) x and f 1 ( f ( x)) x .
f(f
1
( x))
x 5
f
2
We need to show both sides are true:
x 5
2
5
2
x55 x
f
1
( f ( x))
So our answer is correct.
f 1 2 x 5
(2 x 5) 5 2 x
x
2
2
Section 1.8 Notes Page 3
EXAMPLE: Given f ( x) x 7 find f
1
( x) . Then verify your answer is correct.
Step 1: y x 7
Step 2: x
y7
Step 3: x
2
y7
2
x2 y 7
x2 7 y
Step 4: x 2 7 f 1 ( x)
Now we need to verify our answer is correct.
We need to show f ( f 1 ( x)) x and f 1 ( f ( x)) x .
f(f
1
( x))
f x 7
We need to show both sides are true:
2
( x 7) 7
2
x x
2
EXAMPLE: Given f ( x)
2x 3
find f
x4
1
f
1
f
( f ( x))
1
x7
2
So our answer is correct.
x7 7
x77 x
( x) . You do not need to verify this one.
2x 3
x4
2y 3
Step 2: x
Now to solve this one let’s first multiply both sides by y + 4 to cancel the fraction.
y4
Step 3: x( y 4) 2 y 3 Now we need to solve for y. First distribute.
xy 4 x 2 y 3 Let’s get all the terms that have y in it to one side of the equation.
xy 2 y 4 x 3 Now factor out a y.
y ( x 2) 4 x 3 Now divide both sides by x – 2 and replace y with f 1 ( x) .
4x 3
Step 4: f 1 ( x)
x2
Step 1: y
EXAMPLE: Given f ( x)
3x 5
find f
2x 3
1
( x) . You do not need to verify this one.
3x 5
2x 3
3y 5
Step 2: x
Now to solve this one let’s first multiply both sides by 2y – 3 to cancel the fraction.
2y 3
Step 3: x(2 y 3) 3 y 5 Now we need to solve for y. First distribute.
2 xy 3 x 3 y 5 Let’s get all the terms that have y in it to one side of the equation.
2 xy 3 y 3 x 5 Now factor out a y.
y (2 x 3) 3 x 5 Now divide both sides by 2x – 3 and replace y with f 1 ( x) .
3x 5
The f(x) graph was already symmetric to the line y = x so its inverse is itself!
Step 4: f 1 ( x)
2x 3
Step 1: y
Section 1.9 Notes Page 1
1.9 Distance and Midpoint Formulas; Circles
Distance Formula
The distance formula is used to find the distance between two points x1 , y1 and x2 , y2 .
Let’s first start with two points, (-2,1) and (1,5). First we plot the points. Then we will connect the points with
a line. I will also darken the vertical and horizontal differences of the points. A right triangle is now formed. I
will now label the actual vertical and horizontal distance.
Since we have a right triangle, we can use the Pythagorean Theorem to find the length of the longest side, which
is our distance. The Pythagorean Theorem says that if you have a right triangle, then a 2 b 2 c 2 where c is
the longest side of the triangle. So we will solve the equation:
(3) 2 (4) 2 c 2
9 16 c 2 We are not done. We need to solve for c, so take the square root of both sides.
5c
So the distance between these two points is 5 units.
Instead of plotting the points we can use the distance formula, which still involves the Pythagorean theorem.
If you have points x1 , y1 and x2 , y2 then the distance formula is d
x 2 x1 2 y 2 y1 2
EXAMPLE: Use the distance formula to find the distance between the points (-2, 4) and (1, 6)
It does not matter the order that the points are in. I will let the (-2, 4) be x1 , y1 and (1, 6) be x2 , y2 . We will
now substitute these numbers into the distance formula:
d
1 (2) 2 6 42
d 3 2 2 2 13
You don’t need to turn this into a decimal.
There are different applications the distance formula can be used for One example is circle equations, which we
will look at later. You can also use the distance formula to see if a triangle is isosceles (two sides equal). Once
you know the distance of all the sides, you can plug these into the Pythagorean Theorem to see if it is a right
triangle. If it is a right triangle then both sides will be equal when you use the formula a 2 b 2 c 2 .
Section 1.9 Notes Page 2
Midpoint Formula
The midpoint is the halfway point on a line. If the line is formed by the points x1 , y1 and x2 , y2 , then
x x 2 y1 y 2
The midpoint is: M 1
,
.
2
2
Notice our answer is a point (x, y). This divides the line into two pieces of equal length.
EXAMPLE: Find the midpoint of a line segment containing (2, -3) and (4, 2).
I will let the (2, -3) be x1 , y1 and (4, 2) be x2 , y2 . Plug these into the formula:
2 4 3 2
M
,
2
2
1
M 3,
2
Circles
Standard Form
x h 2 y k 2 r 2 where (h, k) is the center and r is the radius
EXAMPLE: Write the standard form of a circle with a center of (-2, 3) and a radius of 4 and graph.
We will use the standard form formula. Here, h = -2, k = 3, and r = 4. Plug them into the formula.
Equation: x 2 y 3 4 2 To graph, plot the center. The go vertically up 4 and make a point. Then
from the center go down 4 and make a point. Then to the left and to the right. See all graphs on next page.
2
2
EXAMPLE: Write the standard form of a circle with a center of (5, 1) and a radius of 9 and graph.
We will use the standard form formula. Here, h = 5, k = 1, and r = 3. Plug them into the formula.
Equation: x 5 y 1 3 2 To graph, plot the center. The go vertically up 3 and make a point. Then
from the center go down 3 and make a point. Then to the left and to the right. See all graphs on next page.
2
2
EXAMPLE: Find the center and radius and graph: ( x 1) 2 ( y 3) 2 25
We will use the standard form formula. To find the center take the opposite sign of each number inside the
parenthesis. In the first parenthesis there is a +1 so the x coordinate of the center will be -1. In the second
parenthesis there is a -3, so the y coordinate of the center will be 3. So the center is (-1, 3). The radius is
always the square root of the number after the equal sign, so 25 5 . See all graphs on next page.
Section 1.9 Notes Page 3
EXAMPLE: Find the center and radius and graph: ( x 3) y 13
2
2
We will put this into the standard form formula. We can rewrite the second term as: ( y 0) 2 . Now our
equation becomes: ( x 3) 2 ( y 0) 2 13 . To find the center take the opposite sign of each number inside the
parenthesis. In the first parenthesis there is a -3 so the x coordinate of the center will be 3. In the second
parenthesis there is a 0, so the y coordinate of the center will be 0. So the center is (3, 0). The radius is always
the square root of the number after the equal sign, so 13 3.6 . We can’t simplify this anymore. After
plotting the center, go up, down, left, and right 3.6.
EXAMPLE: Find the center and radius and graph: x 2 y 2 10 x 8 y 16 0
This is not in standard form, like the above example so we need to first get it into the proper form. We want to
do this by by completing the square.
x 2 10 x y 2 8 y 16 First put like terms together and move the 16 to the other side.
Now we need to complete the square. Take the number in front of the variable that is not being squared and
divide it by two. Then square this result. You will add this result to both sides of the equation. For example,
there is a 10x, so first divide 10 by 2 and you get 5. Then square this number and you will get 25. Do the same
for the 8y term. Divide -8 by 2 to get -4. Then square the -4 to get 16. We always divide by 2 and then square
it. x 2 10 x 25 y 2 8 y 16 16 25 16 Notice we added 25 to both sides and 16 to both sides.
( x 5) 2 ( y 4) 2 25 Now factor and simplify the right hand side.
Now this is in standard form, so we know the center is (-5, 4) and the radius is 5. See graph on next page.
EXAMPLE: Find the center and radius and graph: 2 x 2 2 y 2 8 x 7 0
In order for complete the square to work, there must be a 1 in front of the squared terms, so I need to divide
7
both sides of the equation by 2 to get: x 2 y 2 4 x 0
2
Now we will put like terms together and moved the fraction to the other side to get
7
The next step is to complete the square. We can only complete the square with the x since
x 2 4x y 2
2
this has a non-squared term. We take the number 4 in front of the x and divide it by 2 and we get 2. Then we
square this and get 4. We will add this to both sides.
7
x 2 4 x 4 y 2 4 Now we need to factor the left side and simplify the right side.
2
x 22 y 02 1 Notice that I rewrote y 2 . It didn’t change the problem, just put in standard form.
2
1
1
2
The radius is always the square root of the number after the = sign.
Center: (-2, 0) Radius:
2
2
2
Section 1.9 Notes Page 4
Example 1 Graph
Example 2 Graph
Example 5 Graph
Example 6 Graph
Example 3 Graph
Example 4 Graph
EXAMPLE: The endpoints of the diameter of a circle is (2, 6) and (-4, -2). Determine the coordinates for the
center of the circle, and find the length of the radius. Then write the standard form of the circle.
This is an application of the two formulas we have learned in this section. To find the center of the circle we
need to find the center, since the center is the midpoint of the radius. The radius is defined as half the length of
the diameter or the distance from the center to a point on the circle. So we have two ways of doing it.
2 4 6 2
Center: M
,
= (-1, 2) We have used the midpoint formula here.
2
2
Radius: r 2 1 6 2 3 4 25 5 Distance formula with points (-1, 2) and (2,6).
You could have also used (-1, 2) and (-4, -2). You could also use (2, 6) and (-4, -2) and take half this distance.
2
2
2
2
Now we put this information into the standard form of a circle formula: ( x 1) 2 ( y 2) 2 25
Section 1.10 Notes Page 1
1.10 Modeling with Functions
In this section we will take a look at some applications working with functions.
1
EXAMPLE: The volume of a right circular cone is V r 2 h . If the height is triple the radius, express the
3
volume V as a function of r.
First we need to figure out what they are asking for. The phrase “express the volume V as a function of r”
means that we need to find an equation called V that has r as the only variable. So we want to find V(r). The
equation they gave us has r and h as the variables, so we want to get rid of the h somehow with something that
has r in it. They tell us that the height is triple the radius, so we know that h = 3r. We can remove the h and
1
replace it with 3r to get: V r 2 (3r ) . The last thing to do is simplify: V (r ) r 3 . The 3 cancels the
3
fraction and both of the r’s make r cubed. Notice we need to write V(r) as our answer since they ask for a the
volume V as a function of r. Now r is the only variable.
EXAMPLE: A farmer has 300 ft. of fencing available to enclose a rectangular field. One side of the field likes
along a river, so only 3 sides require fencing. Express the area , A, as a function of width x.
It’s best to draw a picture for this one. It says to label the width x. I will
label the other side x as shown in the diagram:
x
x
Now we need to find an expression for the perimeter. To find the perimeter
y
we need to add all the sides, which is three in this case. The amount of
fencing was 300 ft, and this is actually the perimeter. Add all three sides and set it equal to 300. You will get:
300 = 2x + y. The question asks us to find the area. From our rectangle the area would be A = xy. We want to
write the area as a function of width x, so all the variables in that equation must be x. We will solve the
perimeter equation for y. If you do that you will get y = 300 – 2x. We just need to replace the y in the area
equation with 300 – 2x. You will get A( x) x(300 2 x) . This answer is correct or you could also multiply
out by the x to get A( x) 300 x 2 x 2 . This is also correct.
EXAMPLE: A rectangular box has a square base. The length of the box is twice the length of one of the sides
of the square base. Express the volume of the box as a function of x where x is one of the sides of the square
base.
Let’s label the length as y. Then since this is twice the length of one of the sides of the square base we have the
equation y = 2x. Let’s also draw a picture and label our sides.
The volume of a box like this is V = lwh. The width and the height are
both x. The length is 2x. So we have V(x) = (2x)(x)(x). Simplifying we
will have our equation: V ( x) 2 x 3
x
y = 2x
x
Section 1.10 Notes Page 2
EXAMPLE: A right triangle has one vertex on the graph y 9 x 2 , (x > 0), another vertex at the origin, and
a third vertex on the positive x-axis. (See graph). Express the area of the triangle as a function of x.
y 9 x2
Here the graph is drawn for us. Let’s look at the triangle. It has a
base from the origin to some value x. So the length of the base of
triangle is x. What about the vertical height? Well this one
depends on where it touches the curve. So the height is y, but we
9 x 2 , so this is the height. We
1
now need the area of a triangle formula. It is A bh . Now we
2
have an expression for y. It is
just need to substitute an x for b and 9 x 2 for h. You will get:
1
A( x) x 9 x 2 which is the answer.
2
EXAMPLE: Let P (x, y) be a point on the graph of y x 2 8 . Express the distance d from P to the point
(0, -1) as a function of x.
For this one we need to use the distance formula. The distance formula is d ( x 2 x1 ) 2 ( y 2 y1 ) 2 . In
order for this to work we need two points. One is (0, -1) and the other is (x, y). The second point we need to
write so that x is the only variable since the question is asking us to find the distance as a function of x. We can
write the second point as x, x 2 8 . Now we put both of these into the distance formula and simplify:
d ( x 2 x1 ) 2 ( y 2 y1 ) 2
d ( x 0) 2 ( x 2 8 (1)) 2
d x 2 ( x 2 7) 2
Here we need to multiply ( x 2 7)( x 2 7) x 4 14 x 2 49
d x 2 x 4 14 x 2 49
d ( x) x 4 13x 2 49 This is our final answer written with function notation.
1
EXAMPLE: The price p and quantity x sold of a certain product obey the demand function p x 100
3
where 0 x 300 . Express the revenue R as a function of x. What is the revenue if 100 units are sold?
In order to do this one we need to first know what the formula is for revenue: Revenue = Price x Quantity.
1
In our problem the price is x 100 and the quantity is x. We will multiply these together to get our revenue
3
1
1
formula: R ( x) x 100 x . After multiplying we get R( x) x 2 100 x . This answers the first
3
3
question. Next they want to know what the revenue is if 100 units are sold. Since the revenue is x we put in
1
100 for x: R (100) 100 2 100(100) . After simplifying we get $6666.67.
3
Section 1.10 Notes Page 3
EXAMPLE: The cost of renting a truck is $20 a day plus 50 cents per mile. What is the maximum number of
miles that can be driven in one day so the cost does not exceed $100?
First we need the equation for the cost. No matter how many miles are driven you must pay $20 for the one day
rental. Added to this is 50 cents per mile. The cost equation is C = 0.5x + 20 where x is the number of miles
driven. This cost can’t exceed, or go over $100. In other words the cost must be less than or equal to $100. So,
0.5 x 20 100 . Now solve for x. You will get x 160 . Therefore you must not exceed 160 miles in one day
in order to keep the costs below $100 for one day.
EXAMPLE: A company is planning to manufacture a certain product. The fixed costs will be $500000 and it
will cost $400 to produce each product. Each will be sold for $600. What is the profit equation and how many
units must be sold for in order to break even?
Profit is defined as the revenue minus the costs. We need to find our revenue and cost equations. The costs
involve a fixed price plus a variable price. The equation is C = 400x + 500000. Since each is sold for $600 then
this is the revenue, which is price times quantity. You will get R = 600x. To get the profit function you need to
subtract the revenue from the cost: P = 600x – (400x + 500000). Simplifying you get: P = 200x – 500000.
When you break even the profit will be zero. Put a zero in for P and solve for x: 0 = 200x – 500000. Solving
this you will get x = 2500 units.
EXAMPLE: On a certain route, an airline carries 7000 passengers per month, each paying $90. A market
survey indicates that for each $1 decrease in the ticket price, the airline will gain 60 passengers. Express the
number of passengers per month, N, as a function of the ticket price, x. Then express the monthly revenue for
the route, R, as a function of the ticket price, x.
Let’s first set up our variables. Let’s let x equal the ticket price as it says in the problem. Now let’s express the
decrease in the ticket price. We start with $90, and since we are decreasing the ticket price, we need to use
subtraction, so our expression is 90 – x. Now we need to find the increase in passenger due to the fare decrease.
The increase in passengers is equal to 60 times whatever the decrease in price is: 60(90 – x). The number of
passengers depends on the ticket price. This is the original number, 7000, PLUS the number added due to the
fare decrease. Here is the corresponding formula: N ( x) 7000 60(90 x) . We can simplify this to:
N ( x) 12400 60 x . This answers the first question.
For the second question we want to know the monthly revenue for the route. Revenue is equal to price times
quantity. In our case it is the ticket price (x) times the number of passengers, N. So, R ( x) x(12400 60 x ) .
This can be also written as: R ( x) 12400 x 60 x 2 .
Section P.7 Notes Page 1
P.7 Equations
Quadratic Functions – Standard Form is f ( x) ax 2 bx c where a 0 .
In this section we will look at a couple ways of finding the x-intercepts of a quadratic equation as defined
above. We will solve them by factoring, square root method, and the quadratic equation. The book covers
completing the square, but we will not focus on that method.
Factoring
EXAMPLE: Find the x-intercept by factoring: f ( x) x 2 7 x 6
Since we are find the x-intercepts the f(x) will be zero. So we are trying to solve: 0 x 2 7 x 6 . For
quadratics in this form with a 1 in front of the squared term we need to find two numbers that multiply to make
the last term but add to make the middle term. So we need to find two numbers that multiply to make 6 but add
together to make 7. The numbers would be 1 and 6. So we factor: 0 ( x 1)( x 6) . Setting both of these
factors equal to zero you will get x = -1 and x = -6. As intercepts we write (-1, 0) and (-6, 0).
EXAMPLE: Find the x-intercept by factoring: f ( x) 3x 2 6 x 24
Since we are find the x-intercepts the f(x) will be zero. So we are trying to solve: 0 3 x 2 6 x 24 . Now it
may seem like we have large numbers, but we should always see if we can factor out any common factors. In
this problem all three numbers are divisible by 3, so 3 is a common factor. Factor this out and you will get
0 3( x 2 2 x 8) . You can divide both sides by 3 to get: 0 x 2 2 x 8 . Now we look for factors of -8 that
add up to -2. These will be 2 and -4. We will get 0 ( x 2)( x 4) . Setting both equal to zero we will get two
answers: x = -2 and x = 4. As intercepts we write (-2, 0) and (4, 0).
EXAMPLE: Factor: f ( x) 6 x 2 x 12
For this problem you have three different ways to solve it, but I will only show two ways. The third method is
trial and error, but if a and c have lots of factors it could take a long time to factor it, which is why I will only
focus on the first two methods.
I will solve the same problem two different ways.
Grouping Method
Multiply the a coefficient (6) by the constant c (-12). You will get -72. Now we need to find a way of rewriting
the middle term b (-1) as factors of -72. Instead of –x we will write 8x – 9x. So now our problem becomes:
6 x 2 8 x 9 x 12 From here we will factor by grouping (factor first two and last two terms separately).
2 x(3x 4) 3(3x 4) Now factor out the common factor of 3x + 4.
(3x 4)(2 x 3)
Section P.7 Notes Page 2
Bottom’s Up (Hervey) Method
This method you probably have not seen before. This method only works once you have taken out any common
factors. You will still do the same step as in the grouping method. Multiply the first number by the last: 6(-12)
= -72. What will also happen is that the first number (6) disappears, and the -12 will turn into our product of 72. Here’s what it will look like:
x 2 x 72
First number is gone and the -12 is now a -72. From here, factor normally.
( x 8)( x 9)
Now the number that disappeared (6) is now written below each number in the parenthesis
8
9
x x So now we formed fractions. These ALWAYS need to be reduced in order to be correct.
6
6
4
3
x x Now that they are reduced, bring the “bottoms up”. Bring the 3 in front of the x and bring the
3
2
2 in front of the x. This is where this process gets its name because the bottom comes up.
(3 x 4)(2 x 3) This is our final answer, same as with grouping.
EXAMPLE: Solve for x: 4 x 2 38 x 70 0
Notice that numbers are large on this one. We must see if there is a common factor first. We find that each
number is divisible by 2, so we can factor out a 2: 2(2 x 2 19 x 35) 0 Now we can factor what is inside the
parenthesis. This doesn’t ask for an intercept, but just to solve for x.
Grouping:
2(35) = 70. Want to find to factors of 70 than add to -19 so we can rewrite the middle term. The numbers that
will work are -5 and -14. Now we rewrite the problem and finish by using factoring by grouping.
2(2 x 2 5 x 14 x 35) 0
2( x(2 x 5) 7(2 x 5)) 0
2(2 x 5)( x 7) 0 So now we set each part equal to zero and solve.
5
x and x 7 . What about the 2 in front? This can’t be zero, so this does not give us another answer.
2
Bottom’s Up (Hervey) Method:
2(35) = 70. Remember the first number (2) disappears and 35 is replaced by 70.
2( x 2 19 x 70) 0 . Now factor as normal.
2( x 5)( x 14) 0 Now divide both numbers by the number that disappeared (8).
5
14
2 x x 0 Reduce any fractions, if possible. We can only reduce the second fraction.
2
2
5
2 x x 7 0 We can actually set this to zero now without bringing the bottom up.
