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Table of Contents
Section 3.5 - Linear Functions as Mathematical Models
2
Section 4.5 - Graphing Linear Equations with Three Variables
5
Section 5.3 - X Intercepts and the Quadratic Formula
7
Section 6.10 - Properties of Logarithms
8
Section 7.3 - Special Products and Factoring
9
Answers
11
2
Section 3.5 - Linear Functions as Mathematical Models
In this chapter, you will learn to apply linear equations to real-world problems.
Objective: Given a situation in which two real-world variables are related by a straight-line graph, be
able to:
a. Sketch the graph.
b. Find the particular equation.
c. Use the equation to predict values of either variable
d. Figure out what the slope and intercepts tell you about the real world
Example:
As you drive home from the soccer game, the number of kilometers you are away from home
depends on the number of minutes you have been driving. Assume that distance varies
linearly with time. Suppose that you are 12 km from home when you have been driving for 7
Minutes, and 12 minutes later, you are 6 km from home.
a. Define variables for distance and time.
i. Since time is a constant, it would be advised to put time (in minutes) as x. That leaves
the kilometers away from home as y.
b. Find the particular equation expressing distance in terms of time.
i. Writing the existing data into pairs would give you (7,12) and (19,6). Since, as the x
value goes up, the y value decreases, it is obvious that you will have a negative slope.
19 − 7
1
Dividing 6−12 gives you -2which is the slope. Now, you use the slope-intercept form to
find the y-intercept.
1
𝑦 − 12 = − 2 (𝑦 − 7)
1
1
1
1
𝑦 = − 2 𝑦 + 3 2 + 12
1
y = − 2 𝑦 + 15 2, y-intercept is 152
c. Calculate your distance from home when you have been driving for 20, 25, and 30 minutes.
Time (in minutes)
Kilometers (away from home)
20
52 km away
25
3 km away
1
1
30
2
d. When were you 9 km from home?
i.
1
1
9 = − 2 𝑦 + 15 2
1
1
ii. - 52 = − 2 𝑦
km away
3
iii. x = 11
iv. 11 minutes
e. What does the distance-intercept equal, and what does it represent in the real world?
i.
1
Distance-intercept = y-intercept = 152
ii. It represents the distance away you were when you first started driving, i.e. how far
you started
f. What does the time-intercept equal, and what does it represent in the real world?
i. Time-intercept = x-intercept = 31
ii. It represents how long it takes to get home.
g. In what domain does this linear equation give you reasonable answers?
i. Within the positive domain
h. What are the units of the slope? Based on these units, what do you suppose the slope
represents in the real world? What is the significance of the fact that the slope is negative?
i. The units of the slope are kilometers away per minutes, which represents the speed.
Since it is negative, that means that the person is getting closer and closer to home.
Practice Problems:
3-5 Problem 1. (Easy) A music store sells 10 CDs for $20, and 40 CDs for $80. Assume that the number
of
dollars varies linearly with the number of CDs. Write the particular equation expressing
dollars in terms of CDs, and use it to predict the price for a box of 100 CDs. Sketch the graph.
3-5 Problem 2. (Medium) Suppose you own a car that is presently 40 months old. From an automobile
dealer’s “Blue Book”, you find that its present trade-in value is $4,800. From an old Blue Book
you
find that its trade-in value 12 months ago was $6,360. Assume that its trade in value
decreases
linearly with time.
a. Write the particular equation expressing trade-in value of your car as a function of its age in
months.
b. You plan to get rid of your car when its trade-in value drops to $2000. How much longer can
you keep the car?
c. By how many dollars does the car “depreciate” (decrease in value) each month? What part of
the mathematical model tells you this?
d. When do you predict that the car will be worthless? What part of the mathematical model
tells you this?
e. According to our linear model, what was the trade in value of your car when it was new?
f. The car actually cost $11,235 when it was new. How do you explain the difference between this
number and the answer to part (e)?
g. Sketch the graph of this function.
4
3-5 Problem 3: (Hard) If the constant B in y = mx + b equals zero, then y is said to vary directly with x.
The
amount of pancake batter you must make varies directly with the number of people who will
come
to breakfast. Suppose that it take 9 cups of batter to serve 15 people.
a. Define the variables and write the equation expressing number of cups in terms of people.
b. How many cups must you prepare for 47 people?
c. About how many people can you serve with 12 cups of batter? 198 cups of batter?
d. Sketch the graph of this function. Through what special point does the graph of a direct
variation function go?
