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The Daily Parabola
Featured story: ​
Inside Turn Left
Industries “Graphing Functions changed
my life” - Ratchel the Satchel
Included in this Issue:​
Variables!
​
Expressions! Functions! Krabby Patties!
and MORE!!
Bonus!! Special Real World
Problems!!! Super Fun!!
In next week’s issue: ​
Linear Programming, Polynomials, and Inverses of
Fractions!
by Jamie Kesten, Jia Seow, Philip Carey, Hannah Yang
Period 1
1-3 Variables and Expressions
Goal​
: Utilize Order of Operations to evaluate variable expressions
Definition​
:
Variable: A variable is a letter which stands for an unspecified number from a given set
Agreement​
: Unless otherwise specified, the domain of a variable will be assumed to be the
set of all real numbers
Definition:
Expression: An expression is a collection of variables and constants connected by operation
signs ( +, -, x, /, etc.) which stands for a number.
Definitions:
Subtraction: x - y means x + (-y)
Division: x/y means x * 1/y
Exponentiation: x^n means n, x’s multiplied together. For example, x³ means x*x*x
Agreement:
Order of Operations:
1. Do any operations inside parentheses first
2. Do any exponentiations next
3. Do multiplication and division in the order in which they occur, from left to right
4. Do addition and subtraction last, in the order in which they occur, from left to right
Definition:
Absolute Value: Distance between the number x and the origin of the number line
| x | = x if x is positive (or 0)
| x | = -x if x is negative
Key Points:
1. When doing a calculation, substitute the same value of x everywhere it appears
2. The answer you get correlates with the order of operations, so follow Order of
Operations!
Practice Problem: ​
6x² - 3x(x - 3y) + 3y(3y + x) - 3x²
Step 1: Factoring: 6x² - 3x² + 9xy + 9y² + 3xy - 3x²
Step 2: Combining Like Terms: 12xy + 9y²
Final Answer: 12xy + 9y²
Inside Turn Left Industries
Turn Left Industries is a company that fights for the normalization of
making 3 left turns to make a right turn. It was started by a high
school girl named Ratchel the Satchel who had to bike to school
everyday, she only knew how to make left turns which often led her to
go the wrong way and fall of her bike. Her company has collected lots
of data on the need for people to make 3 left turns instead of a right
turn and she has presented her data to the public, however it did not
have the expected impact that Ratchel hoped for. After viewing her
presentation, she has decided that her presentation was ineffective due to the fact that she
presented her information in just numbers instead of graphs. She decided to head back to Los
Altos High School and learn how to graph a function, and her new presentation was a success!
Ratchel the Satchel says, “Graphing functions changed my life! It really allows people to
finally see my vision!”
2-2 Graphs of Functions
Overview: Learning to graph a function when given the equation and understanding domain
and range.
Definition:
Domain: The domain of a function is the set of values of the independent variable.
Definition:
Range: The range of a function is the set of values of the dependent variable corresponding to
all values of the independent variable in the domain.
Method:
2​
1. Find the vertex using the equation −b
+ bx + c = y, which finds the
2a where ax​
x-coordinate of the vertex.
2. Plug in values for X and plot points
Tips:
1. Try to isolate the variable firsts
2​
2. If a in ax​
+ bx + c = y is positive the function will open upwards, and if a is negative
the function will open downwards.
Example Problems:
Plot the graph of the function and tell the range:
2​
1. y = 0.4x​
, domain = {real numbers}
Step 1: Find the vertex using
−b
2a
:
−0
2(.4)
2​
Step 2: Plug 0 in for x:
y = 0.4(0)​
Step 3: Plug in values for x:
x
y
=0
Vertex = (0,0)
2
1.6
1
.4
-1
-.4
Step 4: Plot points
Step 5: Find the range:
y≥ 0
Real World Application:
Fall Ing is trying to find how long it will take for him to get to work. He works at Turn Left
Factory which is owned by Ratchel the Satchel. He has 30 minutes to get to work. For every
9 cars on the road it will take him and extra 3 minutes to get work. Assume that it will take
him 15 minutes to get to work if there are no cars on the road.
a. Write the equation for how long it will take him to get to work, using the variables t
for the time it will take to get to work and c for the number of cars on the road.
b. Graph the function
c. If there are 130 cars on the road how long, will Fall Ing make it to work on time?