2
22 x 5 x 7 0 If you did bring the bottom up you would have the same equation as above. Now solve.
5
x and x 7 .
2
Section P.7 Notes Page 3
EXAMPLE: Find the x-intercept by factoring: f ( x) x 36
2
We put a zero in for f(x) to get 0 x 2 36 . Now we can factor using the difference of squares formula, which
is a 2 b 2 (a b)(a b) . So we get 0 ( x 6)( x 6) . Solving it we get (-6, 0) and (6, 0).
Square Root Method
This method involves taking the square root of both sides of the equation. Let’s look at the above problem. We
had 0 x 2 36 . Instead of factoring we can just solve for x. We can add 36 to both sides to get 36 x 2 .
Then we can take the square root of both sides to get 6 x . Then we write it as (-6, 0) and (6, 0).
EXAMPLE: Solve 0 (3x 2) 2 4 by using the square root method.
We could distribute and multiply here, but there is an easier way. We will solve this similarly to the problem
above. First add 4 to both sides. You will get: 4 (3 x 2) 2 . Now take the square root of both sides.
2 3 x 2 . Don’t forget the plus and minus signs whenever you take the square root. Since we get two
numbers we will solve two equations: 2 = 3x – 2 and -2 = 3x – 2 . Solve each one individually and we will get
4
x and x 0 . This is how we can leave our answer since we are not finding intercepts.
3
Quadratic Formula
b b 2 4ac
.
2a
You usually want to use this method when it is too hard to factor or if it doesn’t factor at all. So a, b, and c in
the formula come from the standard form 0 ax 2 bx c .
This last method will work for anything in this form: 0 ax 2 bx c . The formula is: x
EXAMPLE: Solve by using the quadratic formula: 0 x 2 4 x 2
This one doesn’t factor, so the quadratic formula is the only way to solve it. Here a = 1, b = 4, and c = 2. We
will put these into the quadratic formula and simplify:
2
b b 2 4ac 4 4 4(1)(2) 4 16 8 4 8 4 2 2
x
2 2
2a
2(1)
2
2
2
Notice we broke down the
8 by writing
8 4 2 2 2 . Then we divided everything by 2.
EXAMPLE: Solve by using the quadratic formula: 0 4 x 2 x 2
Here we have a = 4, b = -1, c = 2. Put it into the formula:
2
b b 2 4ac 1 (1) 4(4)(2) 1 1 32 1 31
x
. We can’t take the square root of a
2a
2(4)
8
8
negative number, so the answer is undefined.
The book mentions how to solve equations that look like quadratics but have different powers. We will not be
covering this material due to lack of time. Skip those problems from the homework.
Section 2.1 Notes Page 1
2.1 Complex Numbers
In this section we will review complex (or imaginary) numbers. First we will start with the basic definition of
an imaginary number. We will let i 1 . Normally we would not be able to get a value of this, but in this
section we will be working with i and how to simplify expressions involving these kind of numbers. Let’s look
at what happens when we see different powers of i:
i0 1
i1 i
Anything to the power of 0 is always 1.
Anything to the power of 1 is itself.
i 2 1
i 3 i
i4 1
This is true because i 2 i i 1 1 1 1
This is true because i 3 i 2 i 1 i i
This is true because i 4 i 3 i i i 1 1 (1) 1
2
Notice we get back to 1. This whole pattern will kept repeating. So every the pattern repeats every fourth
power of i. We can use this principle to answer the following:
EXAMPLE: Simplify: i 38 .
To do these kind of problems, always divide the power by 4 since our pattern above repeats after every four.
You want to get the remainder. If we divide 38 by 4 we will get 9 with 2 left over. Whatever the remainder is
will become the new power. So i 38 i 2 1 . So our conclusion is that i 38 simplifies to – 1.
EXAMPLE: Simplify: i 359 .
Again we divide the power (359) by 4. You will get 89 with 3 left over. Since 3 is our remainder, then 3
becomes our new power: i 359 i 3 i . So our conclusion is that i 359 simplifies to –i.
Standard Form of an Imaginary (Complex) Number
a + bi
Here the a is always the real number part and b represents the imaginary, or complex part. If a question asks
you to write your answer is standard form you must always put the real number part first and the i part second.
EXAMPLE: Add and simplify: (9 12i ) (5 2i ) . Write your answer in standard form.
Here you need to add the like terms. The real numbers are like terms and the i terms are also like terms. Add
the real numbers and you will get 14. Add the i terms and you will get 7i. Now just write your answer in
standard form: 14 + 10i.
EXAMPLE: Subtract and simplify: (8 5i ) (2 6i ) and write your answer in standard form.
We will first distribute the negative and we will get: 8 5i 2 6i . Now add the like terms. We can add 8 and
2 to get 10 and we can add -5i and -6i to get -11i. Now write in standard form: 10 – 11i. Again the real
number part must come first and the i part second.
Section 2.1 Notes Page 2
EXAMPLE: Multiply and simplify: (2 5i )(3 4i ) . Write your answer in standard form.
For this one we need to use the FOIL method. This means we multiply the First, Outer, Inner, and Last terms.
You will get: 6 8i 15i 20i 2 . We can simplify this to: 6 7i 20i 2 . From our list at the beginning of the
section, we know that i 2 1 . So we will replace the i 2 in the above expression with -1. Then you will have:
6 7i 20(1) . Now simplify and write in standard form. Your answer is 26 7i .
EXAMPLE: If z 1 2i and w 3 4i , find z w and write your answer in standard form.
First we write our problem as (1 2i )(3 4i ) . Just multiply this out using the FOIL method: 3 4i 6i 8i 2 .
After simplifying you have: 3 2i 8i 2 . Now put in a -1 for i 2 and you will have 3 2i 8(1) which is
11 2i in standard form.
EXAMPLE: Simplify and write your answer in standard form:
We can rewrite this problem as:
in an i for 1 . Then we have
4 16 .
4 16 1 . Then we can separate this into: 4 16 1 . We can put
4 16 i . We can take the square roots to get: 2 4i .
EXAMPLE: Divide and simplify:
4 2i
. Write your answer in standard form.
3i
In order to do this we need to multiply the top and bottom by something so that I don’t have an i in the
denominator. The conjugate will work here, which is 3 – i.
4 2i 3 i
3i 3i
So now that we have it set up we need to multiply across the top and bottom.
12 4i 6i 2i 2
9 i2
Across the top is the same process as the last example. On the bottom is a difference of
squares: (a b)(a b) a 2 b 2 . Now we need to simplify the top and bottom.
12 10i 2i 2
9 i2
Again we need to replace the i 2 with -1.
12 10i 2(1)
9 (1)
Now simplify.
10 10i
10
This is not written in standard form. We can divide both things on top by 10.
1 i
This is written in standard form and is our answer.
Section 2.1 Notes Page 3
EXAMPLE: Divide and simplify:
4 49
25 9
. Write your answer in standard form.
First we need to get rid of the roots. You will get:
2 7i
.
5 3i
Now we need to multiply the top and bottom by the conjugate of the bottom, which is 5 + 3i.
2 7i 5 3i
5 3i 5 3i
Now multiply across the top and bottom of the fraction.
10 6i 35i 21i 2
25 9i 2
We can simplify the numerator.
10 41i 21i 2
25 9i 2
Now put in a -1 for i 2 .
10 41i 21(1)
25 9(1)
Now simplify
11 41i
34
This is not in standard form, so divide each thing in the top by 34.
11 41
i
34 34
Now this is in standard form even though we can’t simplify the fractions anymore.
EXAMPLE: Simplify and write your answer in standard form:
12
4 2 .
First we need to change all of these into imaginary numbers: For the 12 we can break this down by writing
4 3 1 2 3 i 2i 3 . Usually we put the i in front of the square root to avoid confusion of whether the i
is under the square root or not. Changing everything to imaginary numbers we get: 2i 3 2i 2 . Now we
distribute: 4i 2 3 2i 6 . We can replace the i with -1: 4 3 2i 6 .
2
EXAMPLE: Solve the equation using the quadratic formula and write in standard form: 0 3 x 2 4 x 6 .
To review, the quadratic formula is: x
b b 2 4ac
for ax 2 bx c 0 . Here a = 3, b = -4, and c = 6.
2a
(4) (4) 2 4(3)(6) 4 16 72 4 56 4 2i 14 4 2i 14 2
14
x
i
. Here, we
2(3)
6
6
6
6
6
3
3
can write 56 as 4 * 14 * 1 4 * 14 * 1 2i 14 . We needed to divide everything individually by 6
since they wanted us to write this in standard form.
Section 2.2 Notes Page 1
2.2 Quadratic Functions
This section is all about quadratic functions, which give U shaped graphs called parabolas. First we need to
define a couple of terms involving parabolas:
At the bottom of this graph we have the vertex. This is either the
lowest or highest point on a parabola. In the standard form
y ax 2 bx c if a > 0 then the graph opens up and the vertex is
the lowest point on the graph. If a < 0 then the graph opens down
and the vertex is the highest point on the graph.
Another new term is the axis of symmetry. This is a vertical line
that always goes through the x-coordinate of the vertex. Since this
is a vertical line it will start with x = . This is considered a fold
line since the parabola can be folded on top of itself.
Vertex form: y a ( x h) 2 k . If an equation is written in this form, then the vertex is (h, k). If the a value is
negative, then the graph opens down. If the a value is positive, then the graph opens up.
EXAMPLE: Find the vertex, axis of symmetry, intercepts and graph of y 2( x 3) 2 8 .
In order to find h, we can rewrite the formula as: y 2( x (3)) 2 8 .
Now we know h = -3. We also know that k = 8. So the vertex is (-3, 8).
The axis of symmetry goes through the x-coordinate of the vertex, so this
equation is x = -3. To find the y-intercept, put in a zero for x in our original
equation: y 2(0 3) 2 8 . Solving this we get y = -14. The y-intercept
is written as (0, -10). For the x-intercept, put in a zero for y:
0 2( x 3) 2 8 . To solve this, subtract 4 from both sides: 8 2( x 3) 2 .
Now divide both sides by -2: 4 ( x 3) 2 . Now take the square root of both
sides to get: 2 x 3 . This gives us two separate equations to solve:
x + 3 = 2 and x + 3 = -2. Solving each individually we get x = -1, -5.
We write the x-intercept as (-1, 0) and (-5, 0). Plot all these points to get the graph.
So now what do we do if the quadratic equation is not written in vertex form? The book shows on page 290
that we can start with the standard form of a quadratic and use the completing the square method to find a
formula to find the vertex if an equation is not written in vertex form. I will just give the formula here.
Given y ax 2 bx c , then the x-coordinate of the vertex is x
into the original equation.
b
. To find the y-coordinate put the x back
2a
Section 2.2 Notes Page 2
EXAMPLE: Find the vertex by using the formula: y 3 x 6 x 1 . Then find the axis of symmetry.
2
Here, a = 3, b = 6, and c = 1.
(6)
1 . So now we know that x = -1. Now we need to put -1 in for x in the
2(3)
original equation y: y 3(1) 2 6(1) 1 . After simplifying we get y = -2. We write the vertex as (-1, -2).
This will be an actual point on the parabola. The a in this problem is greater than zero, therefore the vertex will
be a minimum.
First we use the formula: x
The axis of symmetry is equal to the x-coordinate of the vertex. To find it, just put x = and then the xcoordinate of the vertex. For this problem the axis of symmetry is x = -1.
EXAMPLE: Find the vertex by using the formula: y 2 x 2 5 x 1 . Then find the axis of symmetry.
Here, a = -2, b = -5 and c = 1. Using the vertex formula we get: x
(5)
5
. We get a fraction. This
2(2)
4
5
5
doesn’t matter. We still put this in for x in the y equation: y 2
1 . To square a fraction
5
4
4
25 5
you need to square the top and bottom number you will get: y 2 5
1 . Now multiply across
16 4
25 25
the top and across the bottom when multiplying fractions: y
1 . We need common
8
4
denominators.
2
25 50 8
33
Now we can add the fractions to get: y
. So the vertex is
8
8 8
8
5
than zero, the vertex will be a maximum. The axis of symmetry is x .
4
y
5 33
, . Since a is less
4 8
Now we will do problems that involve graphing the parabola. You will need to find the x and y intercepts and
the vertex first. Then you can plot the points and get the graph.
EXAMPLE: Find the intercepts, vertex, axis of symmetry, domain, range, and the graph of y x 2 6 x 5 .
First we will find the x-intercept. Put in a zero for y. You will get 0 x 2 6 x 5 . In order to solve this you
must factor. You will get 0 ( x 1)( x 5) . Solving you will get x = 1 and x = 5, or (1, 0) and (5, 0) in
intercept form. To find the y-intercept, put in a 0 for x. You will get y = 5, or (0, 5) in intercept form. Now we
need to find the vertex. We will use the vertex formula so find the x coordinate of the vertex:
b (6)
x
3 . So now we will put in a 3 for x in the original equation to get the y-value of the vertex.
2a
2(1)
y (3) 2 6(3) 5 so y = -4. The vertex is (3, -4). Now we just need to plot the points to get the graph.
Section 2.2 Notes Page 3
Notice that the number in front of the x 2 is greater than zero, so
the parabola opens up and the vertex is the lowest point on the
graph.
Here the domain (x values) is: , .
The range (y values) is 4,
EXAMPLE: Find the intercepts, vertex, axis of symmetry, domain, range, and the graph of y x 2 5 .
When we solve 0 x 2 5 to find the x-intercept we can use the square root method. Add 5 to both sides. You
will get x 2 5 . Take the square root of both sides to get x 5 , or 5 ,0 in intercept form. To find the
y-intercept, put in a 0 for x. You will get y = -5, or (0, -5) in intercept form. Now we need to find the vertex.
0
b
0.
We will use the vertex formula so find the x coordinate of the vertex. Here the b is zero: x
2a 2(1)
So now we will put in a 0 for x in the original equation to get the y-value of the vertex: y 0 2 5 5 so the
vertex is (0, -5). To graph we will plot the points we found. To plot 5 ,0 we can change this into a decimal
so we know where this is on our x-axis. The decimal form is: 2.24,0 .
The number in front of the x 2 is greater than zero, so the parabola
opens up and the vertex is a minimum.
Here the domain (x values) is: , .
The range (y values) is 5, .
EXAMPLE: Find the intercepts, vertex, axis of symmetry, domain, range, and the graph of y 6 4 x x 2 .
Putting in a zero for x will give us the y-intercept, which is (0,6). To find the
y-intercept, put in a 0 for y. You will get: 0 x 2 4 x 6 . We need to use the quadratic formula since this one
does not factor. Notice I rewrote the problem with descending powers. Now we know a = 1, b = -4, and c = 6.
If you put this into the quadratic formula, you will get:
(4) (4) 2 4(1)(6) 4 16 24 4 8 4 2i 2
2 i 2 . Notice this is not a real
2(1)
2
2
2
number. This means the graph does not have any x-intercepts, so we know it does not cross the x-axis. Now
we need to find the vertex. We will use the vertex formula so find the x coordinate of the vertex.
x
Section 2.2 Notes Page 4
b (4)
2 . So now we will put in a 2 for x in the original equation to get the y-value of the vertex:
2a
2(1)
y 6 4(2) (2) 2 6 8 4 2 so the vertex is (2, 2). Since we have the vertex then we can also find the
axis of symmetry, which is x = 2. To graph we will plot the points we found, which is (0, 6) and (2, 2). Now
we need to find a third point. This is where we can use the axis of symmetry. We can draw a vertical line
through the graph once we plot our two points. Let’s look at the point (0, 6). This is a distance of 2 units away
from the axis of symmetry. We know that there is another point two units to the right of the axis of symmetry
that has the same y-value as (0, 6). We will get the point (4, 6). Notice that this graph is completely above the
x-axis, meaning it does not have any REAL x-intercepts, however it does have two complex intercepts: 2 i 2
x
The domain is , .
The range is 2, .
Modeling with Quadratic Functions
This section is almost the same as section 1.6. We will still be setting up functions from word problems, only
now we will go one extra step. These problems involve finding maximum or minimum values. ALL of the
problems in this section require you to find the vertex once you have your equation.
To find a maximum or minimum value you must find the vertex.
EXAMPLE: Suppose you have 900 ft of fencing and you want to enclose a rectangular field that borders a
river, so only three sides require fencing. What is the largest area that can be enclosed?
Start by drawing a picture and labeling the sides. We only have three sides to this rectangle:
x
x
First we want to find a formula for the perimeter. It is 900 2 x y .
Solving for y we get y 900 2 x . Now substitute 900 – 2x for y in the
area formula A xy . You will get A( x) x(900 2 x) , or
A( x) 900 x 2 x 2 . Since it is asking you for the largest area, we need to
find the vertex. We will use the vertex formula.
900
225 . So we know that x = 225. To find the y value we will use the formula y 900 2 x and put
2(2)
in 225 for x. You will get y 900 2(225) 450 . So y = 450. The question is asking us to find the largest
area. We will multiply x and y to get the answer. Our area will be 225 multiplied by 450, which is 101250 ft 2 .
x
Section 2.2 Notes Page 5
EXAMPLE: Among all pairs of numbers whose difference is 24, find a pair whose product is as small as
possible. What is the minimum product?
We need to set up two equations on this one. The word DIFFERENCE means subtraction, so one equation is
x – y = 24. The word product means multiply, so another formula is P = xy. We need to solve the first
equation for either x or y and then substitute this into the second equation. If we add y to both sides in the first
equation we will get x = y + 24. Now put this into the second equation in place of x. We will get P = (y + 24)y.
Distributing will give us p y 2 24 y . Now we use the vertex formula. It doesn’t matter that we have y
24
12 . So we know that y = -12. To find the
instead of x. Just means we will find the y first: y
2(1)
second number, we will use the formula x = y + 24. This means x = -12 + 24 = 12. So our pair of numbers is
12 and -12. If we multiply these together we will get the product: 12(-12) = -144.
EXAMPLE: A rain gutter is to be made of aluminum sheets that are 20 inches wide. As shown in the figure,
the edges are turned up to form right angles. What depth will provide maximum cross-sectional area resulting in
allowing the most water to flow?
If the edges are bent up, it will form a rectangle: x
The sides bent up are x. To get the bottom we
can see that the original piece of metal was 20
inches long. The middle part is 20 – 2x. So
now we have both sides of the rectangle. We
will multiply the sides together to get our area.
This is A( x) x(20 2 x) , or A( x) 20 x 2 x 2 .
x
20 – 2x
20
5.
2(2)
This problem is not asking for dimensions like the previous examples. It just asks for the depth, which is x.
So we need to make the depth 5 inches to allow the maximum water to flow.
Since the problem says the word maximum, we need to once again use the vertex formula: x
EXAMPLE: Suppose the height of an object shot straight up is given by h 512t 16t 2 where h is measured
in feet and t is in seconds. Find the maximum height and the time at which the object hits the ground.
This problem is easier that the previous ones since we don’t need to first figure out the formula. You want to
find the vertex. This will give us the time at which the maximum height will occur. Use the vertex formula:
512
16 . So we know it takes 16 seconds for the object to reach its maximum height. To find the
t
2(16)
height, put in a 16 for each t in h 512t 16t 2 . You will get: h 512(16) 16(16) 2 4096 ft .
The second part of the problem asks us to find the time at which the object hits the ground. This occurs when
the height is zero. So put in a zero for h and solve for t: 0 512t 16t 2 . We will solve by factoring.
Section 2.2 Notes Page 6
0 512t 16t
0 16t (32 t ) Solving this you will get. t 32 seconds. You also get x = 0 as a solution.
2
EXAMPLE: Find the coordinates of the point on the line y 3x 1 that is closest to the point (4, 0).
This time we have the word “closest” which means minimum distance. Since the word “distance” is used then
we must have to use the distance formula. The distance formula is d ( x 2 x1 ) 2 ( y 2 y1 ) 2 . In order for
this to work we need two points. One is (4, 0) and the other is (x, y). The second point we need to write so that
we only have one variable. We can replace the y with 3x + 1. The second point will be (x, 3x + 1) . Now we
put both of these into the distance formula and simplify. Our two points: (4, 0) and (x, 3x + 1).
d ( x 2 x1 ) 2 ( y 2 y1 ) 2
Start with the distance formula.
d ( x 4) 2 (3x 1 0) 2
Plug in our points. Now simplify.
d ( x 4) 2 (3 x 1) 2
Now we need to distribute.
d x 2 8 x 16 9 x 2 6 x 1
We have just distributed, so now we can simplify.
d 10 x 2 2 x 17
This is as far as we can go. Now we can find the vertex of the function inside the radical. So we can ignore the
radical and still get the correct vertex for the whole function. We will use the vertex formula to find x. You
(2) 1
1
. Now that we have x we need to find the y. We will put
in for x in y = 3x + 1.
will get: x
2(10) 10
10
13
1
1 13
. The coordinates are , . This means
You will get: y 3 1 . If you finish this you will get
10
10
10 10
that this point, on the line y 3 x 1 , is the closest one to the point (4, 0).