5
Section 4.5 - Graphing Linear Equations with Three Variables
You probably have seen linear equations with two variables, such as 3x+8y = 12. To find the intercepts
of this equation, set the other variable to zero, and solve for the remaining variable. An intercept is
the value of a variable when the other variable(s) is/are set to zero. By this method, the x-intercept
of this graph would be 4, and the y-intercept would be 1.5.
Now, we will explore equations with three variables, like 3x+8y+5z = 30. To find each intercept, let the
other two variables equal zero, thus letting you solve for one variable. Repeat for each variable, and
graph each intercept onto a three-dimensional graph, connecting each of the points.
The graph of a linear equation with three variables is a plane in space. The graph includes the xy, yz,
and xz traces, with are obtained by setting the third variables equal to zero, and finding the ordered
pair of the two variables’ intercepts. For example, for 3x+8y+5z = 30, the xy trace would be (10, 3.75),
because the x-intercept is 10, and the y-intercept is 3.75.
To find each intercept of 3x+8y+5z = 30
X: 3x+8(0)+5(0) = 30
3x=30
x-int=10
Y: 3(0)+8y+5(0) = 30
8y=30
y-int=3.75
Z: 3(0)+8(0)+5z = 30
5z=30
z-int=6
Graph on next page
6
Problems: Find each intercept and graph on an xyz plane.
1.
2x+4y+3z=18
2. 4x-2y-z=14
3. x+y-z=8
7
Section 5.3 - X Intercepts and the Quadratic Formula
In this chapter you will be able to find the vertex, x intercept, y intercept, and the symmetric point of
a quadratic equation and be able to graph it as well.
These kind of problems can be useful in real world situations because for example if you have an
quadratic equation that tells you at what speed you burn a certain amount of fuel, you can use what
you know to find the vertex to find the optimal speed to drive at for long distances.
Problem 1: x2 +2x-4, Find the vertex, x intercept, y intercept, and the symmetric point
Problem 2: 2x2 +4x-1, Find the vertex, x intercept, y intercept, and the symmetric point
Example:
a.
First recognize your three terms, a: coefficient of x2 b: coefficient of x and c: the constant.
b
.The x coordinate vertex of a parabola can be written as -b/2a. To find the y coordinate plug in
-b/2a for x and solve for y.
c.
To find the x intercept you must find what number x will be if y is zero to do that you have to
use the quadratic formula, x= -b+or-√𝑦2 − 4𝑦𝑦 /(2a) , you Should get two solutions.
d.
The y intercept is when x=0 so it is C and the point of symmetry is the point where the graph
hits the y coordinate of the y intercept for the second time.
8
Section 6.10 - Properties of Logarithms
In this chapter, you will learn about three of the different properties of logarithms:
1.
Logarithm of a Product
a. logb (xy) = logb x + logb y
b. In words: “The log of a product equals the sum of the logs of the two factors.”
2. Logarithm of a Quotient
a. logb (x/y) = logb x - logb y
b. In words: “The log of a quotient equals the log of the numerator minus the log of the
denominator.”
3. Logarithm of a Power
a. logb (xn) = n logb x
b. In words: “The log of a power equals the exponent times the log of the base.”
These will become very useful as you encounter more complicated logarithms, and it is important that
you memorize them.
Example: Express log2 17 - 2 log2 5 as a single logarithm of a single argument.
1. = log2 17 - log2 52
a. Use the Logarithm of a Power property to write 2 log2 5 as log252
2. = log2 17/25
a. Use the Logarithm of a Quotient to write log2 17 - log2 52 as log2 17/25
3. = log2 0.68
Problem 1: Find the given logarithm without using the log key on your calculator. Assume that log 2 =
0.301, log 3 = 0.477, and log 5 = 0.699.
1. log 30
2. log (3 x 1012)
Problem 2: Write the expression as a single logarithm of a single argument.
1. log 9 + log 8 - log 10
2. 4 log8 8 - log8 6
9
Section 7.3 - Special Products and Factoring
Objective: In this chapter you will learn how to distribute polynomials and how to factor and
conjugate bi/trinomials.
This has modern day and real world relevance because for example if
Problem 1: (2x2-12x+16)(2x2-14x+2)
Example:
a.