Solutions :
a. t = 13 c + 15
b.
c. Plug in 130 for c: t =
1 (130)+ 15
3
t = 43 13 + 15
t = 58 13
58 13 > 30
No he will be late.
Additional Practice:
1. Plot the graph of the function:
a. y = x+4, domain = {positive numbers}
b. y =
−1
x
, domain = {all real numbers}
4-6 Systems of Linear Equations with Three or More Variables
Goal: Be able to find the single ordered triple that satisfies a system of three linear equations
with three variable.
Key Points:
1. Eliminate one variable from the original system of three equations to get a system of
two equations with two variables
2. Solve for one of these variables through linear combination
3. Plug the value into the 2-system equation to find the value of the other variable
4. Use the 2 values to find the value of the third variable in an original equation
Ex: 7x + 5y - 3z = 16
3x - 5y + 2z = -8
5x + 3y - 7z = 0
Step 1: Get a 2 variable System through Linear Combination:
7x + 5y -3z = 16
3x - 5y + 2z = -8
+-----------------------
10x - z = 8
(3x - 5y + 2z = -8) x3 -------------> 9x - 15y + 6z = -24
(5x + 3y - 7z = 0) x5 --------------> 25x + 15y - 35z = 0
+---------------------34x - 29z = -24
Step 2: Solve the 2 variable system
(10x - z = 8) 29 -----------------> 290x -29z = 232
(34x - 29z = -24) -1 ------------> -34x + 29z = 24
+------------------256x = 256 -----> x = 1
Step 3: Plug value x into the 2 system equation to find z
10(1) - z = 8
z=2
Step 4: Plug in both values back into the 3 system equation to find y
7(1) + 5y - 3(2) = 16
5y = 15
y=3
Step 5: Give values of x, y, and z
{1,3,2}
Real World Applications:
The movie theater sold a total of 8500 movie tickets. Their earnings were $64,600. Tickets
can be bought in 3 ways: matinee admission costs $5, student admission is $6 all day, and
regular admissions are $8.50. How many of each type of ticket was sold if twice as many
student tickets were sold as matinee tickets?
Step 1: Define your variables
m = number of matinee tickets sold
s = number of student tickets sold
r = number of regular tickets sold
Step 2: Create your equations
r + s + m = 8500
8.5r + 6s + m = 64600
s = 2m
Step 3: Solve for a 2 variable system
(r + s + m = 8500) -6 ----> -6r - 6s - 6m = -51000
8.5 + 6s + 5m = 64600 ---> 8.5r + 6s + 5m = 64600
+------------------------2.5r - m = 13600
r + s + m = 8500
-s + 2m = 0
+------------------r + 3m = 8500
Step 4: Solve the 2 variable system:
(2.5r - m = 13600) 3 ------> 7.5r - 3m = 40800
r + 3m = 8500 --------------> r + 3m = 8500
+---------------------8.5r = 49300
r = 5800
Step 5: Plug ‘r’ into the 2 variable equation
5800 + 3m = 8500
3m = 2700
m = 900
Step 6: Plug in ‘r’ and ‘m’ into the 3 variable equation
5800 + s + 900 = 8500
s = 1800
Step 7: Write a sentence explaining your answer
The amount of tickets sold were 900 for the matinee, 1800 student tickets, and 5800 regular
tickets.
5-7 Quadratic & Linear Functions as Mathematical Models
Goal: find a mathematical model, quadratic or linear, for a real world situation
Find the particular equation of a quadratic equation from points on a graph to solve real
world problems.
The Krabby Patty has 3 prices for drinks.