EXAMPLE: The daily revenue, R, achieved by selling x boxes of candy is figured to be R( x) 9.5 x 0.04 x 2 .
The daily cost, C, of selling x boxes of candy is C ( x) 1.25 x 250 . How many boxes must be sold to
maximize the profit? What will be the maximum profit?
First we need to find the profit function. Remember that Profit = Revenue – Costs.
P ( x) R( x) C ( x)
P ( x) 9.5 x 0.04 x 2 (1.25 x 250)
P ( x) 9.5 x 0.04 x 2 1.25 x 250
Remember to distribute the negative. Now simplify.
2
P ( x) 0.04 x 8.25 x 250
We have our function, so now apply the vertex formula.
8.25
103.125
x
We need to round up to 104 boxes. This is how many must be sold in order to
2(0.04)
maximize the profit. Now we need to find out what the maximum profit will be. This is done by putting 104
into for x in our profit equation: P ( x) 0.04(104) 2 8.25(104) 250 . We work this out to get our answer:
Section 2.2 Notes Page 7
P ( x) 0.04(10816) 858 250 432.64 858 250 $175.36 . So our maximum profit will be $175.36
when we sell 104 boxes.
Section 2.3 Notes Page 1
2.3 Polynomial Functions and Their Graphs
Polynomial Function:
f ( x) a n x n a n 1 x n 1 ... a1 x ao
The n in the formula above is called the degree, and this is the largest exponent of the polynomial. A
polynomial can only have whole number exponents (no negatives, fractions or decimals). A polynomial must
also be a smooth line with no breaks or corners.
The a n is always in the term with the degree. In other words, it is the number in front of the x with the highest
power. More on this later…
EXAMPLE: Indicate whether the following are polynomials. If they are, indicate the degree and the a n .
a.) f ( x) 3x 2 x 3
x2
3
This is a polynomial since the exponents are all whole numbers. The degree is the highest power that you see,
which in this case is 3. The a n is -2 since it is the number that comes in front of the x 3 term.
b.) f ( x) x 5
This is not a polynomial because a root is a fractional power.
c.) f ( x)
5
x2
This is not a polynomial since this can be written as f ( x) 5 x 2 . Polynomials can never have a negative
exponent.
d.) f ( x) 6
This is a polynomial. This can also be written as f ( x) 6 x 0 . So the degree is zero since that is the highest
power. The a n is 6.
e.) f ( x) ( x 2)( x 5)
We can multiply this out to get f ( x) x 2 3x 10 . This is a polynomial of degree 2. The a n is 1.
Turning point – a point in which the graph changes directions. This happens at a peak or valley.
If n is the degree of a polynomial then the polynomial can have at most n – 1 turning points.
EXAMPLE: Up to how many turning points can y x 2 x 5 have?
Since the degree is 5, then the polynomial can have at most 5 – 1, or 4 turning points. It is important to keep in
mind that it will not have exactly 4 turning points. It means that 4 is the most turning points possible.
Section 2.3 Notes Page 2
EXAMPLE: Which of the following can be a degree 3 polynomial?
A
B
C
D
Graph A has four turning points, which is too many for a degree 3 polynomial. A degree 3 polynomial can have
AT MOST 3 – 1 turning points. Graph B has a break in the graph, and this can’t be a polynomial since they
must be smooth continuous curves. Graph C has a corner or “cusp” which is the official math terminology.
Polynomials can’t have cusps. So choice D is the only one that can be a degree 3 polynomial.
The text looks at the graphs of y x 4 and y x 5 . We will skip this portion. You do not need to do any of the
homework problems that have you do transformations with these graphs
If r is an x-intercept of a graph y, then x – r is a factor.
EXAMPLE: Form a degree 3 polynomial whose zeros are -2, 0, 2.
In order to get the polynomial we must first find its factors. According to the rule above x minus the zero is a
factor. So we can form three factors using our zeros: y ( x (2))( x 0)( x 2) . This can be rewritten as:
y x( x 2)( x 2) . You can multiply this out to verify it is a degree 3 polynomial. You will get:
y x( x 2 4)
y x3 4x
Notice that the highest power is 3, so the degree is 3.
EXAMPLE: Form a degree 3 polynomial whose zeros are -3, and 1.
We can form our two factors to get: y ( x (3))( x 1) which is also y ( x 3)( x 1) . What is wrong with
this answer? The problem is that if we multiply out our answer we only get a degree 2. We can make it a
degree 3 by doing either y ( x 3) 2 ( x 1) or y ( x 3)( x 1) 2 . This gives us the same zeros but it will also
allow our degree to be 3.
The 2 that I added to either of those factors is called the multiplicity. Multiplicity is basically the power on
each factored piece. Usually you will indicate what the zero is and then classify its multiplicity.
EXAMPLE: Indicate the zeros and multiplicities of each zero from y 2 x 3 ( x 1) 2 ( x 2) 4 .
We set the first factor equal to zero we get 0. This has a multiplicity of 3.
We set the second factor equal to zero and we get 1. This has a multiplicity of 2.
We set the third factor equal to zero and we get -2. This has a multiplicity of 4.
Section 2.3 Notes Page 3
If the zero has an even multiplicity, the graph will touch the x-axis at that zero but it will not pass through.
If the zero has an odd multiplicity the graph will cross the x-axis at that zero.
Behavior at each zero:
This tells you what the graph looks like as it crosses the x-axis. It will resemble a function. You can find the
equation by doing the following: put the zero into the function every place except the factor that gave you the
zero. Let’s look at some examples so this will make more sense.
EXAMPLE: Find the zeros and the behavior at each zero for f ( x) x( x 5)( x 3) .
The first zero is x = 0. We want to find the behavior. Since I got a zero from the x in front, I will put zero into
x – 5 and x + 3, but not in for x. You will get: f ( x) x(0 5)(0 3) . Simplifying you will get: f ( x) 15 x ,
which means that as the graph crosses the x-axis at x = 0 the graph will resemble the line f ( x) 15 x .
The second zero is x = 5. We will put a 5 in for x in the first and last factor but not in for x – 5 since this factor
originally gave us the zero. It will look like: f ( x) 5( x 5)(5 3) . Simplifying you will get: f ( x) 40( x 5)
or f ( x) 40 x 200 .
The third zero is x = -3. We will put a -3 for x in the first two factors, but not the last one since x + 3 originally
gave us the zero. You will get f ( x) 3(3 5)( x 3) . This equals f ( x) 24( x 3) or f ( x) 24 x 72 .
End behavior
This is what the graph will do when x is really big or really small. This is where you will need to know what
the a n is. Depending on what the degree is and what the a n is the graph will do the following:
The n is the degree. So this chart is giving us all the
possibilities of how the graph will end.
Section 2.3 Notes Page 4
Now let’s put all of this together and look at some graphs. The quizzes and tests I will give you will have the
correct number of blanks for the zeros so you know how many you should have. I will ask you to find the
zeros, multiplicities, behavior at each zero, turning pts, y-intercept and the graph. Let’s look at some examples.
EXAMPLE: Find the zeros, multiplicities, behavior at each zero, turning pts, y-intercept and the graph of
f ( x) x( x 5)( x 3) .
This was the same equation we used to find the behavior at each zero, so that is done. The multiplicity on each
zero is one in this case because 1 is the exponent on each factored piece. For the y-intercept, put in a zero for x.
When you do you will get y = 0, so the y-intercept is (0, 0). What is our degree? We can multiply this out to
get: f ( x) x 3 2 x 2 15 x and we see that the degree is 3. This means it can have at most 3 – 1 or 2 turning
points. From our previous example we have that the behavior at 0 is f ( x) 15 x , the behavior at 5 is
f ( x) 40 x 200 and the behavior at -3 is f ( x) 24 x 72 .
Now here is how we will graph this. First plot each of the zeros. Then right at each zero make a sketch of the
behavior equations we found. For example at zero the behavior is f ( x) 15 x which is a line slanting to the
left with a large slope. At 5 we have a line slanting to the right with a steep slope. And at -3 we have another
line slanting to the right with a large slope. After making these sketches you will get:
Our equation is f ( x) x 3 2 x 2 15 x which tells us our degree is odd and the a n > 0 . This means that from
our end behavior models the graph will go down and to the left and will go up and to the right. Since we know
this we can connect these lines to get our graph (the graph is scaled by 2).
Your graph will be a sketch. You don’t need to know exactly
how high or how low the graph goes. I am only looking for the
general shape.
Section 2.3 Notes Page 5
EXAMPLE: Find the zeros, multiplicities, behavior at each zero, turning pts, y-intercept and the graph of
y 2( x 3) 2 ( x 4) 2 .
The quiz or test will have the correct number of blanks for the number of zeros. This one only has two zeros, so
there are two blanks. Lets first find the zeros and multiplicities. We will get 3 and -4 as our zeros. The
exponent associated with each factor is 2. Therefore each multiplicity is 2. If we add the multiplicities this will
give us the degree, which is 4. Once we know the degree we can subtract one and this will be the max turning
points. We can also find out the y-intercept by putting in a 0 for x: y 2(0 x) 2 (0 4) 2 , so y = 288. Our yint is (0, 288).
We will now find the behavior at each zero:
For x = 3, we will put a 3 in for x in ( x 4) 2 but NOT in the factor ( x 3) 2 . You will get
y 2( x 3) 2 (3 4) 2 . After simplifying you will get y 98( x 3) 2 .
For x = -4 we will put a -4 in for x in ( x 3) 2 but NOT in the factor ( x 4) 2 . You will get
y 2(4 3) 2 ( x 4) 2 which simplifies to y 98( x 4) 2 . Let’s fill in all the blanks so all our information is in
one place.
zero:___3___ Multiplicity:___2___ Behavior at zero: y 98( x 3) 2
zero:___-4__ Multiplicity:___2___ Behavior at zero: y 98( x 4) 2
y-int:__(0, 288)_ Degree:______4_____ Max turning pts:___3____
To graph this we will first plot the x-intercepts. Then we can plot the y-intercept. Then we need to make a
sketch of our behavior at zero equations. These will both be parabolas, or U shaped graphs.
We found that our degree is even and the a n > 0 because we just multiply
our coefficients (number in front of each x). We have 2 on the outside and
each x in the parenthesis has a 1 in front of it, so 2*1*1 = 2, which is
greater than zero. Our end behavior models the graph will go up and to the
left and will go up and to the right. Since we know this we can connect
these lines to get our graph (the graph is scaled by 20).
Section 2.3 Notes Page 6
EXAMPLE: Find the zeros, multiplicities, behavior at each zero, turning pts, y-intercept and the graph of
y 2 x 3 ( x 2) .
The quiz or test will have the correct number of blanks for the number of zeros. This one only has two zeros, so
there are two blanks. Lets first find the zeros and multiplicities. We will get 0 and -2 as our zeros. The
multiplicity of the zero, x = 0 is 3 and the multiplicity of the zero x = -2 is 1. Again these are the exponents
associated with each factor. If we add the multiplicities this will give us the degree, which is 4. Once we know
the degree we can subtract one and this will be the max turning points. We can also find out the y-intercept by
putting in a 0 for x: y 2(0) 3 (0 2) , so y = 0. Our y-intercept is (0, 0).
We will now find the behavior at each zero:
For x = 0, we will put a 0 in for x in ( x 2) but NOT in the factor 2 x 3 . You will get y 2 x 3 (0 2) . After
simplifying you will get y 4 x 3 .
For x = -2 we will put a -2 in for x in 2 x 3 but NOT in the factor ( x 2) . You will get y 2(2) 3 ( x 2)
which simplifies to y 16( x 2) , or y 16 x 32 Let’s fill in all the blanks so all our information is in one
place.
zero:___0___ Multiplicity:___3___ Behavior at zero: y 4 x 3
zero:___-2__ Multiplicity:___1___ Behavior at zero: y 16 x 32
y-int:__(0, 0)__ Degree:____4___ Max turning pts:___3____
To graph this we will first plot the x-intercepts. Then we can plot the y-intercept. Then we need to make a
sketch of our behavior at zero equations. Notice the behavior at x = 0 is a cubic graph reflected over the
horizontal axis and at x = -2 we have a line slanting to the right.
We found that our degree is even and the a n < 0 because we just multiply
our coefficients (number in front of each x). We have -2 on the outside
and the x in the parenthesis has a 1 in front of it, so -2*1 = -2, which is
less than zero. Our end behavior models the graph will go down and to the
left and will go down and to the right. Since we know this we can connect
these lines to get our graph.
Section 2.3 Notes Page 7
EXAMPLE: Find the zeros, multiplicities, behavior at each zero, turning pts, y-intercept and the graph of
y x 3 ( x 1.5) 2 ( x 1.5) 2 .
We will have three zeros for this one. You will get 0, 1.5, and -1.5. The multiplicity of the zero, x = 0 is 3 and
the multiplicity of the zero x = -1.5 and 1.5 are both 2. If we add the multiplicities this will give us the degree,
which is 7. Once we know the degree we can subtract one and this will be the max turning points, which is 6.
We can also find out the y-intercept by putting in a 0 for x: y 0 3 (0 1.5) 2 (0 1.5) 2 , so y = 0. So (0,0).
We will now find the behavior at each zero:
For x = 0, we will put a 0 in for x in ( x 1.5) 2 and ( x 1.5) 2 but NOT in the factor x 3 . You will get
y x 3 (0 1.5) 2 (0 1.5) 2 . After simplifying you will get y 5.0625 x 3 .
For x = -1.5 we will put a -1.5 in for x in x 3 and ( x 1.5) 2 but NOT in the factor ( x 1.5) 2 . You will get
y (1.5) 3 (1.5 1.5) 2 ( x 1.5) 2 which simplifies to y 30.375( x 1.5) 2
For x = 1.5 we will put a 1.5 in for x in x 3 and ( x 1.5) 2 but NOT in the factor ( x 1.5) 2 . You will get
y (1.5) 3 ( x 1.5) 2 (1.5 1.5) 2 which simplifies to y 30.375( x 1.5) 2 . Let’s fill in the blanks:
zero:___0___ Multiplicity:___3___ Behavior at zero:
y 5.0625 x 3
zero:___1.5__ Multiplicity:___2___ Behavior at zero: y 30.375( x 1.5) 2
zero:__-1.5___ Multiplicity:___2___ Behavior at zero: y 30.375( x 1.5) 2
y-int:__(0, 0)__ Degree:____7___ Max turning pts:___6____
To graph this we will first plot the x-intercepts. Then we can plot the y-intercept. Then we need to make a
sketch of our behavior at zero equations. Notice the behavior at x = 0 is a cubic graph. At x = -1.5 the graph
looks like an upside down parabola. At x = 1.5 it is a regular parabola.
We found that our degree is odd and the a n > 0 because we just multiply
our coefficients (number in front of each x). We have 1 on the outside and
the x in the parenthesis has a 1 in front of each, so 1*1*1 = 1, which is
greater than zero. Our end behavior models the graph will go down and to
the left and will go up and to the right. Since we know this we can
connect these lines to get our graph.
Section 2.4 Notes Page 1
2.4 Dividing Polynomials; Remainder and Factor Theorems
We will be doing a quick review of long division since we need to know this when working with rational
functions. To divide using long division we do the same steps as if we are working with numbers.
EXAMPLE: Divide by using long division: 3453 / 13.
265
13 3453
When we first started we ask ourselves how many times can 13 go into 34. It goes in twice.
Then we multiplied the 13 by 2 to get 26 and this was written under the 34. We then subtracted.
You will always subtract when doing long division. We got 8. Then we brought down the 5 and
repeated the process. We ask how many times does 13 go into 85. We get 6. We continue this
process and then we end up with a remainder of 8. We would write our answer as the following:
3453 / 13 = 265 + 8/13. We always write our remainder over what we are dividing by.
26
85
78
73
65
8
EXAMPLE: Divide by using long division: 2 x 2 3x 35 ( x 5)
We will do this the same way we did the number example. First we need to set it up:
2x 7
x 5 2 x 3 x 35
First we ask ourselves how many times x goes into 2x 2 . We ignore the + 5 at the
2 x 2 10 x
7 x 35
7 x 35
0
moment. We know x goes into 2x 2 an amount of 2x. We then multiply x + 5 by
2x to get 2 x 2 10 x . We write this on the next line and then we subtract. Then
we get -7x – 35. We now ask how many times does x go into -7x. The answer is
-7. We then multiply x + 5 by -7 to get -7x – 35. We subtract and get a
remainder of zero.
2
EXAMPLE: Divide by using long division: 3x 3 x 5 ( x 1)
We notice that this one is missing a term. You want to make sure that each term is accounted for. We need to
add in a 0x 2 as a place keep.
3x 2 3x 4
x 1 3x 3 0 x 2 x 5
Remember you are always subtracting when doing long division.
3x 3 3x 2
3x 2 x
3x 2 3x
4x 5
4x 4
1
This one has a remainder. Here is how you want to write your answer: 3 x 2 3 x 4
1
x 1
Section 2.4 Notes Page 2
EXAMPLE: Divide by using long division: x 3 x x 2 ( x 2)
3
2
2
First we need to make sure the terms are written in descending order. So we wan to write this as:
3x 3 x 2 x 2 ( x 2 2) . The next thing we notice is that we are missing a term in x 2 2 . The x term is
missing so we must add a 0x to make it x 2 0 x 2 .
3x 1
x 0 x 2 3x x x 2
2
3
2
Remember you are always subtracting when doing long division.
3x 2 0 x 2 6 x
x 2 5x 2
x 2 0x 2
5x
This one also has a remainder. Here is how you want to write your answer: 3x 1
5x
.
x2 2
EXAMPLE: Divide by using long division: 3x 4 2 x 1 ( x 1)
We notice that this one is missing terms. We need to put in a 0x 3 and 0x 2 .
3x 3 3x 2 3x 5
x 1 3x 4 0 x 3 0 x 2 2 x 1
Remember you are always subtracting when doing long division.
3x 4 3x 3
3x 3 0 x 2
3x 3 3x 2
3x 2 2 x
3x 2 3x
5x 1
5x 5
6
We have a remainder once more. Here is how you want to write your answer: 3x 3 3x 2 3 x 5
6
x 1
Section 2.4 Notes Page 3
Synthetic Division is alternative to long division. It is used when dividing by something in the form x – or x +
something. In all the problems in this section there will be a one in front of the x.
EXAMPLE: Divide using synthetic division: x 3 7 x 2 13 x 15 x 2
In order to do this we must first set it up. In the problem above we have x + 2. You want to reverse the sign of
the number after the x and place it in a half box like shown below. In this case we have a negative 2 since we
reversed the sign. After this number you want to write the coefficients of your equation afterwards. A
coefficient is just the number that comes in front of the variable. In synthetic division there are no variables.
You are only working with numbers. Under this line skip a space and then draw a line. Your set up should
look like this:
-2 | 1 -7 -13 15
__________
Here is how you do synthetic division. You always drop down the number right after the box like the
following:
-2 | 1 -7 -13 15
__________
1
Now we will multiply the 1 by -2 and we will write this result under the -7.
-2 | 1 -7 -13 15
__-2_______
1
In synthetic division we always add. We will add -7 and -2 to get -9.
-2 | 1 -7 -13 15
__-2_______
1 -9
Now multiply -9 by -2 to get 18. This number gets written under the -13.
-2 | 1 -7 -13 15
__-2_ 18___
1 -9
Now add -13 and 18 to get 5.
-2 | 1 -7 -13 15
__-2_ 18___
1 -9 5
Multiply 5 by -2 to get -10. Now write this under the 15.
-2 | 1 -7 -13 15
__-2_ 18 -10_
1 -9 5
Now add 15 and -10 to get -5.
-2 | 1 -7 -13 15
__-2_ 18 -10_
1 -9 5 5
Now we need to write our answer (see next page).
-2 | 1 -7 -13 15
__-2_ 18 -10_
1 -9 5 5
Section 2.4 Notes Page 4
The last line above, which shows 1 -9 5 5 are the coefficients for our answer. The last 5 is our remainder,
which we write over x + 2. We always start our answer with a power one less than what we began with. Since
5
the original started with x 3 then our answer will start with x 2 . Write our answer as: x 2 9 x 5
.
x2
EXAMPLE: Divide using synthetic division: 3x 2 2 x 3 x 4 4 x 3
Before we set this problem up we need to make sure we have descending powers. We also need to make sure
no powers are missing. If they are then we need to insert a zero as a place keeper. We will rewrite this as:
x 4 2 x 3 3x 2 0 x 4 x 3 . This one the x term was missing so this is a zero place keeper.