To multiply two polynomials you have to make sure every term in one polynomial is
multiplied by each term in the other polynomial
b.
For example, you could move left to right and multiply the first term in polynomial one by
each term in polynomial term.
c.
Then the second term in polynomial one by every term in polynomial two and finally you
would multiply the third term in polynomial one by the every term in polynomial two.
d.
Then after you have what should be 9 terms you simplify it. You can also multiply it right to
left so it is vice versa.
Problem 2: 10x2-52x+21, factor the trinomial completely
Example
There are two ways to factor a quadratic function ax2+bx+c. One is by guessing and checking, one is
by using the quadratic formula. The simpler the problem the more sense it makes to use the first
method. However if it is more complex I would suggest using the Quadratic Formula.
Method 1.
a.
For this style you would first recognize your three terms, A: coefficient of x2 B: coefficient of x
and C: the constant.
b.
When the answer is factored you have to look at what two numbers A can be split into. For
example, if the number is 5 the only options are 5 and 1. However if A is 48 then it could be 12 and 4,
16 and 3, etc. So you have to plan out for each scenario.
c.
Once you have all your guesses on what a could be possible split into write it out as for
example… (3x+?)(4x+?) or (2x+?)(6x+?). Now those were examples that would lead to 12x A being x of
course.
10
d.
Then guess numbers where the question mark would be that multiplied together would make
C. Then multiply with the other two numbers that you guess it could be ex. 2x and 6x, 4x and 3x. Keep
on guessing until you get the the value of B. Don't forget to factor out the LCM out of your answer.
e.
If you get (3x+2)(4x+7) as you answer your solutions would be -7/4 and -⅔
because you write it as the denominator as the x coefficient and the numerator as the
constant with the sign flipped.
Method 2:
a.
For this style you would first recognize your three terms, a: coefficient of x2 b: coefficient of x
and c: the constant.
b.
You would then plug in these values into the quadratic formula: x= -b+or-√𝑦2 − 4𝑦𝑦 /(2a) When
you are solving it and you find that the √𝑦2 − 4𝑦𝑦 is either not a whole number or contains i then
there is no solution to the problem.
c.
However if you do get a positive integer when you solve √𝑦2 − 4𝑦𝑦 then you solve it out. You
should have two real numbers.
d.
Once you have them you write it out as the denominator as the x coefficient and the
numerator as the constant. For example 35/46 and 12/70 as your solution can be written as,
(46x-35)(70x-12).
e.
Don't forget to factor out the LCM out of your answer
11
Answers:
3.5
Problem 1
y = 2x
100 CDs = $200
Problem 2
a. y = -130x + 10000
i. x = age in months
ii. y = trade-in price
b. 2000 = -130x + 10000
i. x = 61.54 months
ii. In around 21.54 months
c. -130 dollars, slope
d. when y = 0, x-intercept
i. 76.923 months after purchase
e. 10,000, y-intercept
f. Taxes, Car fees, Contract
Problem 3
a. x = number of people, y = cups of batter needed
12
i. Equation: y = ⅗ x
b. y = ⅗ (47) = 20.8 cups of batter.
c. 12 = ⅗ x, x = 20 people
i. 198 = ⅗ x, x = 330 people
d. Passes through origin (0,0)
4.5
1.
13
2.
3.
5.3
Problem 1. Vertex =(-1,-5)
`
Y int=-4
X int =-1 + or - (20^.5)/2
Point of Symmetry= -2
14
Problem 2
Vertex= (1,5)
X int =-1 + or - (2^.5)/2
Y int= -1
Point of Symmetry= (-2,-1)
15
6-10
Problem 1: Find the given logarithm without using the log key on your calculator. Assume that log 2 =
0.301, log 3 = 0.477, and log 5 = 0.699.
1. log (3 x 1012)
Solution:
1. log (3 x 1012)
2. = log 3 + log 1012
3. = log 3 + 12 log 10
4. = log 3 + 12 log 5 + 12 log 2
5. = 0.477 + 12 (0.699) + 12 (.301)
6. = 12.477
6-10 Problem 2: Write the expression as a single logarithm of a single argument.
1. 4 log8 8 - log8 6
Solution:
1. 4 log8 3 - 2 log8 9 = log8 34 - log8 92 = log8 81/81 = log8 1 = 0
7-3
Problem 1: 4x4-52x3 +202x2 -448x +32
Problem 2. (5x-21)2(x-1)