Thimble (2oz) 1.35
Sip (8oz) 3.50
Taste (24oz) 4.25
a. Assume that the price is a quadratic function of the number of ounces. Write the
particular equation expressing price in terms of ounces.
b. If the Krabby Patty made a drink called Small that was 32 oz what would you predict
the price would be? Does this answer make sense?
c. Suppose they wanted another drink that cost 3.55. How many ounces would it be?
d. In what domain do you think this quadratic function will give reasonable values for
ounces?
e. What does c represent in this equation?
a. Let: o=the number of ounces
p=the price ​
in cents
2​
General equation: p=ax​
+bx+c
​
Ordered pairs: (2, 135), (8, 350), (24, 425)
Substituting the three ordered pairs:
4a+2b+c=135
64a+8b+c=350
576a+24b+c=425
Solving this equation we get:
a=-1.4157
b=49.9905 c=40.6818
2​
-1.4157o​
+49.9905o+40.6818=p
b. substitute 32 in for ounces
we get 190.701 which means that it would cost 1.91
No because it is cheaper than the drinks smaller than it.
c. we put 355 in for p and solve to get 27 ounces
d. x<18
e. the price of the cup and service charge, does not include price of liquid
Practice Problems:
1. When a volleyball is served, it goes up into the air, reaches a maximum height, then
goes back down. This is a quadratic function.
a. let t=number of seconds
b. let d= distance above the ground
c. The ball was served at 8 feet above the ground. One second later the ball is 14
feet above the ground. 2 seconds later the ball is 10 feet up. What is the
equation?
d. what does the vertex represent?
e. A volleyball net is 7 feet 4 inches. It is 1.5 seconds away from the endline. Will
the ball go over the net?
f. The length of a volleyball is 59 feet long. It takes 2.4 seconds for the ball to get
there Assume that the volleyball was served at the endline exactly. Will the
ball land in? (lines are in)
6-4 Exponentiation for Rational Exponents
Goal: Learn the definitions of exponentiation for rational exponents by evaluating powers and
by simplifying expressions involving rational expressions.
Exponentiation for Negative Exponents: ​
the
xa
xb
If
and a = b, then
xa
xb
expression x−n =
1
xn
= 0.
Exponentiation for 0 Exponent: ​
x⁰=
1, provided x =/ 0
Exponentiation for Reciprocal Exponents:
1
xn
= √n x
Definitions of the expression √n x :
n - root index
x - radicand
√ - radical sign
Exponentiation for Fractional Exponents (where a
​​
and ​
b​
are integers):
x b = ( √x a ) a
b
Sample Problems:
90x 19y 27
​ith positive exponents.
w
270x 33y 23
73
2. Simplify​
: 7 0 1. Simplify​
:
3. Express
4
√z 3 with a fractional exponent.
4. Am I. Wright attempted to simplify the product of
27x 39y 26
3x 3y 13
13 2
= 24x 1 y 27x 39y 26
3x 3y 13
:
= 24x13y 2
Is her answer correct? If not, show the steps to how you got the correct solution.
Answers:
1-3:
1. 12xy + 9y²
2-2:
1. a.
b.
4-6:
1. x = 1, y = 3, z = 2
2. m = 900, s = 1800, r = 5800
5-7:
2​
c. -5x​
+11x+8
d. the peak of the serve
e. yes
f. no
6-4:
1.
2.
3.
90x 19y 27
(2•3 2•5)y 4
33 23 =
(2•3 3•5)x 14
270x y 3
3
70 = 1. 7 7 0 = 71 = 73
4x 4
3x
3
√z 3x = z 4x 4 = z4x 3 4. Her answer is incorrect.