In the problem above we have x – 3. You want to reverse the sign of the number after the x and place it in a
half box like shown below. In this case we have a positive 3 since we reversed the sign. After this number you
want to write the coefficients of your equation afterwards. Under this line skip a space and then draw a line.
Notice below we still wrote the 0 to hold the place. Your set up should look like this:
3 | 1 -2 3 0 -4
__________
Start by dropping the 1 below the line we just drew.
3 | 1 -2 3 0 -4
__________
1
Now multiply 1 by 3 and write this result under -2.
3 | 1 -2 3 0 -4
__3________
1
Now ADD the -2 and 3 to get 1. We always add with synthetic division.
3 | 1 -2 3 0 -4
__3________
1 1
Now multiply the 1 by 3 and write this under the 3.
3 | 1 -2 3 0 -4
__3_3______
1 1 6
Now multiply 6 by 3 to get 18. Write this under 0.
3 | 1 -2 3 0 -4
__3_3_18_____
1 1 6
Add and repeat the process to the end. You will get:
3 | 1 -2 3 0 -4
__3_3_18_54___
1 1 6 18 50
The last line shows the coefficients for our answer. We always start our answer with a
power one less than what we began with. Since the original started with x 4 then our answer will start with x 3 .
50
.
The 50 above is our remainder, which we write over x – 2. Write our answer as: x 3 x 2 6 x 18
x3
EXAMPLE: Divide using synthetic division: 2 x 6 x 14 x x 2
5
Section 2.4 Notes Page 5
3
We also need to make sure no powers are missing. If they are then we need to insert a zero as a place keeper.
We will rewrite this as: 2 x 5 0 x 4 6 x 3 0 x 2 14 x 0 x 2 . We had the x 4 , x 2 and the constant term
missing. You need all three of these zeros as place keepers. We are dividing by x – 2. You want to reverse the
sign of the number after the x and place it in a half box like shown below. In this case we have a positive 2
since we reversed the sign. Your set up should look like this:
2 | 2 0 -6 0 -14 0
_____________
Start by dropping the 2 below the line we just drew.
2 | 2 0 -6 0 -14 0
_____________
2
Multiply the 2 by 2 and write this under the 0.
2 | 2 0 -6 0 -14 0
_ 4___________
2 4
Add 0 and 4 to get 4. Now multiply the 4 by 2 to get 8. Write this under -6.
2 | 2 0 -6 0 -14 0
_ 4_ 8_________
2 4
Add -6 and 8 to get 2. Then multiply the 2 by 2 to get 4. Write this under the 0.
2 | 2 0 -6 0 -14 0
_ 4_ 8_4________
2 4 2
Add 0 and 4 to get 4. Now multiply the 4 by 2 to get 8. Write this under -14.
2 | 2 0 -6 0 -14 0
_ 4_ 8_4__8____
2 4 2 4
Add -14 and 8 to get -6. Now multiply the -6 by 2 to get -12. Write this under 0.
2 | 2 0 -6 0 -14 0
_ 4_ 8_4__8_ -12__
2 4 2 4 -6
Now add 0 to -12 to get -12. This is our remainder.
2 | 2 0 -6 0 -14 0
_ 4_ 8_4__8_ -12__
2 4 2 4 -6 -12
Now we need to write our answer. The last line shows the coefficients for our answer. We always start our
answer with a power one less than what we began with. Since the original started with x 5 then our answer will
start with x 4 . The -12 above is our remainder, which we write over x – 2. Write our answer as:
12
.
2x 4 4x 3 2x 2 4x 6
x2
Section 2.4 Notes Page 6
Remainder Theorem: If f (x) is divided by x – c, then the remainder is f (c)
Factor Theorem: If f (c) 0 then x – c is a factor of f (x) . Also, if x – c is a factor of f (x) then f (c) 0 .
EXAMPLE: Use the Remainder Theorem to find the remainder when f ( x) 2 x 3 4 x 2 5 x 3 is divided by
x – 2 . Then use the Factor Theorem to determine whether x – 2 is a factor of f (x ) .
We will use synthetic division for this problem. I will just show the result of synthetic division here. For a full
explanation please see section A.5 notes.
2 | 2 -4
4
2 0
5 -3
0 10
5 7
The remainder is 7. Since this is not zero we know right away that x – 2 is not a factor by the Factor Theorem.
When we did synthetic division the number in the box was 2. Let’s find f (2) . From a previous section we find
that f (2) 2(2) 3 4(2) 2 5(2) 3 which equals f (2) 16 16 10 3 7 . Notice that this is the remainder
that we got when we did synthetic division. This is what the Remainder Theorem says. This is an alternative
way of evaluating the function at 2 without as much calculations.
EXAMPLE: Use the Remainder Theorem to find the remainder when f ( x) 2 x 4 12 x 3 6 x 2 5 x 75 is
divided by x + 5. Then use the Factor Theorem to determine whether x + 5 is a factor of f (x) .
We start with synthetic division:
-5 | 2 12 6 -5 75
-10 -10 20 -75
2 2 -4 15 0
This time we get a remainder of zero. The Factor Theorem tells us that x + 5 is a factor of f (x) since it divided
in evenly. The Remainder Theorem tells us that f (5) 0 .
EXAMPLE: Solve the equation 2 x 3 3 x 2 11x 6 0 given that -2 is a zero of f ( x) 2 x 3 3x 2 11x 6 .
We know that if we do synthetic division, we should get a remainder of zero:
-2 | 2 -3
-4
2 -7
-11 6
14 -6
3 0
We do get a zero for a remainder. This also tells us 2 x 2 7 x 3 is a factor of f(x). Because of this, we can
factor 2 x 2 7 x 3 and set it equal to zero to find our other solutions. Factoring will give (2 x 1)( x 3) 0 .
1
We set both of these other factors equal to zero to get x and x 3 . If you do not remember how to factor
2
polynomials, you should review P.5. You can also see me during office hours and I will be happy to review this
for you.
Section 2.5 Notes Page 1
2.5 Zeros of a Polynomial Functions
The first rule we will talk about is Descartes’ Rule of Signs, which can be used to determine the possible times a
graph crosses the x-axis and has a zero. A sign change is a place in f where you have + to – or – to +.
Descartes’ Rule of Signs
1.) The number of positive real zeros of f equals the number of sign changes in f (x) or less an even integer.
2.) The number of negative real zeros of f equals the number of sign changes in f ( x) or less an even integer.
“Less an even integer” means we keep subtracting 2 from the number of sign changes until we get to zero or an
negative number. Suppose that the number of sign changes was 6. Then we keep subtracting 2 to get the other
answers. We will get 6, 6-2, 6-2-2, 6-2-2-2 which is 6, 4, 2, 0. If we have 5, then our answers are 5, 3, or 1.
EXAMPLE: Use Descartes’ Rule of Signs to find the number of possible positive and negative zeros of
f ( x) 5 x 6 3x 4 4 x 3 6 x 2 x 4 .
Positive Zeros:
We need to count the number of sign changes in f ( x) 5 x 6 3x 4 4 x 3 6 x 2 x 4 .
|____| |____|
|___| |___|
The brackets above indicate the place where we have + to – or – to +. In this case we have 4 sign changes. To
write our answer, we start with 4 and then keep subtracting 2 to get the other answers since it is less an even
integer. So the number of positive real zeros is 4 or 2 or 0. You keep subtracting 2 until you get a zero or a
negative number.
Negative Zeros:
In order to find the negative zeros we must look at f ( x) :
f ( x) 5 x 3 x 4 x 6 x x 4
f ( x) 5 x 6 3 x 4 4 x 3 6 x 2 x 4
|_____|
|_____|
6
4
3
2
Put in a –x for x in the equation for f.
As we notice above there are two sign changes in f ( x) . The number of sign changes will be 2 or 0. Again
we keep subtracting 2 from our answer until we get to 0 or a negative number.
EXAMPLE: Use Descartes’ Rule of Signs to find the number of possible positive and negative zeros of
f ( x) 5 x 3 2 x 2 x 3 .
Positive Zeros:
We need to count the number of sign changes in f ( x) 5 x 3 2 x 2 x 3 .
|____| |___| |__|
We see that there are 3 sign changes, so as our answer we would write: 3 or 1.
Section 2.5 Notes Page 2
Negative Zeros:
In order to find the negative zeros we must look at f ( x) :
f ( x) 5 x 2 x x 3
f ( x) 5 x 3 2 x 2 x 3
3
2
We notice there are no sign changes in f ( x) . This means there are no negative sign changes (0).
Rational Zeros Theorem
Let f ( x) a n x n a n 1 x n 1 ... a 0 be a polynomial. Then the number of possible real zeros of f is:
factors of a0
factors of a n
To review a factor is a number that evenly divides into something. For example, the factors of 6 are 1, 2 and 3.
EXAMPLE: List the possible real zeros of: f ( x) 3x 3 7 x 2 22 x 8
Using the Rational Zeros Theorem we will write the factors of 8 over the factors of 3:
1, 2, 4, 8
Now divide each number on top by each on the bottom. You will get the list of zeros:
1, 3
1
2
4
8
1, 2, 4, 8, , , , . These do not need to be in any special order.
3
3
3
3
This list represents all the possible places the graph could cross the x-axis.
EXAMPLE: Given f ( x) x 3 8 x 2 11x 20 , a.) Use Descartes’ Rule of Signs to find the number of
possible positive and negative zeros. b.) Use the Rational Zero Theorem to find the list of possible zeros.
c.) Find the zeros using synthetic division.
a.) We need to find the number of sign changes of f ( x) x 3 8 x 2 11x 20 .
|_____|
There is only one sign change, so the number of positive real zeros is 1.
For the negative zeros we need to look at : f ( x) x 8 x 11 x 20 You will get:
f ( x) x 3 8 x 2 11x 20
|_____| |_____|
3
2
There are two sign changes. Therefore the number of negative real zeros is 2 or 0.
Section 2.5 Notes Page 3
b.) We need to use the Rational Zeros Theorem to find our list of possible zeros.
1, 2, 4, 5, 10, 20
Now divide each number on top by each on the bottom. You will get:
1
1, 2, 4, 5, 10, 20 . This is our list of possible zeros.
c.) Now we need to see which one is a zero. In order to do this, pick a zero to test and use synthetic division
with the original equation. If we get a zero for the remainder we know that this is a zero. For example, let’s
first test the x-value -1. We need to set up and do the synthetic division:
-1 | 1 8
-1
1 7
11 -20
-7 -4
4 -24
We don’t get a zero for the remainder, so we know -1 is not one of our answers.
Let’s try a different number. We will now test x = 1.
1 | 1 8 11 -20
We do get a zero, so x = 1 is one of our answers. We could go through this
1 9 20
process for all the other possible zeros in our list, but since we found one we can
1 9 20 0
find the other ones in an easier way.
The last row of our synthetic division was 1 9 20. This means our new equation is x 2 9 x 20 . We can
set this equation equal to zero to get the other solutions. You will get x 2 9 x 20 0 . Factoring this we will
get ( x 4)( x 5) 0 , so our other answers are x = -4 and x = -5. So our answer for this one would be:
x = -5, -4, and 1. This confirms Descartes’ Rule of Signs. We have one positive zero and two negative zeros.
EXAMPLE: Given f ( x) 3x 4 11x 3 x 2 19 x 6 , a.) Use Descartes’ Rule of Signs to find the number of
possible positive and negative zeros. b.) Use the Rational Zero Theorem to find the list of possible zeros.
c.) Find the zeros using synthetic division.
a.) We need to find the number of sign changes of: f ( x) 3x 4 11x 3 x 2 19 x 6
|_____|
|____|
There are two sign changes, so the number of positive real zeros is 2 or 0.
For the negative zeros we need to look at : f ( x) 3 x 11 x x 19 x 6 You will get:
f ( x) 3x 4 11x 3 x 2 19 x 6
|_____|
|_____|
4
3
2
There are two sign changes. Therefore the number of negative real zeros is 2 or 0.
b.) We need to use the Rational Zeros Theorem to find our list of possible zeros.
1, 2, 3, 6
Now divide each number on top by each on the bottom. You will get:
1, 3
1
2
3
6
1, 2, 3, 6, , . I did not put , because this is the same as 1, 2 . We already have it.
3
3
3
3
Section 2.5 Notes Page 4
c.) Now we need to see which one is a zero. We need to start testing zeros. We will start with x = -1 again.
-1 | 3 -11 -1 19 6
-3 14 -13 -6
we can
3 -14 13 6 0
2 | 3 -11 -1 19 6
6 -10 -22 -6
3 -5 -11 -3 0
We do get a zero here. This will leave us with an x cubed equation which we
will not be able to factor very easily. We need to find another zero so that
reduce this cube to a square.
After more trail and error we get x = 2 is another zero.
Let’s look at the first row of numbers we got when we used x = -1. We got 3 -14 13 6 . Since we know
x = -2 is also a zero, let’s use synthetic division with this row and the second zero we found, x = -2:
2 | 3 -14 13 6
6 -16 -6
3 -8 -3 0
This will leave us with 3x 2 8 x 3 which we can set equal to zero to find the
other answers. (We also could have done synthetic division with the other row
above: 3 -5 -11 -3 and used x = -1. Does not matter which one you do).
Now we can solve 3x 2 8 x 3 0 . Factoring we will get: ( x 3)(3x 1) 0 . Solving this we will get
x = 3, x = -1/3. Putting this all together our final answer for part c will be x = -1, -1/3, 2, and 3. This confirms
our Descartes’ Rule of Signs. We got two positive zeros and two negative zeros.
EXAMPLE: Given f ( x) x 4 4 x 3 3x 2 4 x 4 , a.) Use Descartes’ Rule of Signs to find the number of
possible positive and negative zeros. b.) Use the Rational Zero Theorem to find the list of possible zeros.
c.) Find the zeros using synthetic division.
a.) We need to find the number of sign changes of: f ( x) x 4 4 x 3 3x 2 4 x 4
|____|
There is only one sign change, so the number of positive real zeros is 1.
For the negative zeros we need to look at : f ( x) x 4 x 3 x 4 x 4 You will get:
f ( x) x 4 4 x 3 3 x 2 4 x 4
|____| |____| |_____|
4
3
2
There are three sign changes. Therefore the number of negative real zeros is 3 or 1.
b.) We need to use the Rational Zeros Theorem to find our list of possible zeros.
1, 2, 4
Now divide each number on top by each on the bottom. You will get: 1, 2, 4 .
1
c.) Now we need to see which one is a zero. We need to start testing zeros. We will start with x = -1 again.
-1 | 1
4 3 -4 -4
-1 -3 0 4
1 3 0 -4 0
We do get a zero here. This will leave us with an x cubed equation which we
will not be able to factor very easily. We need to find another zero so that we can
reduce this cube to a square.
Section 2.5 Notes Page 5
1| 1
1
4 3 -4 -4
1 5 8 4
5 8 4 0
We find that x = 1 is another zero.
Let’s look at the first row of numbers we got when we used x = -1. We got 1 3 0 -4 . Since we know
x = 1 is also a zero, let’s use synthetic division with this row and the second zero we found, x = 1:
1| 1
3
1
4
1
0
4
4
-4
4
0
This will leave us with x 2 4 x 4 which we can set equal to zero to find the
other answers. (We also could have done synthetic division with the other row
above: 1 5 8 4 and used x = -1. Does not matter which one you do).
Now we can solve x 2 4 x 4 0 . Factoring we will get: ( x 2)( x 2) 0 . Solving this we will get
x = -2. Putting this all together our final answer for part c will be x 1, 2 . But what about Descartes’ Rule
of Signs? We got only two negative zeros and one positive zero. Descartes’ told us that we should have three
negative zeros. When we solved ( x 2)( x 2) 0 this actually gave us a double root, -2 and -2. So
technically the answers are x 1, 2, 2 which would satisfy Descartes’. We don’t need to write a repeated
root more than once.
EXAMPLE: Given f ( x) 2 x 4 x 3 35 x 2 113x 65 , a.) Use Descartes’ Rule of Signs to find the number of
possible positive and negative zeros. b.) Use the Rational Zero Theorem to find the list of possible zeros.
1
c.) Find the zeros using synthetic division given f (5) 0 and f 0 .
2
a.) We need to find the number of sign changes of: f ( x) 2 x 4 x 3 35 x 2 113x 65
|_____|
|____|
There are two sign changes, so the number of positive real zeros is 2 or 0.
For the negative zeros we need to look at : f ( x) 2( x) 4 ( x) 3 35( x) 2 113( x) 65 You will get:
f ( x) 2 x 4 x 3 35 x 2 113x 65
|_____|
|______|
There are two sign changes. Therefore the number of negative real zeros is 2 or 0.
b.) We need to use the Rational Zeros Theorem to find our list of possible zeros.
1
5
13
65
1, 5, 13, 65
Dividing gives us 1, 5, 13, 65, , , , .
2
2
2
2
1, 2
c.) We were given the first two zeros so we don’t need to do trial and error on this one. We will do synthetic
division twice to get a quadratic. It doesn’t matter which zero you start with. I will start with x = 5.
5| 2
2
1 -35 -113 65
10 55 100 -65
11 20 -13 0
Section 2.5 Notes Page 6
½| 2
2
11
1
12
20
6
26
-13
13
0
Again it does not matter what zero you start with.
Since we started out with a fourth power, doing synthetic division twice brought this down to a quadratic. We
now have to solve 2 x 2 12 x 26 0 . We notice there is a common factor of 2. We can divide both sides of
the equation by 2 to make our calculations easier. You will get x 2 6 x 13 0 . This cannot be factored, so
we must use the quadratic formula:
x
6 6 2 4(1)(13) 6 36 52 6 16 6 4i
3 2i .
2(1)
2
2
2
We do not get a real number as a result, but we still need to include this in our answer.
We write our answer as x =
1
, 5, 3 2i .
2
Does Descartes’ Rule take into consideration the complex numbers? NO! Descartes’ only gives you the
possible REAL negative and positive zeros. In part a it said we would either get 2 positive real zeros or no
positive real zeros. This was the same for the negative zeros. We should get either two or none of them. In our
case we have two positive zeros and no negative zeros, so this does fit Descartes’ Rule of Signs.
Section 2.6 Notes Page 1
2.6 Rational Functions and Their Graphs
Rational Function: a function with a variable in the denominator.
To find the y-intercept for a rational function, put in a zero for x.
To find the x-intercept for a rational function, set the numerator equal to zero
EXAMPLE: Find the x and y intercepts for y
2 x 12
x2
2(0) 12 12
6 . So our
02
2
point is (0, -6). Next we will find the x-intercept. To do this we must set the numerator (top) equal to zero.
You will get: 2 x 12 0 . Solving this you will get x = 6. So our point is (6, 0).
First we will find the y-intercept. We will put in a zero for x. You will get: y
Asymptote: describes the behavior of a graph as x or y approaches infinity. There are two types of asymptotes.
There is a vertical and horizontal asymptote as show in the picture below. The vertical asymptote has an
equation that starts with x = since this is a vertical line. The horizontal asymptote has an equation y = since
this is a horizontal line. Now we will show how you find these algebraically.
To find the vertical asymptote:
set the denominator equal to zero and solve for x.
Suppose we wanted to find the vertical asymptote of our
2 x 12
previous example, y
. To do this we need to set
x2
the denominator equal to zero, so we will have x + 2 = 0.
The equation will be x = -2. You would write your answer
in this form. You don’t need parenthesis. That is only for
intercepts.
To find the horizontal asymptote:
First we need to define some variables. Let’s look at the general form from a ration expression:
f ( x)
a n x n a n 1 x n 1 ....
bm x m bm 1 x m 1 ....
Let n be the highest power (degree) of the numerator.
Let m be the highest power (degree) of the denominator.
Let a n be the number that comes in front of the x with the highest power in the numerator.
Let a m be the number that comes in front of the s with the highest power in the denominator.
Section 2.6 Notes Page 2
In order to determine the horizontal asymptote we need to look at the n and m.
1.) If n < m then the equation of the horizontal asymptote is automatically y = 0.
a
2.) If n = m then the equation of the horizontal asymptote is y n .
bm
3.) If n > m then there is no horizontal asymptote. There is an oblique asymptote:
You find the oblique asymptotes by using long division. More on this later.
2 x 12
, if we were asked to find the equation
x2
of the horizontal asymptote we need to see which rule applies. Here the
highest power on top is the same as the highest power on the bottom, so
rule 2 applies. We know a n 2 and bm 1 , so the horizontal asymptote is
2
y or just y = 2.
1
In our original equation y
1 x2
EXAMPLE: Find the intercepts and asymptotes but DO NOT GRAPH: y 2
x 5x 6
(1 x)(1 x)
. This can’t be simplified
( x 2)( x 3)
anymore, so now let’s find the y-intercept. To do this we need to put in a 0 for x. When you do that you will
(1 0)(1 0) 1
1
. So the y-intercept is 0, . To find the x-intercept we need to set the top equal to
get y
(0 2)(0 3) 6
6
zero. So we have (1 x)(1 x) 0 . Solving this we get x 1 , so our x-intercepts are 1,0 .