correct solution:
27x 39y 26
3x 3y 13
36 13
= 9x 1y = 9x36y 13
4
y = 3x 14
The ​
Transcendental​
Magazine
by ​
Daniel Hu​
,​
Dana Popovsky​
,​
Eric Warmoth​
, and​
Ananya Subramanian (Period 1)
Feature Article: ​
The Adventures of Mr. Nguyen: Modeling linear functions
“So good, it ​
transcends​
my expectations!” ­ Anonymous “This is the ​
e​
πtome of a magazine!” ­ Anonymous “100% an A+ magazine. I might even give extra credit.” ­ Mr. Nguyen 1 3.5 ­ The Adventures of Mr. Nguyen: Modeling linear functions In this section, our goal is to sketch the graph, find the equation, predict future values, and interpret the slope and intercepts of a real­world linear function given certain information. Concept: If you are given two values of x and two corresponding values of y, the best way to find the equation is usually to use point­slope form, namely y − y1 = m(x − x1) , where m is the slope and y −y
(x​
is a point on the line. To find the slope, ​
m​
, simply do x22−x11 , substituting the two points 1, y
​ ​
1)
​​
given in for ​
(x1​
, y1​
)​
and ​
(x2​
, y2​
)​
. Then, substitute one of the points given in for x​
and y​
in the ​
​
​
​
1​
1​
original point­slope equation. To graph the equation, plot the two points given on the coordinate plane and connect them with a line. If you are told something like “you start with ​
a​
pigs and you gain ​
b​
pigs per hour,” and you are asked to find the equation of such a description, then the slope would be ​
b​
, and the y­intercept would be ​
(0, a)​
. To get a better understanding of this, we shall turn to a couple of examples. Example 1 Mr. Nguyen, the superhero math teacher who never gets tired, was flying through space. His distance from Earth, in km, is a linear function of the amount of time he has spent flying, in hours. After one hour of flying, Mr. Nguyen ends up 2780 km from Earth. After three hours, he is 4300 km from Earth. (a) Define variables for distance and time, and find the particular equation of the model expressing distance in terms of time. Solution: We let d = distance from Earth in km and t = time spent flying. To find the equation of this, use point­slope form. The slope is 4300 − 2780
= 760 , and we choose the point 3 − 1
(1, 2780) because we can. Thus, the equation is y − 2780 = 760(x − 1) . (b) Predict how far Mr. Nguyen is from Earth after 5 hours and 10 hours. Solution: Plugging the two values into our equation, we get y = 760(5 − 1) + 2780 = 5820 and y = 760(10 − 1) + 2780 = 9620 . This means that after 5 hours, Mr. Nguyen is 5820 km from Earth, and after 10 hours, Mr. Nguyen is 9620 km from Earth. (c) If Mr. Nguyen is 2,2540 km from Earth, then for how long has he been flying? Solution: Instead of substituting for x, we can substitute for y to solve for x. Letting y = 2254, we get 22540 − 2780 = 760(x − 1) ⇒ 19760 = 760(x − 1) ⇒ x − 1 = 26 ⇒ x = 27 , so if Mr. Nguyen is 2,2540 km from Earth, he has flown for 27 hours. Note: I use the notation ⇒ to represent going through the steps; do not confuse it with an “if, then” statement. Others will probably use different notation. (d) What does the slope represent in the real world? 2 Solution: The slope represents how much the distance, in km, increases for every hour, or km/h . This is one of the ways to express speed, so that means that the slope represents speed in this case. Therefore, we can interpret the graph to say that Mr. Nguyen flies at a constant speed of 760 km/h. (e) What does the distance­intercept represent in this case? Solution: The distance­intercept is when the time is 0, so the distance­intercept represents how far away from Earth he was originally. Letting x = 0 in our equation, we can find y to be 2020, so the distance­intercept is (0, 2020). (f) Draw the graph of this equation using an appropriate domain, labeling all intercepts and at least two points on the graph. (g) Is a linear model reasonable for this situation? Solution: No, because everybody has to get tired eventually, and people still need to use the facilities at