First you want to factor to see if it can be simplified further: y
Now let’s find the asymptotes. To find the vertical asymptote we need to set the bottom equal to zero. So we
have ( x 2)( x 3) 0 so x = 2 and x = 3. We will leave our answer in this form. To find the horizontal
asymptote let’s look at our original equation. The highest power on the top is the same as the highest power on
1
the bottom, so we use rule 2 again. Our a n 1 and bm 1 , so the horizontal asymptote is y
so y = -1.
1
x 2 3x
EXAMPLE: Find the intercepts and asymptotes but DO NOT GRAPH: y 2
2x 4x3
x( x 3)
x3
. You always want
2
2 x (1 2 x) 2 x(1 2 x)
to simplify if possible. So now let’s find the y-intercept. To do this we need to put in a 0 for x. When you do
you will get a zero in the denominator, which is undefined. So there is no y-intercept. To find the x-intercept
we need to set the top equal to zero. So we have x 3 0 . Solving this we get x = 3, so this is (3, 0).
First you want to factor to see if it can be simplified further: y
Now let’s find the asymptotes. To find the vertical asymptote we need to set the bottom equal to zero. So we
1
have 2 x(1 2 x) 0 so x 0 and x We will leave our answer in this form. To find the horizontal
2
asymptote let’s look at our original equation. The highest power on the top is less than the highest power on the
bottom, so we use rule 1. This says that the horizontal asymptote is automatically y = 0.
Section 2.6 Notes Page 3
EXAMPLE: Find the asymptotes but DO NOT GRAPH: y
3 x 10 x 6
x2
2
This does not ask us to find intercepts, so we will just find the asymptotes. To find the vertical asymptote we
need to set the bottom equal to zero. So we have x 2 0 so x 2 . We will leave our answer in this form.
To find the horizontal asymptote let’s look at our original equation. The highest power on the top is more than
the highest power on the bottom, so now we have rule 3. This tells us there is no horizontal asymptote, but
there is an oblique asymptote. We need to find this. In order to do that we must use long division like we did in
a previous section
3x 4
x 2 3 x 10 x 6
2
3x 2 6 x
4x 6
4x 8
-2
From doing the long division we get 3x + 4. This is the equation of the oblique
asymptote. We always ignore the remainder. We just write y = 3x + 4.
Now we will look at some special graphs. These are: y
y
1
x
y
1
1
and y 2 . The normal graphs look like:
x
x
1
x2
We can still graph by using transformations like we did in previous sections.
EXAMPLE: Graph y
1
using transformations.
x3
The x – 3 says we need to move the graph of y
1
three places to the right. We will get
x
Section 2.6 Notes Page 4
EXAMPLE: Graph y
1
using transformations.
x 6x 9
First we will factor: y
1
1
which can be written as y
. The x + 3 tells us that we need to
( x 3)( x 3)
( x 3) 2
move y
2
1
three places to the left. The negative tells us we need to flip the graph horizontally.
x2
EXAMPLE: Graph y
x 1
using transformations.
x
First I will divide everything by x to get y
This tells us to move the graph of y
1
x 1
or y 1
x x
x
1
up one unit and then
x
flip it horizontally.
Now we will look at the graphs of rational functions.
EXAMPLE: Find the intercepts, asymptotes, and graph of y
x 1
.
x2 9
First we will find the x-intercept by setting the top equal to zero: x + 1 = 0 so x = -1. We write (-1, 0).
To find the y-intercept, put in a zero for x: y
1
0 1
1
. You will get y and we write 0, .
2
9
9
0 9
To find the vertical asymptote, set the bottom equal to zero: x 2 9 0 We get x 3 .
For the horizontal asymptote, we notice the highest power on the top is less than the highest power on the
bottom. From the previous section we know that the horizontal asymptote is automatically y = 0.
Section 2.6 Notes Page 5
Now we are ready for the graph. First plot the intercepts. Then use dotted lines to draw in the asymptotes. The
y-axis is one asymptote, so we have a horizontal dotted line at y = 0. The other asymptotes are x 3 . These
will be vertical lines going through 3 and -3 on the x-axis.
Now we need to graph. On the left and right sides of the graph we
have two possible ways the graph can be drawn as the picture
below shows. Notice that on the ends the graph will follow one
asymptote, turn and follow the other asymptote. This is always the
case with rational expressions:
We need to chose whether the graph is above or below the x-axis on each end. In order to do this we need to
choose a test points. We need to choose a point that is less than negative 3 since we want to know what the
graph is doing to the left of vertical asymptote, x = -3. I will choose -4. We will put our test point into the
4 1
3
original equation: y
. We get y . Since this number is negative I know that the graph must
2
7
(4) 9
be below the x-axis. Now let’s test a point to the right of the vertical
asymptote, x = 3. I will choose 4 since this is greater than 3. Once
4 1
again we will put 4 in for x in the original equation: y
(4) 2 9
5
We get y which is positive so this tells me the graph is above
7
the x-axis. So now we know the graph looks like the following:
Now we need to take care of the middle part of the graph. In between the two vertical asymptotes a rational
graph will always look like one of the following:
Section 2.6 Notes Page 6
If we look at where our intercepts are that will help us choose one of the following. The only graph above that
would fit the intercepts we are already plotted would be the third graph, so the following will be the our
completed graph:
EXAMPLE: Find the intercepts, asymptotes, and graph of y
2x 2 4x
.
x 2 2 x 15
2 x( x 2)
. We should always factor something like this to see if anything
( x 3)( x 5)
cancels. Nothing does on this one so we will proceed to find the information. First we will find the x-intercept
by setting the top equal to zero: 2 x( x 2) 0 . We will get x = 0 and x = -2. We will write our answer as (0,
0) and (-2, 0). Since (0, 0) is an x-intercept then automatically this is our y-intercept as well.
Let’s factor this first: y
To find the vertical asymptote, set the bottom equal to zero: ( x 3)( x 5) 0 We get x = 3 and x = -5.
For the horizontal asymptote, notice that the highest power on top is the same as the highest power on the
a
2
bottom, so we know that y n . Here the a n 2 and bm 1 , so the horizontal asymptote is y 2 .
bm
1
Now we are ready for the graph. First plot the intercepts. Then use dotted lines to draw in the asymptotes.
There will be a horizontal line through y = 2. The other asymptotes are x = 3 and x = -5. These will be vertical
lines going through 3 and 5 on the x-axis.
Now we are ready to draw the graph. Let’s look at the part of the
graph to the left of x = -5. We need to decide whether the graph
will be above or below the horizontal asymptote. We actually do
not need to use test points for this. Notice that our only intercepts
are between the two vertical asymptotes. This means this is the
only play the graph crosses the x-axis. So when we look at the part
of the graph where x is less than -5 and since there are no
intercepts there I know the graph has to be above the horizontal
asymptote. The same reasoning can be said about the part of the
graph where x is greater than 3. There are no x-intercepts in the
part of the graph where x is greater than 3, so I know this part of
the graph will also be above the horizontal asymptote. Now let’s
look between the two vertical asymptotes. Remember there are
only four types of graphs that could appear here:
Section 2.6 Notes Page 7
By the way the intercepts are located this tells me that only the first two graphs would work. How can we tell
which one it is? We need to pick a test point. It can be any number between -5 and 3. I will test x = -3. I will
2(3) 2 4(3)
7
put this in for x in the original equation: y
. Since this number is negative that
2
(3) 2(3) 15 30
means when x is -3 the graph should be below the x-axis. This is only going to happen with the second graph
above (upside down parabola). So now we can finish our graph:
EXAMPLE: Find the intercepts, asymptotes, and graph of y
x
.
x 4
2
x
. Now we will find the x-intercept by setting the top equal to zero:
( x 2)( x 2)
x 0 . We are done. We will write our answer as (0, 0).
Let’s factor this first: y
Again since (0, 0) is an x-intercept then automatically this is our y-intercept as well.
To find the vertical asymptote, set the bottom equal to zero: ( x 2)( x 2) 0 We get x 2 .
For the horizontal asymptote, notice that the highest power on top is less than the highest power on the bottom,
so we know that the horizontal asymptote is automatically y = 0.
Section 2.6 Notes Page 8
Now we are ready for the graph. We only have one point to plot this time. Then we have our dotted lines to
draw in the asymptotes. There will be a horizontal line through y = 0. The other asymptotes are x 2 . These
will be vertical lines going through -2 and 2 on the x-axis.
Now we need draw the graph. Right now not much
information is given. We need to use test an x value that is
less than -2 and greater than -2 to see if the graph is above or
below the x-axis. First I will test x = -3. We will put this in
3
3
for x in the original equation: y
. We get a
2
5
(3) 4
negative number so I know the graph is below the x-axis at
this point. I also need to test a point greater than 2. I will
3
3
choose x = 3. You will get y 2
. So I know that
3 4 5
the graph must be above the x-axis. What about the part of
the graph between the two vertical asymptotes? We need to
do test points here as well to determine which of the four
models the graph resembles. I will test x = 1 and x = -1.
1
1
This tells us that when x = 1 the graph is below the x-axis.
3
1 4
1
1
. So we know the graph is above the x-axis at x = -1.
Now if we test x = -1 you will get y
2
(1) 4 3
This tells us that the only graph that will work in the middle
is the third one.
We will not test those values: y
2
EXAMPLE: Find the intercepts, asymptotes, and graph of y
x2 4
.
x
( x 2)( x 2)
. Now we will find the x-intercept by setting the top equal to zero:
x
( x 2)( x 2) 0 . We get x 2 . well can write this as 2,0 .
Let’s factor this first: y
For the y-intercept you would put in a zero for x. If you do then you will be dividing by zero, so there is no yintercept in this case.
To find the vertical asymptote, set the bottom equal to zero: x = 0. We only have one vertical asymptote here.
Section 2.6 Notes Page 9
For the horizontal asymptote, notice that the highest power on top is more than the highest power on the bottom,
so we know that this is no horizontal asymptote. However we need to find the oblique asymptote by doing long
division:
x0
x x 2 0 x 4 We need to put in a 0x term since it is missing. After we do the division we end up with
x2
an x + 0. This is the equation of the line that the graph will follow. We need to set up
our graph by plotting the intercepts and the asymptotes. We will have one vertical
asymptote at x = 0 and the oblique asymptote will be the line y = x.
0x
0x
4
We see where the graph will cross the x-axis. This tells us that the graph
will be in these section. To the left of the vertical asymptote the graph
will follow the line y = x and turn turn towards (-2, 0). Then it will follow
the vertical asymptote. To the right of the vertical asymptote the graph
will follow the vertical asymptote, turn through (2, 0) and then follow the
line y = x.
EXAMPLE: Find the intercepts, asymptotes, and graph of y
3 x 2 10 x 8
.
x 2 16
(3x 2)( x 4)
. Notice that we can cancel out the x – 4 from the top and bottom.
( x 4)( x 4)
Whenever this happens you will have what is called a hole in the graph. Whatever part you can cancel you
want to set this equal to zero. We will have x – 4 = 0. Solving it we get x = 4. This is not a vertical asymptote.
There will be a hole in the graph at the x value of 4.
3x 2
So the equation we will now look at to get the information will be y
.
x4
Let’s factor this first: y
Section 2.6 Notes Page 10
2
2
To find the x-intercept by setting the top equal to zero: 3 x 2 0 . We get x , so we have ,0 .
3
3
3(0) 2 1
1
. So 0, .
For the y-intercept you would put in a zero for x. If you do then you will have y
(0) 4 2
2
To find the vertical asymptote, set the bottom equal to zero: x + 4 = 0 and we get x = -4. We only have one
vertical asymptote here since the other term canceled out. We already determined that there will be a hole at the
x value of 4.
To find the horizontal asymptote we see the highest power on top is the same as the highest power on the
a
3
bottom. So the horizontal asymptote is y n . Here the a n 3 and bm 1 , so y 3 .
bm
1
So we have drawn in the asymptotes and the intercepts. We notice
that there are no x-intercepts to the right of x = -4. This tells us
that the graph must be above the horizontal asymptote (y = 3). If
the graph was below this line then it would have crossed the x-axis
but we know this does not happen. To the right of the x = -4 we
notice the graph does cross the x-axis so we know the graph is in
this lower portion and not above y = 3. Of course if you are unsure
where the graph should be you can always use tests points. You
test a point less than x = -4 and greater than x = -4. When we draw
the graph we need to make sure that we put a hole when x = 4.
Section 2.7 Notes Page 1
2.7 Polynomial and Rational Inequalities
This section is looking a sections of graphs that are above and below the x-axis.
EXAMPLE: Solve and write you answer in interval notation: ( x 5)( x 2) 2 0
We can rewrite the above equation as ( x 5)( x 2)( x 2) 0 . We need to find the critical points. This is what
makes each factor equal to zero. This would be x = 5 and x = -2. Now we want to set up a table with our
critical points and factors. Even though x + 2 is repeated we still need to include BOTH of them on our table:
x–5
x+2
x+2
-2
5
So our table is complete with the factors down the left column and our critical point in order from smallest to
largest along the bottom. Now we need to put in our test values. You need to test a value less than -2, and this
can be any number less than -2. I will use -3. In the middle column I need to pick any number between -2 and
5, so I will use 0. In the last column I need to test a value that is larger than 5, so I will use 6. Put these test
values in for x in all three factors in the left column and put a negative sign if you get a negative number and a
positive if the number is positive. For example when I test -3 I will put this in for x in the expression x – 5 and I
will get -3 – 5, which is -8. Since this is negative I will put a negative sign in that part of the table. Do this for
all the other values and you will get this table:
x–5
x+2
x+2
-
+
+
-2
+
+
+
5
Now is the step where I need to multiply the signs together in each column. In the left column I have three
negatives, so I will multiply those together to get a negative. In the middle column I have one negative and two
positives so when I multiply them I will get a negative. The last column has three positives so when I multiply
them I get a positive. I want to put these results on the table:
x–5
x+2
x+2
-
-2
+
+
-
5
+
+
+
+
The factored form was ( x 5)( x 2)( x 2) 0 which means that whatever we get for x must be less than zero.
This means we are looking for negatives. Since we want negative values, we want to indicate the regions on
our table that resulted in being negative. We notice that everything to the left of -2 was negative and everything
between -2 and 5 was negative. We will write (. 2) (2,5) as our answer. Are you wondering if we can
write our answer as (,5) ? The answer is no. The reason why is because when x = -2 the result will be zero.
In our problem it ends with < 0, meaning that zero is NOT included, so since -2 gives a zero it can’t be
included.
Now let’s look at the graph. You will not need to draw this graph in this section, but seeing a graph will help
you to visualize what it is you are actually doing in this section:
Section 2.7 Notes Page 2
Notice that the graph is below the x-axis until x = 5. Then
the graph goes above the x-axis after x = 5. This
confirms our answer in interval notation.
EXAMPLE: Solve and write you answer in interval notation: 3x 3 15 x 2
We want to rewrite the above equation as 3x 3 15 x 2 0 so that a zero is on the right hand side. We need to
find the critical points. This is what makes each factor equal to zero. If we factor this we will get:
3x 2 ( x 5) 0 . Setting it equal to zero we get x = 0 and x = -5. Now we want to set up a table :
3x 2
x+5
-5
0
So our table is complete with the factors down the left column and our critical point in order from smallest to
largest along the bottom. Now we need to put in our test values. You need to test a value less than -5, and this
can be any number less than -5. I will use -6. In the middle column I need to pick any number between -5 and
0, so I will use -1. In the last column I need to test a value that is larger than 0, so I will use 1. Put these test
values in for x in all three factors in the left column and put a negative sign if you get a negative number and a
positive if the number is positive. For example when I test -6 I will put this in for x in the expression 3x 2 and I
will get 3(6) 2 108 . Since this is positive I will put a positive sign in that part of the table. Do this for all the
other values and you will get this table:
3x 2
x+5
+
-
+
+
-5
+
+
0
Now is the step where I need to multiply the signs together in each column. In the left column I have a positive
and a negative, so I will multiply those together to get a negative. In the middle column I have two positives so
when I multiply them I will get a positive. The last column has two positives so when I multiply them I get a
positive. I want to put these results on the table:
3x 2
x+5
+
-
-5
+
+
+
0
+
+
+
The factored form was 3x 2 ( x 5) 0 which means that whatever we get for x must be greater than or equal to
zero. This means we are looking for positives. Since we want positive values, we want to indicate the regions
on our table that resulted in being positive. We notice that everything between -5 and 0 was positive and
everything to the right of zero is positive. This time 0 can be included since the sign was 0 . We will write
Section 2.7 Notes Page 3
[5, ) as our answer. Are you wondering if we can write our answer as [5,0) (0, ) ? The answer is no.
The reason why is because 0 should be included this time because we have 0 .
( x 2)( x 1)
EXAMPLE: Solve and write you answer in interval notation:
0
x4
This problem is already factored for us. For these rational functions we want to also find the critical points. To
do this you will set all factors in the numerator and denominator equal to zero. You will get x = 2, -1 and 4 as
the critical points. Now we will set up a table:
x–2
x+1
x–4
-1
2
4
So our table is complete with the factors from both top and bottom of the fraction down the left column and our
critical points are in order from smallest to largest along the bottom. Now we need to put in our test values. In
the first column I need to pick a number less than -1, so I will use -2. In the second column I need to pick any
number between -1 and 2, so I will use 0. In the third column I need to test a value that is between 2 and 4, so I
will use 3. In the fourth column I need to test a value that is greater than 4, so I will use 5. Put these test values
in for x in all three factors in the left column and put a negative sign if you get a negative number and a positive
if the number is positive. After you finish that your table should look like:
x–2
x+1
x–4
-
+
-1
+
+
2
+
+
+
4
Now is the step where I need to multiply the signs together in each column. In the first column I have three
negatives, so I will multiply those together to get a negative. In the second column I have two negatives and a
positive so when I multiply them I will get a positive. The third column has two positives and a negative so
when I multiply them I get a negative. The fourth column has three positives, so when I multiply those I get a
positive. I want to put these results on the table:
x–2
x+1
x–4
-
-1
+
+
2
+
+
-
4
+
+
+
+
( x 2)( x 1)
0 which means that whatever we get for x must be less than or equal to
x4
zero. This means we are looking for negatives. Since we want negative values, we want to indicate the regions
on our table that resulted in being negative. Our negative regions include everything less than -1 and also
everything between 2 and 4. So we write our answer as: (,1] [2,4] . BUT WAIT! Do you see an error
in my answer? I am supposed to have brackets in my answer, but I do not want them around the 4 because this
makes the denominator equal to zero and we can’t divide by zero. Therefore I do not want to include the 4 even
though my original problem had brackets. My final correct answer is: (,1] [2,4) .
The factored form was
Section 2.7 Notes Page 4
x2 4
0
EXAMPLE: Solve and write you answer in interval notation:
x3
First we should factor this as:
( x 2)( x 2)
0.
x3
We need to find our critical points, where we set each factor equal to zero no matter if it is on the top or bottom.
You will get x = 2, x = -2, and x = 3. These are the critical values that we will put on our table:
x+2
x–2
x–3
-2
2
3
So our table is complete with the factors from both top and bottom of the fraction down the left column and our
critical points are in order from smallest to largest along the bottom. Now we need to put in our test values. In
the first column I need to pick a number less than -2, so I will use -3. In the second column I need to pick any
number between -2 and 2, so I will use 0. In the third column I need to test a value that is between 2 and 3, so I
will use 2.5. In the fourth column I need to test a value that is greater than 3, so I will use 4. Put these test
values in for x in all three factors in the left column and put a negative sign if you get a negative number and a
positive if the number is positive. After you finish that your table should look like:
x+2
x–2
x–3
-
+
-2
+
+
2
+
+
+
3
Now is the step where I need to multiply the signs together in each column. In the first column I have three
negatives, so I will multiply those together to get a negative. In the second column I have two negatives and a
positive so when I multiply them I will get a positive. The third column has two positives and a negative so
when I multiply them I get a negative. The fourth column has three positives, so when I multiply those I get a
positive. I want to put these results on the table:
x+2
x–2
x–3
-
-2
+
+
2
+
+
-
3
+
+
+
+
( x 2)( x 2)
0 which means that whatever we get for x must be greater than zero.
x3
This means we are looking for positives. Since we want positive values, we want to indicate the regions on our
table that resulted in being positive. Our positive regions include everything between -2 and 2 and also
everything greater than 3. So we write our answer as: (2,2) (3, ) . We don’t need to worry about dividing
by zero like the previous problem. Three is not included anyway.
The factored form was
EXAMPLE: Solve and write you answer in interval notation:
x2
1
x4
Section 2.7 Notes Page 5
The problem with this one is that there is not a zero on the right hand side. The process we have been doing
x2
only works if there is zero on the right side. So we can subtract 1 from both sides to get:
1 0 . But
x4
x 2 x 4
now we need a single fraction on the left side so we need to get common denominators:
1
0.
x 4 x 4
x 2 ( x 4)
x2 x4
0 . Distribute the negative to get:
0.
Combining to one fraction we get:
x4
x4
6
0 . Now for this one we will only have one critical value, which is x = 4.
x4
The top of the fraction can’t be set equal to zero, which is why we only have one critical point here.
After simplifying we will get
6
x–4
4
So all we need to do now is test a value less than 4 and greater than 4. I will test 0 and 5. The top row of the
table will automatically by positive since 6 is positive. There is no variable to put a test value in on the top. On
the bottom row I can put in my test value. Let’s complete the table:
6
x–4
+
-
+
+
4
Now is the step where I need to multiply the signs together in each column. In the left column I have a positive
and a negative, so I will multiply those together to get a negative. In the right column I have two positives so
when I multiply them I will get a positive. I want to put these results on the table:
6
x–4
+
-
4
+
+
+
6
0 which means that whatever we get for x must be greater than zero. This
x4
means we are looking for positives. Since we want positive values, we want to indicate the regions on our table
that resulted in being positive. We notice that everything greater than 4 was positive. This is the only region
that was positive so we write our answer as (4, ) .
The factored form was
EXAMPLE: Solve and write you answer in interval notation:
2 x 2 3x 4
2
x 1
Section 2.7 Notes Page 6
This is another one where we need to get a zero on the right hand side. We first subtract 2 from both sides to
2 x 2 3x 4
2 x 2 3x 4
x 1
get:
2 0 . Now we need to get common denominators again:
2
0.
x 1
x 1
x 1
2 x 2 3x 4 2( x 1)
Combining to one fraction we get:
0 . Now distribute the -2 and you will get:
x 1
2 x 2 3x 4 2 x 2
2x 2 x 6
0 . After simplifying you should get:
0 . Now you can factor the top
x 1
x 1
( x 2)(2 x 3)
using any method. You will get:
0 . Now we will find the critical points now that it is
x 1
3
factored. You will get x = -2, , and -1. Now we can make our table:
2
x+2
2x – 3
x+1
-2
-1
3/2
So our table is complete with the factors from both top and bottom of the fraction down the left column and our
critical points are in order from smallest to largest along the bottom. Now we need to put in our test values. In
the first column I need to pick a number less than -2, so I will use -3. In the second column I need to pick any
number between -2 and -1, so I will use -1.5. In the third column I need to test a value that is between -1 and
3/2, so I will use 0. In the fourth column I need to test a value that is greater than 3/2, so I will use 2. Put these
test values in for x in all three factors in the left column and put a negative sign if you get a negative number
and a positive if the number is positive. After you finish that your table should look like:
x+2
2x – 3
x+1
-
+
-2
+
+
-1
+
+
+
3/2
Now is the step where I need to multiply the signs together in each column. In the first column I have three
negatives, so I will multiply those together to get a negative. In the second column I have two negatives and a
positive so when I multiply them I will get a positive. The third column has two positives and a negative so
when I multiply them I get a negative. The fourth column has three positives, so when I multiply those I get a
positive. I want to put these results on the table:
x+2
2x – 3
x+1
-
-2
+
+
-1
+
+
-
3/2
+
+
+
+
( x 2)(2 x 3)
0 which means that whatever we get for x must be less than or equal to
x 1
zero. This means we are looking for negatives. Since we want negative values, we want to indicate the regions
on our table that resulted in being negative. Our negative regions include everything less than -2 and between 3
1 and 3/2. So we write our answer as: (,2] 1, . Notice that -1 has a parenthesis because this will
2
cause the bottom of the fraction to be zero.
The factored form was
Section 3.1 Notes Page 1
3.1 Exponential Functions
In this section we will be working with exponents, so here is a quick review of the exponent rules:
Laws of Exponents
a s a t a s t
Example: 2 5 2 3 2 53 2 8
as
a s t
t
a
26
Example: 3 2 63 2 3
2
a
Example: 2 3
s t
a st
a b s
as bs
5
2 35 215
Example: 2 3 2 5 35
5
1s 1
Example: 134 1
a0 1
Example: 4 0 1, 0 1
a s
a
b
s
1
as
b
a
Example: 4 2
s
2
Example:
3
2
1
1
2
16
4
2
9
3
4
2
Exponential function: y b x
We will look at a specific exponential function to see its characteristics. To do this we will make a table.
Then we will plot the points. The graph will be a curved line:
Graph of y 2 x
x
y 2x
(x, y)
-2
y 2 2
1 1
22 4
1 1
y 2 1 1
2
2
1
2,
4
1
1,
2
(0, 1)
-1
0
y 20 1
1
y 21 2
2
y 22 4
1, 2
2, 4
Notice from the graph of y 2 x that the y-intercept is (0, 1). This will always be the case for exponential
functions. Also notice that there is a horizontal asymptote at y = 0.
Section 3.1 Notes Page 2
Let’s now look at the graph of y e . To get the letter e on your calculator, look where your LN button is and
probably right above it should be a e^ key. Hit this and then the number you want to raise e to. Let’s make a
table:
x
Graph of y e x
x
y ex
-2
y e 2
0
1
1
0.135
2
7.39
e
1
1
y e 1 1
0.368
2.72
e
y e0 1
1
y e1 2.72
2
y e 7.39
-1
2
x, y
2, .0.135
1, 0.368
(0, 1)
0, 2.72
2, 7.39
This graph still has a y-intercept of (0, 1) and a horizontal asymptote at y = 0. The difference here is that this
graph is steeper. So the larger the number is on the bottom, the steeper the graph is. What if we have a fraction
instead of a number? Let’s find out.
1
Graph of y
2
x
x, y
x
1
y
2
x
-2
1
y
2
2
-1
1
2
y 2
2
1
0
1
y 1
2
0
2
2, 4
1
1, 2
2
4
1
1
0, 1
1
1
1,
2
2
1
2,
4
1
1
1
y
2
2
2
1
1
y
4
2
x
2
1
Notice that a fraction causes the graph to flip about the vertical axis. This is because y
1
2
x
Since the exponent is negative this means we flip the graph of y 2 about the vertical axis.
x
2x .
Section 3.1 Notes Page 3
EXAMPLE: Find the intercepts and graph using transformations: y 2 4
x
To find the y-intercept, we put in a zero for x. You will get y 2 0 4 . This is y 1 4 3 . We write our
answer as (0, -3).
To find the x-intercept, we put in a zero for y. You will get 0 2 x 4 . To solve this we will add 4 to both
sides: 4 2 x . We will solve this by setting the bases equal to each other. We can rewrite this as 2 2 2 x so
we know that x = 2. We write this as (2, 0).
Since there is a -4 on the outside of the function this tells
us to move the graph of y 2 x down 4 units. What we
are actually doing is moving the horizontal asymptote
down 4 units. To graph this we will first draw our shifted
horizontal asymptote. The graph crosses the y-axis exactly
one unit above the horizontal asymptote, which here is (0, -3)
EXAMPLE: Find the intercepts and graph using transformations: y 2 x 1 2 .
To find the y-intercept, we put in a zero for x. You will get y 2 01 2 . This is y 2 1 2 . This is the
1
3
3
same as y 2 . We write our answer as 0, .
2
2
2
To find the x-intercept, we put in a zero for y. You will get 0 2 x 1 2 . To solve this we will add 2 x 1 to
both sides: 2 x 1 21 . Since the bases are now the same, we get x – 1 = 1, so x = 2. We write this as (2, 0).
Since there is a 2 on the outside of the function this tells
us to move the graph of y 2 x up 2 units. The x – 1 in
the exponent tells us to move the graph one unit to the right.
To graph this we will first draw our shifted horizontal
asymptote by using a dotted line. Next we will draw in a
vertical line at x = 1 since the graph shifts one unit to the right.
The negative in front of the 2 x 1 tells us to flip the graph over
the horizontal axis, which is why the graph goes down.
Section 3.1 Notes Page 4
1 x
EXAMPLE: Find the y-intercept and graph using transformations: y 3
2.
To find the y-intercept, we put in a zero for x. You will get y 310 2 which is y = 1. So we write (0, 1).
To graph using transformations we must first have it in the
correct form. You need to factor out a negative one in the
exponent. You will get y 3 ( 1 x ) 2 which is the same
as y 3 ( x 1) 2 . So this tells us we need to move the
graph of y 3 x one unit to the right and down two units.
Then we need to flip the graph over the vertical axis. You
always want to use dotted lines to represent the new set of
axis. Notice again that the graph crosses our new vertical
axis exactly one unit above the new horizontal axis. Also
notice that the graph of y 3 x is slightly steeper than y 2 x
but the general shape is the same. Do not confuse the vertical
dotted line as an asymptote. There are NO vertical asymptotes
in exponential graphs.
EXAMPLE: Find the x-intercept and graph using transformations: y e ( x 2 ) 1 .
To find the x-intercept we put in a zero for y to get: 0 e ( x 2) 1 . We need to add the e ( x 2 ) to both sides.
You will get e ( x 2 ) 1 . The bases are not the same, but we can still solve this. The only way we can get a 1 is
if we raise something to a power of zero. So we need to see what number will give us a zero in the exponent.
This will be when x = -2. So we write (-2, 0).
This is given to us in the correct form, so we can now graph
it using transformations. So this tells us we need to move the
graph of y e x two units to the left and up one unit. So we
draw a horizontal dotted line one unit above the x-axis. Then
we draw a vertical line two units to the left of the y-axis. We
need to flip the graph over the vertical axis AND the horizontal
axis since there are two negatives in this problem. We always
want to use dotted lines to represent the new set of axis. Notice
again that the graph crosses our new vertical axis exactly one
unit below the new horizontal axis.
EXAMPLE: Approximate 5
3
using a calculator. Round your answer to three decimal places.
To enter this on the calculator, first enter 5. Then you want to look for the ^ key. On some calculators you may
see a y x or x y . All of these mean the same thing. Hit this key and then enter the 3 . If your calculator
displays what you are entering in, then you will be able to hit the square root key first and then the 3. Otherwise
you may need to enter a 3 first and then the square root. Once this is done, then press enter again to display the
answer. You should get 16.242 rounded to three places.
Section 3.2 Notes Page 1
3.2 Logarithmic Functions
In a previous section we looked at inverses. In order to find a inverse we need to switch x and y.
Suppose we wanted to find the inverse of our exponent function, y a x . First we need to switch x and y. We
will get x a y . How do we solve for y? This is where we need logarithms, which are a way to solve for an
exponent.
With logarithms there are two forms: exponential and logarithmic.
Exponential form: x b y
Logarithmic form: y log b x
Both of these forms are exactly the same. Let’s practice changing between the two forms.
EXAMPLE: Change 7 log m 5 into exponential form.
Here y = 7, b = m and x = 5. If we put these into the exponential form we get m 7 5 .
EXAMPLE: Change log c 6 8 into exponential form.
Here b = c, y = 8 and x = 6. If we put these into the exponential form we get c 8 6 .
EXAMPLE: Change q 1.4 5 into logarithmic form.
Here x = q, b = 1.4 and y = 5. If we put these into the logarithmic form we get 5 log1.4 q .
EXAMPLE: Change 2 d 8 into logarithmic form.
Here x = 8, b = 2 and y = d. If we put these into the logarithmic form we get d log 2 8 .
EXAMPLE: Change x e y into logarithmic form.
Changing this into logarithmic form we get y log e x . This has a special form: y ln x . This called the
natural logarithm. This is the same as a logarithm with a base of e.
We can do logs on our calculator. We have a ln key and also a log key. Our calculator will only do base 10 and
base e logarithms. Let’s practice using the calculator:
Section 3.2 Notes Page 2
EXAMPLE: Find the value of log 5 on a calculator and round to the nearest thousandth.
In your text you will sometimes see logs without bases. If there is no base then this automatically means it is
base 10. So we need to find the numerical value of log10 5 on our calculator. If you have a scientific calculator
that does not display what you are typing in, you will need to enter the 5 and then hit the log key. For all other
calculators, enter log and then 5. You should get 0.699.
EXAMPLE: Find the value of ln 7 on a calculator and round to the nearest hundredth.
Enter 7 and the ln key if you have a scientific calculator the does not display what you are typing in. Otherwise,
first enter ln and then 7. You should get 1.95.
EXAMPLE: Find the exact value of log 4 64 .
This one we can’t do on our calculator like the ones above since we don’t have a base 4. What we can do is
change it into exponential form and solve. First we will let y log 4 64 . Changing into exponential form we
will get 4 y 64 . We want to find which number will make this statement correct. The answer is three, since
(4)(4)(4) = 64. So now we know that log 4 64 3 .
EXAMPLE: Find the value of log 0 on a calculator and round to the nearest hundredth.
We have a problem with this one. If we put this in our calculator we will get an error or undefined.
Let’s try and draw a graph of y log b x . In order to do this, we need to first draw the graph of x b y . Then
we need to draw its inverse. We know that the log function
is the inverse of the exponential function. We also know that
all inverse are reflective about the line y = x. In the graph to
the right you can see both graphs drawn. We see that the point
(0, 1) and (1, 0) are flipped, which it should since they are
inverses. So the graph of y log b x will always cross over
the x-axis one unit away from the vertical asymptote. Let’s
look at the graph of the logarithmic form. The y-axis is a
vertical asymptote. There is no horizontal asymptote on the
logarithmic graph. From the graph we can conclude:
Domain of y log b x is x > 0. Notice that zero is not included
since this is a vertical asymptote. Also notice that it doesn’t
matter what a is.
Let’s look at some problems that ask you to find the domain.
Section 3.2 Notes Page 3
EXAMPLE: Find the domain of y log 3 (6 x) and write your answer in interval notation.
Whatever is inside the parenthesis must be set to be greater than zero since this is the domain: 6 – x > 0. If we
subtract 6 from both sides we get –x > –6. Dividing both sides by -1 we get x < 6. Notice we needed to switch
the sign because we divided by a negative. We write our answer as , 6 .
EXAMPLE: Find the domain of y ln(2 x 7) and write your answer in interval notation.
Even though we have a natural log we will still solve this the same way since a natural log is a just a log with a
7
7
base of e: 2x – 7 > 0. Solving this we will get x . We write our answer as , .
2
2
x 2
EXAMPLE: Find the domain of y log 2
and write your answer in interval notation.
2 x
x2
0 . Recall that if we have a fraction with an inequality, then this requires us to set up a
2x
table with the plus and minus values. Our critical points will be -2 and 2.
We will have
x+2
x–2
–
+
–
-2
+
+
+
2
+
–
–
So we see that our answer is (-2, 2).
EXAMPLE: Graph using transformations: y log 3 ( x 4) and identify the x-intercept.
We need to take the graph of y log b x and shift it 4 places
to the right. This will move the vertical asymptote over 4
places. Notice once again that the graph will cross the x-axis
one unit away from the vertical asymptote. Therefore the
x-intercept is (5, 0).
Section 3.2 Notes Page 4
EXAMPLE: Graph using transformations: y log 2 ( x 3) and identify the x-intercept.
This will shift the graph of y log b x three places to the
left. First we can draw in the vertical asymptote as a dotted
line. The negative in front of the log will flip this graph over
the horizontal axis, which is why it is below the x-axis instead
of above. The x-intercept is always one unit away from the
vertical asymptote, so it is (-2, 0).
EXAMPLE: Graph using transformations: y ln(1 x) and identify the x-intercept.
This is equivalent to y log e (1 x) which will have the
same initial shape as the previous two examples. Before
we can use transformations we first need to get the x to
come first. We will get y log e ( x 1) . Now we factor
out a –1: y log e ( x 1) . This tells us we need to
move graph 1 place to the right, so this is where we will
draw our vertical asymptote. The negative inside the log
will flip the graph over the vertical asymptote so now it
will point to the left instead of the right. The x-intercept is
one unit away from the vertical asymptote, which is (0, 0).
EXAMPLE: Graph using transformations: y log ( x 2) and identify the x-intercept.
If there is no base then this is assumed to be base 10, so we
have y log10 ( x 2) . We need to move this graph
two places to the left, so this is where we will draw the
vertical asymptote. The outside negative flips the graph
over the horizontal axis. The inside negative flips the graph
over the vertical axis. The x-intercept still is one unit away
from the vertical asymptote, which is (-3, 0).
Section 3.3 Notes Page 1
3.3 Properties of Logarithms
The following is a list of properties of the logs with corresponding examples. We will look at some examples
of properties #6 and 7 a little later in this section.
Properties of Logarithms
1.) log b 1 0
Example: log 3 1 0 , ln 1 0
2.) log b b 1
Example: log 2 2 1 , ln e 1 , log 10 1
3.) b logb M M
Example; 2 log 2 5 5 , 5 log5
4.) log b b r r
Example: log 3 37 7 , log 2 2 5
5.) log b M r r log b M
Example: log 3 58 8 log 3 5
6.) log b M N log b M log b N
Example: log 2 (3 5) log 2 3 log 2 5
M
7.) log b log b M log b N
N
24
Example: log 3 log 3 24 log 3 6
6
EXAMPLE: Find the exact value using logarithm properties: log 3 3 5 .
This one uses property #4, so the answer is -5.
EXAMPLE: Find the exact value using logarithm properties: ln e 6 .
This one also uses property #4, so the answer is 6.
EXAMPLE: Find the exact value using logarithm properties: log 3 21 log 3 7 .
21
First we will use property #7. This will give us: log 3 . This equals log 3 3 . We will use property #2 to
7
get our answer of 1.
EXAMPLE: Find the exact value using logarithm properties: 5 log5 6 log5 7 .
First we will use property #6. This will give us: 5 log5 ( 67 ) . This equals 5 log5 42 . We will use property #3 to get
our answer of 42.
Section 3.3 Notes Page 2
EXAMPLE: Express log 9 x 3 x 5 as a sum or difference of logarithms. Express powers as factors.
2
We will first use property #6 to break this apart. You will get log 9 x 2 log 9 3 x 5 . The square root can be
1
2
written as a ½ power: log 9 x log 9 (3 x 5) . Now use property #5 to bring the powers down in front of the
1
logs since it wants us to express powers as factors: 2 log 9 x log 9 (3x 5) . This is our answer.
2
2
EXAMPLE: Express ln
( x 5) 4
as a sum or difference of logarithms. Express powers as factors.
x3
We will first use property #7 to break this apart. You will get ln( x 5) 2 ln x 3 . Now use property #5 to bring
the powers down in front of the logs since it wants us to express powers as factors: 2 ln( x 5) 3 ln x .
( x 5) 5 3 x 2
EXAMPLE: Express log 4
as a sum or difference of logarithms. Express powers as factors.
( x 1) 4
We will first use property #7 to break this apart. You will get log 4 ( x 5) 5 3 x 2 log 4 ( x 1) 4 . Now we can
use property #6 to break up the first log. You will get: log 4 ( x 5) 5 3 x 2 log 4 ( x 1) 4 . We can rewrite the
1
3
cube root as a 1/3 power: log 4 ( x 5) ( x 2) log 4 ( x 1) 4 . Now use property #5 to bring the powers down
1
in front of the logs since it wants us to express powers as factors: 5 log 4 ( x 5) log 4 ( x 2) 4 log 4 ( x 1) .
3
5
1
x 2 5x 6 4
EXAMPLE: Express ln
as a sum or difference of logarithms. Express powers as factors.
3
( x 2)
We will first use property #5 to bring down the ¼. You will get
see if anything can be canceled:
1 x 2 5x 6
ln
. Now factor the inside to
4 ( x 2) 3
1 ( x 2)( x 3)
ln
. We will use property #6 and #7 to break apart this log:
4 ( x 2) 3
1
ln( x 2) ln( x 3) ln( x 2) 3 . Now we use property #5 again with that third term:
4
1
ln( x 2) ln( x 3) 3 ln( x 2). You can either leave your answer as this or you could distribute:
4
1
1
1
ln( x 2) ln( x 3) ln( x 2) 3 .
4
4
4
Section 3.3 Notes Page 3
2 x ( x 4)
as a sum or difference of logarithms. Express powers as factors.
5( x 3) 5
4
EXAMPLE: Express log 5
When we use property #6 and 7 to break this apart, everything on top will have a positive sign in front of it and
everything on the bottom will have a negative sign in front of it. After using the properties we will get:
log 5 2 log 5 x 4 log 5 ( x 4) log 5 5 log 5 ( x 3) . Notice that the term 2x 4 can also be broken up into 2 x 4 .