least a little bit. Therefore, a constant trend does not work. Unless, of course, Mr. Nguyen is a superhero that never gets tired … oh, wait … Example 2 Mr. Nguyen is back for his superhero business, but now he is charging precious money for his heroic deeds. If you want to ask him for help, he charges a $50 service fee and $20 an hour. 3 (a) Defining variables for money and time, find the particular equation of the model expressing cost (dollars) in terms of time (hours). Solution: The cost­intercept of this equation is when the time is 0, which is given to be 50, and the slope, which is cost per hour, is 20. Therefore, letting c being the cost in dollars and t being the time in hours, the equation of this model is c = 20t + 50 . (b) How much will Mr. Nguyen charge you for saving a kitten from a tree, taking him 10 hours? Solution: c = 20 * 10 + 50 = $250 (c) If Mr. Nguyen charges you $850 for stopping a gigantic asteroid from hitting Earth and devastating humanity, then how long did it take him? If Mr. Nguyen charges you $1250 for chasing away a mouse from your house, then how long did it take him? Solution: 850 = 20t + 50 ⇒ 20t = 800 ⇒ t = 40 Therefore, if he charges you $850, it took him 40 hours to stop the asteroid. 1250 = 20t + 50 ⇒ 20t = 1200 ⇒ t = 60 Therefore, if he charges you $1250, it took him 60 hours to chase away the mouse. (d) Draw the graph of this equation using an appropriate domain, labeling all intercepts and at least two points on the graph. 4 Problems Problem 1 In the examples, we said that Mr. Nguyen took 60 hours to chase away the mouse, but the mouse is on the loose again. The mouse is 50 km away from the White House and approaching quickly, while Mr. Nguyen is 80 km away from the White House. After three minutes of chasing, the mouse was 35 km away from the White House and Mr. Nguyen was 55 km away. If the mouse reaches the White House before Mr. Nguyen can stop him, then the entire world is doomed! Assume that distance from the White House is a linear function of time. (a) Defining variables for distance and time, find ​
two​
equations for the model, one expressing the mouse’s distance from the White House (km), and the other expressing Mr. Nguyen’s distance from the White House (km), in terms of time after Mr. Nguyen started flying (minutes). (b) Five minutes after Mr. Nguyen started flying, how far away is the mouse and how far away is Mr. Nguyen from the White House? Express in terms of kilometers. (c) If the mouse is 15 km from the White House, then how far away is Mr. Nguyen? (d) If Mr. Nguyen is 35 km from the White House, then how long has he been flying? (e) Sketch the graph of both lines. Label all intercepts and at least ​
two​
points on the graph. (f) What do the slopes of the lines represent about the moving objects? (g) How much faster is Mr. Nguyen going than the mouse? (h) Will he catch up in time to save the world from destruction? Why or why not? Real World Application Distance vs. Time graphs allows us to apply data for modeling speed. Distance vs. Time is often used to calculate speed and acceleration, an important thing to pay attention to especially when beginning to drive a car for the first time. Knowing the speed and acceleration of an object can help one calculate force or energy. For example, a linear graph shows a constant speed because it increases at a constant rate. Distance vs. Time can also help you calculate average speed by taking the individual values of the abscissa and the ordinate at a certain point. Speed vs. Time graphs allow us to use mathematical models to answer questions in real life to figure out the distance an object will travel. It is also helpful in everyday life for simple things like creating a plan; for example, using speed vs. time graphs, you can easily determine how much time you need to complete a certain task. Maybe you are calculating how many problems you have time to solve in the hour that you have set aside for studying based on your problem­solving speed. Not only