We can do one more thing with log 5 5 . This is using property #5 to bring down the powers:
log 5 2 4 log 5 x log 5 ( x 4) log 5 5 5 log 5 ( x 3) . We can do one more thing with log 5 5 . We can use
property #2 to get our final answer: log 5 2 4 log 5 x log 5 ( x 4) 1 5 log 5 ( x 3) . It doesn’t matter the
order in which you write the terms.
EXAMPLE: Express
1
log 6 x 2 log 6 ( x 2) as a single logarithm without negative exponents.
2
These problems are having us do the opposite steps of what we were doing in the previous powers. In the
previous problems the last step we did was use property #5 to bring down the powers, so now this will be our
1
first step. We will bring up the powers using property #5: log 6 x 2 log 6 ( x 2) 2 . Now we will use property
1
#6 to combine these together into one log: log 6 x 2 ( x 2) 2 , or log 6 x ( x 2) 2 .
EXAMPLE: Express ln( x 2 1) 2 ln( x 1) as a single logarithm without negative exponents.
First use property #5 to bring up the power on the second term: ln( x 2 1) ln( x 1) 2 . Now use property #7 to
x2 1
. We can factor the numerator using the difference of squares:
( x 1) 2
x 1
( x 1)( x 1)
. Now we can reduce to get our answer: ln
.
ln
2
x 1
( x 1)
write these as a single log: ln
EXAMPLE: Express 2 log 3 x log 3 ( x 2 2) log 3 4 3 log 3 x log 3 8 as a single logarithm without
negative exponents.
I will rewrite this with the positive terms first and the negative terms following:
log 3 ( x 2 2) 3 log 3 x log 3 8 2 log 3 x log 3 4 . What I will do first is to distribute the negative from the
last two terms: log 3 ( x 2 2) 3 log 3 x log 3 8 (2 log 3 x log 3 4) . Now that the negative has been factored
out I will use property #5 to bring up the powers: log 3 ( x 2 2) log 3 x 3 log 3 8 (log 3 x 2 log 3 4) . I will now
use property #6 to condense these: log 3 8 x 3 ( x 2 2) (log 3 4 x 2 ) . Now we can use property #7 to write these
as a single log: log 3
8 x 3 ( x 2 2)
. Finally we can reduce this: log 3 2 x( x 2 2) .
4x 2
Section 3.3 Notes Page 4
Change-of-Base Formula
In the last section we mentioned that the only type of logs we can do on our calculator is base 10 and base e.
This formula will allow us to find the numerical value of a log with ANY base.
log a M
log b M
log b a
In this formula the b can be any base we want.
EXAMPLE: Use the Change-of-Base Formula to calculate log 5 8 to two decimal places.
In this problem M = 8 and a = 5. Since we want to do this on the calculator we want b to be either 10 or e since
log 8
1.29 . Notice we are using base 10 here.
these are the only bases we can do on our calculator: log 5 8
log 5
ln 8
You could also have done: log 5 8
1.29 . Both will give you the same answer.
ln 5
EXAMPLE: Use the Change-of-Base Formula to calculate log 3 8 log 8 9 to two decimal places.
log 8 log 9
. (You could have also used natural
log 3 log 8
log 9
log 9
logs here (ln). We can cancel the log 8 from the top and bottom to get
log 3 . You
. Notice
log 3
log 3
cannot divide 9 by 3 since it is inside the log. We calculate log 9 and log 3 separately and divide to get our
answer of 2.
We can rewrite this using the formula twice: log 3 8 log 8 9
Section 3.4 Notes Page 1
3.4 Exponential & Logarithmic Equations
In this section we will look at how to solve equations involving logarithms or exponents.
Equal Bases Property:
If a u a v then u = v.
EXAMPLE: Solve: 4 x 2 64 .
In order to solve this, we must make both the bases the same. Since there is a 4 on the left hand side, I want to
write 64 as 4 raised to some power. It is known that 4 3 64 so we can now rewrite our equation:
4 x 2 4 3 . The property above states that if the bases are the same then we can set the exponents equal to each
other. If we do this we will have x – 2 = 3. Solving this we get x = 5.
1
EXAMPLE: Solve:
2
1 x
4.
We need to get both the bases to be the same. Both sides can be written with a base of 2. If we flip the fraction
of ½ over we will change the sign of the exponent. We can also write 4 as 2 2 . Our equation is now:
(1 x )
2
2 2 . Now both bases are 2. We can set the exponents equal to each other. You will get –(1 – x) = 2.
1
Solving this you will get x = 3.
EXAMPLE: Solve: 27 x 2 9 2 x .
Again we need the same base. 27 can’t be written as 9 to a power, so we need something smaller. Each of
these can be written with a base of 3:
3
3 x2
3
3 x 6
32
34 x
2x
We are raising a power to another power and when you do, multiply the exponents.
Now the bases are the same so let’s set the exponents equal: 3x + 6 = 4x. Solving for x
we get x = 6.
EXAMPLE: Solve: 3 4 y 1 3 .
1
2
4 y 1
1
2
3 . The bases are the same,
We can change the right hand side to be 3 . Then our equation becomes 3
1
so we can set the exponents together. You will get 4 y 1 . To solve this, multiply both sides by 2 to clear
2
3
the fractions. You will get 8 y 2 1 . Then solve for y. The answer is y .
8
EXAMPLE: Solve: 2
x x 2
Section 3.4 Notes Page 2
8.
Since we are raising a power to another power we need to multiply the exponents. I also want to change the 8
2
into 2 3 so we can have the same base on both sides. Now our equation becomes 2 x 2 x 2 3 . Since the bases
are the same we can now set the exponents equal to each other. The equation becomes x 2 2 x 3 . In order to
solve this you must set it equal to zero and solve. We have x 2 2 x 3 0 . Factoring we get ( x 1)( x 3) 0
and so x = -1 and x = 3.
EXAMPLE: Solve: 3 x 7 .
For this problem, we can’t use the Equal Bases Property since no matter what I do I can’t get the same base
here. In order to solve this you can take either the natural log or regular log of both sides:
ln 3 x ln 7
x ln 3 ln 7
ln 7
x
.
ln 3
Now use property #5 to bring down the x.
Divide by sides by ln 3 to solve for x.
Note, you answer could also have been x
log 7
, which is the same answer.
log 3
EXAMPLE: Solve: e x 5 4 .
You definitely want to take the natural log of both sides so we can cancel out the e.
ln e x 5 ln 4
x 5 ln 4
x ln 4 5
This is the same as log e e x 5 ln 4 . We can use property #4 to simplify this:
The ln and e cancel, so now we can solve for x.
This is our final answer.
If you are wondering if we can do subtract the 5 from 4 then the answer is no. These are not like terms.
EXAMPLE: Solve: 2 x 1 51 2 x .
ln 2 x 1 ln 51 2 x
( x 1) ln 2 (1 2 x) ln 5
x ln 2 ln 2 ln 5 2 x ln 5
x ln 2 2 x ln 5 ln 5 ln 2
x(ln 2 2 ln 5) ln 5 ln 2
ln 5 ln 2
x
ln 2 2 ln 5
First take the natural log of both sides and use property #5 to bring down powers.
Now distribute.
Get all the terms with x in it on one side of the equation.
Now factor out an x.
Solve for x.
We can leave our answer in this form. You do not need a decimal.
Are you allowed to cancel the ln 2 from top and bottom? The answer is NO. You can only do this if all the
terms on the top and bottom are being multiplied. Therefore this is as far as we can go with simplifying.
Section 3.4 Notes Page 3
EXAMPLE: Solve: 3 3 72 0 .
2x
x
This one will factor. In order to do this, let’s make a substitution. We will let u 3 x and u 2 3 2 x . The
always the middle term, ignoring the sign in front of it, so u is 3 x .
u 2 u 72 0
We have substituted. Now it is easier to factor.
(u 9)(u 8) 0
Now put back the u, which is u 3 x .
(3 x 9)(3 x 8) 0
Set each one equal to zero and solve each one individually.
3x 9 0
3x 9
3 x 3 2 so x = 2.
3x 8 0
3 x 8
ln 3 x ln 8 The problem with this one is that -8 is not in our domain, so this
is not one of our answers. The only answer is x = 2.
EXAMPLE: Solve: 2 2 x 7 2 x 12 0 .
This one will factor. In order to do this, let’s make a substitution. We will let u 2 x and u 2 2 2 x . The
always the middle term, ignoring the sign in front of it, so u is 3 x .
u 2 7u 12 0
We have substituted. Now it is easier to factor.
(u 3)(u 4) 0
Now put back the u, which is u 2 x .
(2 x 3)(2 x 4) 0
Set each one equal to zero and solve each one individually.
2x 3 0
2x 3
ln 2 x ln 3
2x 4 0
2x 4
Solving we get x = 2.
ln 3
ln 3
So x ln 2 ln 3 , and after dividing we get x
. So our two answers are x 2, x
ln 2
ln 2
EXAMPLE: Solve: log 5 (4 x 5) 2 .
To solve this one, we first want to change from logarithmic form to exponential form. You will get
5 2 4 x 5 . So we have 25 4 x 5 , in which 20 = 4x, so x = 5.
EXAMPLE: Solve: 2 log 7 x log 7 16
First we want to use property #5 from the previous section to bring up the 2. You will get: log 7 x 2 log 7 16 .
Now we need to get rid of the logs. We can do this be changing from the log form to exponential. It doesn’t
matter which log you start with. I will change the log on the left side of the equation. Changing from log form
to exponential you will get: 7 log7 16 x 2 . This simplifies to 16 x 2 , so by solving this we get x = 4 and x = -4.
What happens if we put x = -4 back into the original equation? That’s right, you will get an error. The answer
of x = -4 is not in the domain, so we need to discard it. So our only answer to this one is x = 4.
MAKE SURE YOU ALWAYS CHECK YOUR ANSWERS TO SEE IF IT FITS THE DOMAIN!! IF NOT,
YOU WILL NEED TO REJECT THIS ANSWER.
Section 3.4 Notes Page 4
EXAMPLE: Solve: log 8 ( x 7) 1 log 8 (2 x)
The strategy with these problems is to first get all the logs on one side of the equation. Next the logs will be
combined into one by using the log properties. Last we will change it into exponential form. Let’s first get all
the logs on one side of the equation: log 8 ( x 7) log 8 (2 x) 1 . Now we can use property #6 to combine
these into a single log: log 8 ( x 7)(2 x) 1 . Now we need to change this into exponential form:
( x 7)(2 x) 81 . We need to multiply on the left side of the equation: x 2 5 x 14 8 . Now set this
equal to zero: x 2 5 x 6 0 . Now multiply both sides by -1 to get: x 2 5 x 6 0 . Now factor:
( x 6)( x 1) 0 . We get x = -6 and x = 1 as our answers. Now let’s check to make sure both of these are in
our domain. Put -6 into the original equation to get log 8 (6 7) 1 log 8 (2 6) . We get:
log 8 (1) 1 log 8 (8) . Since both of the numbers inside the log are positive, that means it is in our domain.
Now let’s try x = 1: log 8 (1 7) 1 log 8 (2 1) . You will get log 8 (8) 1 log 8 (1) . The numbers inside the
log are positive, so we know that x = 1 is in our domain. Our answers are x = -6 and x = 1.
EXAMPLE: Solve: log 2 ( x 11) log 2 ( x 7) 5
log 2 ( x 11)( x 7) 5
2 ( x 11)( x 7)
32 x 2 18 x 77
0 x 2 18 x 45
0 ( x 3)( x 15)
x 3, x 15
5
First combine into one log.
Change into exponential form
Multiply and simplify
Set it equal to zero
Factor
These are our answers. Now we need to make sure they are in domain.
If we put -3 into the original we get log 2 (3 11) log 2 (3 7) 5 which is log 2 8 log 2 4 5 This is okay
since both 8 and 4 are in the domain. If we put in -15 we get log 2 (15 11) log 2 (15 7) 5 which results
in log 2 (4) log 2 (8) 5 . We can’t have negative numbers inside a log, therefore -15 is not one of our
answers. Our only answer for this problem is x 3 .
EXAMPLE: Solve: log 2 ( x 3) log 2 ( x 5) 1
log 2 ( x 3) log 2 ( x 5) 1
x 3
log 2
1
x 5
x3
21
x5
2( x 5) x 3
2 x 10 x 3
x 7
First combine into one log. This time we will turn it into a fraction
We used property #7. Now change into exponential form.
We can solve this by cross multiplying.
Now solve for x.
If we put -7 into the original we get log 2 (7 3) log 2 (7 5) 1 which is log 2 (4) log 2 (2) 1 We can’t
have a negative inside the log, so we reject the answer x = -7. Since our only answer did not work, the answer
to the problem is “no solution” or undefined.
Section 3.4 Notes Page 5
EXAMPLE: Solve: log 3 ( x 1) 2 log 3 (2 x 1)
log 3 ( x 1) log 3 (2 x 1) 2
x 1
log 3
2
2x 1
x 1
32
2x 1
x 1
9
2x 1
9(2 x 1) x 1
18 x 9 x 1
10
x
.59
17
Get all the logs on one side of the equation.
First combine into one log. This time we will turn it into a fraction
We used property #7. Now change into exponential form.
We can solve this by cross multiplying.
Now solve for x.
If we put this into the original problem we get log 3 (.59 1) 2 log 3 (2(.59) 1) .
There are positive number inside the logs, so x = 10/17 is our answer.
EXAMPLE: Solve: log 2 ( x 3) log 2 x log 2 ( x 2) 2
x( x 3)
log 2
2
x2
x( x 3)
22
x2
2
x 3x
4
x2
x 2 3x 4( x 2)
x 2 3x 4 x 8
x2 7x 8 0
( x 1)( x 8) 0
x = -1, 8
Combine into one log. Positive logs on top, negative logs on bottom.
We changed from log form into exponential form.
Now cross multiply.
Now solve for x. Because it’s a quadratic, we need to set it equal to zero.
Now factor and set each factor equal to zero.
If we put these into the original problem, we find that the only answer is x = 8.
Section 3.5 Notes Page 1
3.5 Exponential Growth and Decay; Modeling Data
There are many applications to exponential functions that we will focus on in this section. First let’s look at the
exponential model.
Exponential Growth / Decay Model
f (t ) Ao e kt
f (t ) = current population, Ao = original population, k = growth or decay constant
t = time measured in any unit.
EXAMPLE: At the start of the experiment there are 125 cells. Three hours later there are 235 cells.
a.) What is the exponential growth formula? b.) How many cells are present after 5 hours? c.) How long
(rounded to the nearest hour) will it take the population to reach 442 cells?
We are going to use the exponential growth model above. First we need to identify what information we are
given. Since there are 125 cells at the start of the experiment we know Ao 125 . It says “after three hours”, so
we know that t = 3. It says that after 3 hours there are 235 cells, so f (t ) 235 since this is a current
population. The only thing we don’t know is the growth constant k. We need to find this. First we will
substitute the information we identified.
Part A.
235 125e 3k
We have substituted the information so now we need to solve for k.
235 125e 3k
125
125
First divide both sides by 125.
1.88 e 3k
We need to get rid of the e so we will take the natural log of both sides.
ln 1.88 ln e 3k
The ln e from the right side will cancel leaving us with 3k.
ln 1.88 3k
Now divide both sides by 3.
k
ln 1.88
0.2104
3
You can always round k to 4 decimal places.
Note, if you decide to use the whole number for k instead of rounding, this is okay. If you don’t round anything
then you might get slightly different answers than I got, but you will be close enough. You will still be correct
if you are within a few decimal places. We are ready to answer part a. This asks us for the growth function.
When we write this we DO NOT put anything in for t or f(t). Our answer to part a is f (t ) 125e 0.2104t
Now for part B. It asks for how many cells are present after 5 hours. We know that t = 5. We will put this into
our growth formula: f (5) 125e 0.2104( 5) . To solve this we need to first multiply the two numbers in the
exponent: f (5) 125e1.052 . To solve this we need to find out what e1.052 is. If you have a graphing calculator,
hit the second key and then the ln key to get e^. Then enter 1.052. If you have a scientific calculator you will
need to enter 1.052 first and then hit the second key and then ln. Either way you should get 2.8633721. Now
enter this into our formula: f (t ) 125(2.8633721) 358 cells. You can round it to the nearest cell.
Section 3.5 Notes Page 2
Now for Part C. The 442 is the f (t ) . We know everything except the t.
442 125e 0.2104t
First divide both sides by 125.
3.536 e 0.2104t
Take the natural log of both sides.
ln 3.536 ln e 0.2104t
The ln e cancels, leaving you with 0.2104t.
ln 3.536 0.2104t
Divide both sides by 0.2104
t 6 hours.
EXAMPLE: At the start of the experiment there are 1000 bacteria. After 4 hours the population doubled. What
is the exponential growth formula? How many bacteria are present after 6 hours?
Since the population doubled, we know that the new population will be 2000, therefore we know Ao 2000 .
The time is NOT 6 hours. We need to use the time it took the population to double, which is 4 hours, so t = 4.
At the start of the experiment there are 1000 bacteria, so f (t ) 1000 . We need to put this into the formula and
solve for k:
2000 1000e 4 k
First divide both sides by 1000.
2 e 4k
Now take the natural log of both sides.
ln 2 ln e 4 k
The ln e will cancel, leaving us with 4k on the right side.
ln 2 4k
Divide both sides by 4 and round to 4 decimal places.
k 0.1733
Now we need to know the growth formula for this problem. It is: f (t ) 1000e 0.1733t . We need to know the
population after 6 hours, so t = 6. We will put this into our growth formula: f (t ) 1000e 0.1733( 6) . To solve this
we need to first multiply the two numbers in the exponent: f (t ) 1000e1.0398 . To solve this we need to find out
what e1.0398 is first by using our calculator. You will get 2.8286512. Now enter this into our formula:
f (t ) 1000(2.8286512) 2828 bacteria.
EXAMPLE: Complete the table:
2000 pop
Projected 2011 Projected Growth
(in millions) Pop (millions)
Rate
53.4
0.0108
Let’s look at the table and define some given variables. Since we are going from 2000 to 2011, we know that
t 11 . We are given the growth rate, so we know k 0.0108 . We also know the initial population in 2000, so
A0 53.4 . We just need to find the final population. Let’s plug in our given values into the growth formula.
and use our calculator: f (t ) 53.4e 0.0108(11) 53.4e 0.1188 53.4(1.126...) 60.1 million.
Section 3.5 Notes Page 3
Half-life Problems
The next problems will focus on types of decay. We will first look at half-life problems. Half-life is the
amount of time it takes half of a substance to decay. The problems we will do will contain this information. We
will still use the same exponential model, except that half-life problems have a special formula to solve for k.
Decay constant for a half-life problem: k
ln 2
.
half life
EXAMPLE: The half-life of strontium-90 is 28 years. What is the decay function? How much of a 10 gram
sample is left after 11 years?
We will not use the 90 in this problem. This is just the name of the isotope. We are given the half-life is 28
ln 2
years, so we can put this into the formula to find our k: k
0.0248 . Our k value should be
28
negative since we are working with a decay problem. We know that the initial amount is 10 grams. Here is our
decay function: f (t ) 10e 0.0248t . Notice whenever we find the growth function we do not put anything in for t
or f(t).
To answer the second question, we will now put in a 11 for t: f (t ) 10e 0.0248(11) . First multiply the two
numbers in the exponent: f (t ) 10e 0.2728 . Then we do the e part on our calculator and you should now have:
f (t ) 10(0.76124502) 7.61 grams . You can round your answer to 2 decimal places.
EXAMPLE: The half-life of radium-226 is 1620 years. How much of a 2 gram sample is left after 1000 years?
We will not use the 226 in this problem. This is just the name of the isotope. We are given the half-life is 1620
ln 2
years, so we can put this into the formula to find our k: k
4.27868E 4 . Your scientific
1620
calculator may display what I put above. If it has an E next to it then this means the calculator is displaying the
answer in scientific notation. You need to move the decimal 4 places to the left, so k = -0.0004. We know that
the initial amount is 2 grams. Here is our decay function: f (t ) 2e 0.0004t .
To answer the second question, we will now put in a 1000 for t: f (t ) 2e 0.0004 (1000 ) . First multiply the two
numbers in the exponent: f (t ) 2e 0.4 . Then we do the e part on our calculator and you should now have:
f (t ) 2(0.67032004) 1.34 grams .
EXAMPLE: An artifact found contains approximately 1.23% of the original amount of carbon-14. The half
life of carbon-14 is 5600 years. Approximate the age of the artifact.
ln 2
0.0001 . We are not given an amount for Ao . We are only
5600
given that we have 1.23% of the original amount left over. If the original amount is Ao , then 1.23% of this is
0.0123 Ao . This will be our current population, so f (t ) 0.0123 Ao . Now we substitute:
We can find the growth constant: k
Section 3.5 Notes Page 4
0.0123 A0 Ao e
0.0001t
Divide both sides by N o .