do speed vs. time graphs make you seem to everyone like you know your algebra, they can help you organize your schedules more effectively too! 5 4.5 ­ Follow the Trace: Linear Equations with Three or More Variables In this section, our goal is to determine what the graph of a linear equation with three variables looks like, and be able to sketch the graph. Concept: When dealing with an equation that has three variables, first place all the variables on one side of the equation, and the constant on the other, example: 9x+7y+3z=20. Any solutions must have three answers, and the order in your answer should determine what value they represent, example: (5,6,3), means that x=5, y=6, and z=3. This is called an ​
ordered triple​
. Example 1 The graph of a linear function is a plane in space. Using the equation of 2x + 3y + 4z = 12, The ​
xy­trace​
is the set of ordered pairs (x,y) that are obtained by making z=0. Similarly, the xz­trace is (x,z) when y=0, and the yz­trace is (y,z) when x=0. 6 The ​
x­intercept​
is the value of x when y=0 and z=0. Similarly, the y­intercept and z­intercept are those values when the other two variables equal 0. When drawing a trace, find the 3 intercepts, and those will be the 3 points of a triangle formed by the 3 traces. From the x­intercept to the y­intercept will be the xy­trace, and the same for the other traces and intercepts. Example 2 Again using the equation, 2x + 3y + 4z = 12, graphing the three traces and intercepts on an xyz­plane: Problems Problem 1 Sketch a graph of each of the following equations by drawing its three traces. a. 4x ­ 3y + 6z = 12 b. 6x ­ 9y + 18z = 18 c. 4x + 16y ­ 8z = 16 d. x ­ y + z = 2 Problem 2 7 For the equation 6w ­ 3x + 9y + 12z = 36 a. Find the four intercepts b. Find the equation of the xyz­trace 5.3 ­ X Marks the Spot: The Treasure of the Quadratic Formula In this section, our goal is to determine the x­intercept(s) and vertex of a quadratic equation using the quadratic formula. Concept: 2​
The general form of a quadratic equation is y = ax​
+ bx + c, where a, b, and c are real numbers and a ≠ 0. Determining the X­intercept Finding the y­intercept is easy; plug in 0 for x and solve for y. But in order to determine the x­intercept, you must first set y equal to 0 and from there either factor the polynomial and solve or solve using the ​
quadratic formula​
. The quadratic formula is: −b±√b2−4ac
x=
2a
By plugging a, b, and c into this equation, you can now solve for x! There’s just one problem…what if there are no x­intercepts?? Let’s say you have this equation: x = 4±√8−20 You’re now stuck, because as we all know, the square root of ­20 is not a real number! 2​
We call the quantity under the radical sign, b​
­ 4ac, the ​
discriminant​
. If the discriminant is negative, the solutions are imaginary, If the discriminant is positive, the solutions are real numbers, and If the discriminant is a perfect square, and if a, b, and c are rational, the solutions are rational. Determining the Vertex Once you have a quadratic equation, determining the vertex is simple. The x­coordinate of the vertex is b
− 2a
The ​
axis of symmetry​
is located at this coordinate. Using this, plug this value into the original equation for x and solve for y to find the y­coordinate of the vertex! Example 1 8 Determine whether or not the following equation produces real solutions. If so, find the vertex, x­intercepts, axis of symmetry, and symmetric point of the equation. 2​
y = x​
+ 12x + 13 2​
Solution: Use the discriminant: ​
12​
­ 4(1)(13) = 92, ​
which is a positive number. Thus we know the equation produces real solutions. ● Finding the vertex: We know the x­coordinate of the vertex is ​
­b/2a, ​
or​
­12/2, ​
which is ­6. Plugging in 6 as x in the original equation, we get ​
y = ­23. ​
So, the vertex is at ​
(­6, ­23)​
. ● This means the axis of symmetry is at ​
x = ­6​
. When x = 0, y = 13. This means the point of symmetry, the point symmetrical to the y­intercept, is ​
(­12, 13) ● To find the x­intercepts, substitute 0 for y and solve for x. x=