0.0123 e 0.0001t
Now take the natural log of both sides
ln 0.0123 ln e 0.0001t
The ln e will cancel
ln 0.0123 0.0001t
Divide both sides by -0.0001.
t 43981.56 years
This will be the approximate age of the artifact.
Other types of decay problems include Newton’s Law of Cooling:
Newton’s Law of Cooling:
T C (To C )e kt
T = final temperature after cooling
C = temperature of the surrounding atmosphere
To = initial temperature before cooling
k = cooling constant
t = time
EXAMPLE: A pizza removed from the oven has a temperature of 450 F . It is left sitting in a room that has a
temperature of 70 F . After 5 minutes the temperature of the pizza is 300 F . What is the temperature of the
pizza after 20 minutes? When will the temperature of the pizza be 140 F ?
Let’s first define our variables. The initial temperature, To is 450 F . The temperature of the room is the
atmosphere’s temperature, so C = 70 F . The time, t, is 5 minutes. The final temperature after cooling is
300 F , so this is T . We have everything we need to solve for the cooling constant, k.
300 70 (450 70)e 5 k
300 70 380e 5 k
230 380e 5 k
0.6053 e 5 k
ln 0.6053 ln e 5 k
ln 0.6053 5k
k 0.1004
We substitute into the formula.
We simplify inside the parenthesis. Now subtract 70 from both sides.
Divide both sides by 380.
Take the natural log of both sides
The ln e will cancel.
Divide both sides by 5.
We have our k value rounded to 4 decimal places.
Our cooling formula is now: T 70 (450 70)e 0.1004t , or T 70 380e 0.1004t . Now to answer the first
question we need to put in 20 for t. You will get:
T
T
T
T
70 380e 0.1004( 20 )
70 380e 2.008
70 380(.13420755)
121 F
First simplify multiply in the exponent.
Now calculate e 2.008 .
Now simplify to get the answer.
Section 3.5 Notes Page 5
To answer the second question, now we are given the final temperature after cooling, T , which is 140 F .
140 70 380e 0.1004t
70 380e 0.1004t
0.1842 e 0.1004t
ln 0.1842 ln e 0.1004t
ln 0.1842 0.1004t
t 17 min
This time we need to solve for t. Subtract 70 from both sides.
Divide both sides by 380.
Take the natural log of both sides.
The ln e will cancel.
Divide both sides by -0.1004.
EXAMPLE: A thermometer reading 72 F is placed in a refrigerator where the temperature is a constant
38 F . After 2 minutes the thermometer reads 60 F . What will it read after 7 minutes? How long will it take
before thermometer reads 39 F ?
Let’s first define our variables. The initial temperature, To is 72 F . The temperature of the refrigerator is the
atmosphere’s temperature, so C = 38 F . The time, t, is 2 minutes. The final temperature after cooling is
60 F , so this is T . We have everything we need to solve for the cooling constant, k.
60 38 (72 38)e 2 k
60 38 34e 2 k
22 34e 5 k
0.6471 e 2 k
ln 0.6471 ln e 2 k
ln 0.6471 2k
k 0.2177
We substitute into the formula.
We simplify inside the parenthesis. Now subtract 38 from both sides.
Divide both sides by 34.
Take the natural log of both sides
The ln e will cancel.
Divide both sides by 2.
We have our k value rounded to 4 decimal places.
Our cooling formula is now: T 38 (72 38)e 0.2177 t , or T 38 34e 0.2177t . Now to answer the first
question we need to put in 7 for t. You will get:
T
T
T
T
38 34e 0.2177 ( 7 )
38 34e 1.5236
38 34(0.21792305)
45.41 F
First simplify multiply in the exponent.
Now calculate e 1.5236 .
Now simplify to get the answer.
To answer the second question, now we are given the final temperature after cooling, u (t ) , which is 140 F .
39 38 34e 0.2177t
1 34e 0.2177 t
0.0294 e 0.2177 t
ln 0.0294 ln e 02177t
ln 0.0294 0.2177t
t 16.2 min
This time we need to solve for t. Subtract 38 from both sides.
Divide both sides by 34.
Take the natural log of both sides.
The ln e will cancel.
Divide both sides by -0.2177.
We will be skipping the material covering Logistic Growth Models and Modeling Data.
Section 7.1 Notes Page 1
7.1 Systems of Linear Equations in Two Variables
In this section we will solve systems of linear equations, which can be solved using substitution and elimination
methods. These are basically equations of lines. You will have three cases with the answers. The first and
most common is that the lines will cross at a point and you will have a solution. The second situation is if the
lines are parallel. Parallel lines never cross, so there will be no solution. The third situation is if one line is on
top of another line. In this case all the points on one line are the same as all the points on the other line, so there
are infinite solutions.
EXAMPLE: Is the point (1, 2) a solution to the system:
3x y 5
x 5y 9
?
Since we have the point (1, 2) this means that x = 1 and y = 2. We will plug these into each equation. If we get
a true statement for each equation, then (1, 2) is the solution:
3(1) + (2) = 5
5=5
This is a true statement.
-(1) + 5(2) = 9
9=9
This is true also. Since we got a true statement for each equation, we know that (1, 2) is a
solution to this system. What this means is that these two lines will intersect at (1, 2).
EXAMPLE: Is the point (0, 1) a solution to the system:
2x 3y 3
5x 2 y 3
?
We know that x = 0 and y = 1 from our point. Once again we will plug these into our equations
-2(0) + 3(1) = 3
3=3
This is a true statement.
5(0) + 2(1) = 3
2=3
This is not true. Therefore we know that (0, 1) is not a solution to our system.
We will now look at two ways of getting a solution to a system: Substitution method and Elimination Method.
Substitution Method
In this method you will take one equation and solve for either x or y. Then you will substitute this into the other
equation.
EXAMPLE: Solve the system using the substitution method:
3x y 11
.
2 x 5 y 16
You can solve for any variable in any equation to begin this problem. You want to choose the variable that will
be the easiest to solve for. I will solve for y in the first equation since there is no number in front of it. If there
was a number in front of the y I would need to divide and that would cause fractions. So I will take the
Section 7.1 Notes Page 2
equation 3x – y =11 and solve for y. You will get y = 3x – 11. Now we need to substitute this into the other
equation for y:
2 x 5(3 x 11) 16
Now solve for x.
2 x 15 x 55 16
13 x 55 16
13 x 39
x3
Now that we have x we need to now solve for y. If we go back to our equation for y we have y = 3x – 11. We
can put a 3 in for x to get y: y = 3(3) – 11. So y = –2 . So for this problem our answer is x = 3 and y = -2.
This means these two lines will cross at the point (3, –2).
EXAMPLE: Solve the system using the substitution method:
3 x 2 y 9
2 x 4 y 6
.
Once again you want to solve for either x or y, whatever is the easiest. I will solve for x in the second equation.
If you take 2x – 4y = –6 and solve for x you will get: x = 2y – 3. Now we can put this into the first equation in
place of x:
3(2 y 3) 2 y 9
6 y 9 2 y 9
8 y 9 9
8y 0
y0
Now solve this for y.
Now that we have y we need to now solve for x. If we go back to our equation for y we have x = 2y – 3. We
can put a 0 in for x to get x: x = 2(0) – 3. So x = –3. So for this problem our answer is x = –3 and y = 0. This
means these two lines will cross at the point (–3, 0).
For both of those previous examples we were able to solve for x or y without fractions. Unfortunately this will
not always happen. Sometimes you might get a fraction when you solve for x or y. This is the limitation of the
substitution method. The elimination method will allow us to solve these type of problems easier.
Elimination Method
In this method we will multiply one or both of the equations by something so that when we add the equations
together one of the variables will be eliminated.
EXAMPLE: Solve the system using the elimination method:
2 x 3 y 4
5 x 6 y 37
.
In this problem we can choose to eliminate the x or y. Suppose we wanted to eliminate the y terms. I want to
multiply one or both equations by a number so that when I add the equations together the y terms will cancel. I
notice that if I multiply the top equation by 2 I will get a 6y. When I add 6y and -6y they will cancel. To find
out what to multiply the equations by you need to find the smallest number that both 3 and 6 will divide into. In
this case it is 6, so I want to get one equation to have a 6 and the other a -6.
Section 7.1 Notes Page 3
Suppose you wanted to cancel the x terms. You want to find the smallest number that both 2 and 5 divide into.
This number would be 10. I want one x term to be 10 and the other to be -10. If I wanted to eliminate the x
terms I would multiply the first equation by -5 and the second equation by 2. I will choose to eliminate the y
terms since I only need to multiply the top equation by 2:
4 x 6 y 8
5 x 6 y 37
Here I multiplied the whole first row by 2. Now I want to add these equations together.
9 x 45
x 5
When I added the equations together, the y terms were eliminated. Now solve for x.
To solve for y, we can put -5 in for x in either the first or second equation. I will put this into the first equation:
2(5) 3 y 4
10 3 y 4
3y 6
y2
Now solve for y.
So our answer is x 5 and y 2 . The point of intersection would be (–5, 2).
EXAMPLE: Solve the system using the elimination method:
3x 4 y 11
5 x 3 y 11
.
You can either eliminate the x or the y. This time I will eliminate the x terms. There is a 3 and a 5. The
smallest number that both 3 and 5 will divide into will be 15. In order for them to cancel I must multiply one of
the equations by a negative. It doesn’t matter which one you multiply by a negative. I will multiply the top
equation by a –5 and I will multiply the bottom equation by 3:
15 x 20 y 55
15 x 9 y 33
Now add the equations together:
11 y 22
y 2
Solve for y.
Now we need to put –2 in for y in either the top or bottom equation. I will use the bottom equation this time:
5 x 3(2) 11
5 x 6 11
5 x 5
x 1
Solve for x.
So our answer is x 1 and y 2 . The point of intersection would be (–1, –2).
Section 7.1 Notes Page 4
2
20
x y
3
3
EXAMPLE: Solve the system using any method:
.
5
2 x y 17
2
Before we use either method, let’s first get rid of the fractions. We can multiply the first equation by 3 since
this is the common denominator. We can also multiply the second equation by 2 to get rid of the equations.
2 x 3 y 20
4 x 5 y 34
4 x 6 y 40
4 x 5 y 34
The elimination method would be easier to solve. Multiply the top equation by –2:
Now add the equations together.
y 6
When you put this back to get x, you can use the non-fraction equations we found. I will use 2x – 3y = 20 and
put in –6 for y: 2 x 3(6) 20 . Solving this we get x = 1. So our answer is x = 1 and y = –6, or (1, –6).
EXAMPLE: Solve the system using any method:
3x y 4
.
6x 2 y 7
This could be solved by either substitution or elimination. I will solve it by substitution. We can solve the first
equation for y. You will get y 3x 4 . We want to substitute this into the bottom equation.
6 x 2(3 x 4) 7
6x 6x 8 7
87
We didn’t do anything wrong here. We ended up with a false statement, which means this system has no
solution. These lines are parallel, so they don’t cross.
EXAMPLE: Solve the system using any method:
3x 2 y 4
.
6 y 9 x 12
First let’s rearrange the variables so that the x and y terms line up:
3x 2 y 4
9 x 6 y 12
9 x 6 y 12
Now I can multiply the first equation by -3.
Now add the equations together. We get 0 = 0. We get a true statement. Both of these
9 x 6 y 12
lines are the same. In order to write our answer, we need set-builder, which is what the book tells us to do. We
will use the top equation since there are no common factors. Then we will put this into the set-builder notation
like the following: {( x, y ) | 3x 2 y 4} .
Section 7.1 Notes Page 5
EXAMPLE: A party mix is made by adding nuts that sell for $2.50 per kg to a cereal mixture that sells for $1
per kg. How much of each should be added to get 30 kg of a mix that will sell for $1.70 per kg.
For this problem we need to find our two equations and then use either substitution or elimination to solve it.
The best way to do these kind of problems is to set up a table:
Number of Price per Value
Kilograms Kilogram
Nuts
Cereal
Mixture
We will let x = the number of kg of nuts. Then y = the number of kg of cereal. We have 30 kg of the mixture.
We know the price per kg of nuts is $2.50. The price per kg of cereal is $1.00. The price per kg for the mixture
is $1.70. Let’s put all of this information on the table. For the value column I just multiplied the number of kg
by the price per kg.
Number of Price per
Value
Kilograms Kilogram
Nuts
x
2.50
2.5x
Cereal
y
1.00
y
Mixture
30
1.70
30(1.70) = 51
We will set up two equations. One will be for the number of kg and the other will be for the value. I know that
Nuts + Cereal = Mixture, so we have:
x y 30
2.5 x y 51
Now we have our system that we can solve by either substitution or elimination. I will solve this by
elimination. I can multiply the first row by -1:
x y 30
2.5 x y 51
Now add the equations together.
1.5 x 21
x 14 kg.
Solve for x.
We can put this into the first equation to find y: 14 y 30 . Solving this we get y = 16 kg. Therefore we
must mix 14 kg of nuts and 16 kg of cereal.
EXAMPLE: A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60%
sodium-iodine solution to obtain 50 ml of a 30% sodium-iodine solution. How many ml of each should be
mixed?
Once again a table would work best here. We will let x = 10% solution, y = 60% solution. We are given that
there must be 50ml of the 30% mixture. We will put this information on the table. To get the amount of
sodium-iodine we must multiply the number of ml by the percent written as a decimal.
Section 7.1 Notes Page 6
Number of
Percent
Amount of
ml
(as a decimal) Sodium-iodine
10% solution
x
0.1
0.1x
60% solution
y
0.6
0.6y
30% solution mixture
50
0.3
50(0.3) = 15
On of our equations will be from the number of ml. The other equation will come from the amount of sodiumiodine:
x y 50
0.1x 0.6 y 15
I will solve this one by substitution. Take the first equation and solve for y: y = 50 –x. Then put this into the
second equation and solve for x:
0.1x 0.6(50 x) 15
0.1x 30 0.6 x 15
0.5 x 30 15
0.5 x 15
x 30 ml
Solve for x.
Since y = 50 – x, then y = 50 – 30, so y = 20ml.
The chemist must mix 30ml of the 10% solution with 20ml of the 60% solution.
EXAMPLE: An airplane flying into a head wind travels the 1800-mile flying distance between two cities in 3.6
hours. On the return flight, the same distance is traveled in 3 hours. Find the ground speed of the plane and the
speed of the wind, assuming that both remain constant. [Ground speed is the speed of the plane if there were no
wind.]
Let x = the speed of the plane. Let y = speed of the wind. We will be using the formula distance equals rate
times time. With the wind the speed of the plane will by x + y. Into the wind the speed is x – y.
Rate Time Distance
With the wind x + y
3
1800
Into the wind x – y 3.6
1800
To get our equations, we will will use r t d . One equation will be with the wind and the other against the
wind:
3 x 3 y 1800
3.6 x 3.6 y 1800
x y 600
I will divide the first equation by 3 to get:
3.6 x 3.6 y 1800
Solving for y we get: y = 600 – x. Put this into the bottom equation.
3.6 x 3.6(600 x) 1800
Solving this we get x = 550 (plane’s speed). Then y = 50 (wind speed).
Section 7.4 Notes Page 1
7.4 Systems of Nonlinear Equations in Two Variables
In the previous section we looked at systems of two equations but all of the equations were linear. In this
section we will still be solving equations using the elimination and substitution method, even though the
equations will not be linear in style.
EXAMPLE: Solve the system:
y x5
x2 y2 5
You will need to decide in these problems which method is easier to use: elimination or substitution. In this
case substitution will be the easiest since one of the variables has already been solved for. We will substitute
the top equation into the bottom one. We will replace the y in the bottom equation and replace it with x – 5.
You will have: x 2 ( x 5) 2 5 . Now everything is in terms of a single variable. We need to expand this to:
x 2 ( x 2 10 x 25) 5 . Now distribute the negative: x 2 x 2 10 x 25 5 . We cancel the x squared terms
to get 10 x 25 5 . Solving this we get x = 3. We need to also find y so we can substitute a 3 for x in the first
equation: y = 3 – 5, so y = -2. As a point this would be written as (3, -2).
You are not expected to know how to graph the second
equation yet but I want to show it to you anyway so you
can see what it is you are actually doing. You are still
finding the intersection of two graphs. The first one is
a line and the second graph is called a hyperbola, which
we will focus on in chapter 9. Notice in this graph that
these intersect at (3, -2) which we found algebraically.
EXAMPLE: Solve the system:
xy 2
x 2y 0
We will use substitution for this one. Let’s solve the bottom equation for x. This way we won’t get fractions.
Solving for x in the bottom equation will give us x 2 y . Now we will substitute this into the top equation.
Replace the x with -2y. Then you will have the equation: (2 y ) y 2 . This will give us 2 y 2 2 . Divide
both sides by -2 to get y 2 1 . After taking the square root of both sides you will get y 1 .
Since we get two different values for y this means we will get two different values for x. We previously solve
for x and we got x 2 y . When y = 1 we get: x = -2(1) = -2. When y = -1 we get x = -2(-1) = 2. So we get
two points as our answer: (-2, 1) and (2, -1). Make sure you put the correct y value with each x value.
Section 7.4 Notes Page 2
x y 0
2
EXAMPLE: Solve the system:
2
2x 2 3y 2 5
It would be difficult to solve for x or y on this one so it will be easier to solve this one using elimination. I can
choose to eliminate either the x or the y. I will eliminate the y. All I need to do is multiply the first equation by
3x 2 3 y 2 0
3. I will get:
. Now add these equations together to get: 5 x 2 5 . Divide both sides by 5 to get
2
2
2x 3y 5
x 2 1 . We will get: x 1 . Now substitute these values back into either the first or second equation. Let’s
use the first equation. Let’s start with x = 1. We will put a 1 in for x in the first equation: 12 y 2 0 . You
will get y 2 1 , so y 1 . Now we will use x = -1. We will put a -1 in for x in the first equation:
12 y 2 0 .
You will get y 2 1 , so y 1 .
So we get four points as our solution: (1, 1), (1, -1), (-1, 1), (-1, -1).
EXAMPLE: Solve the system:
x 2 y 2 10
2 x 2 y 1
There is not a typo in this one. The second equation does not have the y as being squared. Again I think it will
be easier to solve this one using elimination. I will eliminate the x. I need to do is multiply the first equation by
2 x 2 2 y 2 20
-2. I will get:
. Now add these equations together to get: 2 y 2 y 21 . Notice I can’t
2
2 x y 1
combine the two terms on the left side of the equals sign since these are not like terms. I need to set this equal
to zero, so I will add 21 to both sides: 2 y 2 y 21 0 . Now I will multiply both sides by -1 to make it
easier to factor: 2 y 2 y 21 0 . Now factor it using whichever method is easiest. You will get:
7
(2 y 7)( y 3) 0 . Solving this will give us two answers for y: y and y 3 . We can put these into the
2
7
second equation to get x. Let’s start with y . We will put this into the second equation for y:
2
7
7
2 x 2 1 , or 2 x 2 1 . Now solve. We need to subtract seven-halves from both sides:
2
2
9
9
2 x 2 . We can multiply both sides of the equation by one-half to get: x 2 . If you try to take the
2
4
square root of both sides you will get an imaginary number. We do not include imaginary number solutions in
7
this section. We only want to find the actual place the graphs meet. So we know that y is not going to
2
2
give us an x. Let’s now try y = 3. We will put this into the second equation for y: 2x 3 1 . Add 3 to both
sides to get: 2x 2 2 . Divide both sides by 2 to get: x 2 1 . Then we get x 1 .
So we will get two places these graphs intersect: (1, 3) and (-1, 3).
Section 7.4 Notes Page 3
EXAMPLE: Solve the system:
y x
( x 3) 2 y 2 5
This is another one that should be solved by the substitution method since one of our variables is already solved
for. We will substitute x into the bottom equation for y: ( x 3) 2 ( x ) 2 5 . So now the equation
becomes: ( x 3) 2 x 5 . We will expand this: x 2 6 x 9 x 5 . We can add like terms: x 2 5 x 9 5 .
We need to set this equal to zero: x 2 5 x 4 0 . Factoring this we get: ( x 1)( x 4) 0 . We get two
answers for x: x = 1 and x = 4. We can put these into the first equation to get our corresponding answers for y.
When x = 1 we have y 1 , so y = -1. When x = 4 we have y 4 so y = -2. So our two points of
intersection will be (1, -1) and (4, -2).
NOTE: Suppose you used the second equation to solve for y
instead of the first equation. If x = 1 and you put this into the
second equation you would have gotten y 1 . You need to
test this with the first equation. For example on point we got
was (1, 1). If we test this with the first equation we would get
1 1 which is not a true statement so (1, 1) is not a solution.
Look at the graph. We should only have two points of intersection.