x
−b±√b2−4ac
2a
−12±√122−4(1)(13)
=
2(1)
−12±√92 =
2
= 6 + √ 23, 6 ­ √ 23 x
x
x ≈ 10.80, 1.20 Problems Problem 1 Determine whether or not the following equations produce real solutions. If so, find the x­intercepts of the equation. 2​
1. y = 3x​
­ 10x + 6 2​
2. y = ­x​
+ 4x ­ 1 2​
3. y = 10x​
+ 19x +12 Problem 2 Determine the vertex, y­intercept, and point of symmetry of the above problems and use them to graph the equations. Problem 3 Solve for x: 2​
1. x​
­ 10x +25 = 0 2​
2. x​
­ 2x + 2 = 2x Problem 4 (Bonus) Derive the quadratic formula! 2​
ax​
+ bx + c = 0, a, b, and c are constants, and a ≠ 0 Use the “completing the square method” learned in section 5­2 to do this. 9 Challenge Problem! Using the discriminant of the quadratic (a1x + b1)2 + (a2x + b2)2 + (a3x + b3)2 + ... + (anx + bn)2 , prove that (a12 + a22 + a32 + ... + an2)(b12 + b22 + b32 + ... + bn2) ≥ (a1b1 + a2b2 + a3b3 + ... + anbn)2 . 6.10 ­ An Exponential Discovery: Properties of Logarithms In this section, our goal is to understand the three properties of logarithms that come directly from the corresponding properties of exponentiation by transforming and solving equations. By the end of this section you should be able to use those same properties to simplify logarithms. Concept: The first property is called​
​
Logarithm of a Product​
which is “The log of a product equals the sum of the logs of the two factors.” In other words, this means log​
(xy) can also be written is log​
x + log​
y. b​
b​
b​
p​
q
The proof is as follows: Let log​
x = p and log​
y = q. This can be rewritten as b​
= x and b​
b​
b​
p+q​
= y. We can multiply these to obtain b​ = xy, or p+q = log​
(xy), or log​
x + log​
y = b​
b​
b​
log​
(xy), as desired. b​
The second property is called​
​
Logarithm of a Quotient​
​
which is “The log of a quotient equals the log of the numerator minus the log of the denominator.” To clarify, log​
(x/y) = log​
x ­ log​
y. This is similar to the other property. The proof to this b​
b​
b​
identity is similar to the proof of the ​
Logarithm of a Product​
. See if you can prove it yourself, if you are up to the task, that is. The third and last property is called ​
Logarithm of a Power​
​
and it means “The log of a power equals the exponent times the log of the base.” n​ ​
So, if you had log​
(x​
)​
it would be the same as saying n log​
x. Try to prove this as well. b​
b​
Example 1 Express log​
9 + log​
4 as an integer 6​
6​
Solution: Step 1: Use the ​
Logarithm of a Product ​
property to add together the two logarithms. log​
(9*4) = log​
36 6​
6​
2 Step 2: Write 36 as 6​
2​
log​
(6​
) 6​
Step 3: Use the ​
Logarithm of a Power. 2 log​
6 = 2 6​
10 Example 2 x​
Solve 8​
= 75 by taking the log of each member and using the log of a power property Solution: Step 1: Take the log of both sides as said in instructions x​
log 8​
= log 75 Step 2: Use the ​
Logarithm of a Power ​
Property x log 8 = log 75 Step 3: Divide both sides by log 8 x = (log 75)/(log 8) Problems: Problem 1 Given log 2 ≈ 0.301 and log 3 ≈ 0.477, find the given logarithm without using the log key on your calculator. a. log 27 b. log 15 c. log 729 d. log 30 Problem 2 Write the expression as a single logarithm of a single argument a. log​
7 + log​
3 4​
4​
b. 2 log​
5 3​
c. 3 log​
7 + log​
9 ­ log​
5 8​
8​
8​
d. log​
24 ­ 2 log​
3 11​
11​
Problem 3 Using the Richter scale (R = log (I/I​
)) for measuring the magnitudes of earthquakes, find the 0​
magnitude ​
R ​
in the simplest form you can ​
without​
using the log button on your calculator. I​
= 1 0​
a. 2011 Tōhoku earthquake (I = 1,000,000,000) b. 1356 Basel earthquake if the intensity of the Yucatan Peninsula earthquake, who’s magnitude is twice as much as the Basel earthquake’s magnitude, had an I = 10,000,000,000,000. c. 1952 Kamchatka earthquake if another earthquake, whose magnitude is 2 units greater, has an I = 100,000,000,000. Problem 4 Solve the equations by taking the log of each member, then using the Log of a Power Property. 11 a.
b.
c.
d.
x​
2.7​
= 55.1 2x​
6.38​
= 41.3 ­0.2x​
9.67 * 10​
= 15.7 5x ​
13.4 * 10​
= 21.5 7.3 ­ The Winning Factor: Special Products and Factoring In this section our goal is to master how to multiply multiple multivariable polynomials together and to recognize a few special products. You should already be familiar with the following factorizations: (x + y)2 = x2 + 2xy + y2 (x − y)2 = x2 − 2xy + y2 (x + y)(x − y) = x2 − y2 You should be able to recognize these identities both ways ­ that is, you can multiply them out and you can factor them. After all, this ​
is​
assuming that you have passed Algebra I. We will also give a brief review on multiplying multiple polynomials together. After we go through the examples, you will see that the associative property is pretty useful! Example 1 Factor (x + y)2 − (x − y)2 . Solution: This is obviously a difference of squares.
(x + y)2 − (x − y)2 = (x + y + x − y)(x + y − x + y) = (2x)(2y) = 4xy Of course, you can recognize that the terms cancel out, which might be better in this case. (x + y)2 − (x − y)2 = (x2 + 2xy + y2) − (x2 − 2xy + y2) = 4xy Either way, we get the same answer. Example 2 Multiply (x6 − x5y + x4y2 − x3y3 + x2y4 − xy5 + y6)(x + y) . Solution: It looks messy, but terms cancel. x7 − x6y + x5y2 − x4y3 + x3y4 − x2y5 + xy6 + x6y − x5y2 + x4y3 − x3y4 + x2y5 − xy6 + y7 = x7 + y7 Example 3 Factor completely: x8 − 16y8 Solution: Difference of squares. (x4 + 4y4)(x4 − 4y4) = (x4 + 4y4)(x2 + 2y2)(x2 − 2y2) Are we done? Take a closer look at (x4 + 4y4) . We can add in an extra term, 4x2y2 , to factor it further. x4 + 4x2y2 + 4y4 − 4x2y2 = (x2 + 2y2) − (2xy)2 = (x2 − 2xy + 2y2)(x2 + 2xy + 2y2) . Therefore, the final factorization is (x2 − 2xy + 2y2)(x2 + 2xy + 2y2)(x2 + 2y2)(x2 − 2y2) . Example 4 12 Multiply (x − 1)(x − 2)(x − 3)(x − 4) Solution: Unfortunately, we have to manually multiply it out. Shall we commence? (x − 1)(x − 2)(x − 3)(x − 4) = (x2 − 3x + 2)(x2 − 7x + 12) = x4 − 3x3 + 2x2 − 7x3 + 21x2 − 14x + 12x2 − 36x + 24 = x4 − 10x3 + 35x2 − 50x + 24 Thus, the answer is x4 − 10x3 + 35x2 − 50x + 24 Problems Problem 1 Factor x4 + 4x3 + 5x2 + 4x + 1 . Problem 2 Multiply (a − b − c − d)(a2 + (b + c + d)2)(a + b + c + d) Problem 3 Given that ​
a​
, ​
b​
, and ​
c​
are positive integers, the equation 2a2 + 2b2 + 2c2 + 2ab + 2bc + 2ca can be expressed in terms of the sum of three perfect squares, but it can also be expressed in terms of the sum of four perfect squares. Find both expressions. 13 Appendix ­ Solutions to aforementioned problems Section 3.5 Problem 1 (a) We let the mouse’s distance (km) from the White House be ​
m​
, Mr. Nguyen’s distance (km) be ​
n​
, and the time passed (min) be ​
t​
. The distance­intercept of the mouse is 50. We can find the slope: slope = 35−50
3−0 =− 5 , so the overall equation of the mouse is m = 50 − 5t . The 25
distance­intercept of Mr. Nguyen is 80, and his slope is 55−80
3−0 =− 3 , so the overall equation for Mr. Nguyen is n = 80 − 253 t . (b) We can plug in 5 for ​
t​
to find the respective values of ​
m​
and ​
n​
; m = 50 − 5(5) = 25 , and 25
115
n = 80 − 3 (5) = 3 , so the mouse is 25 km away from the white house and Mr. Nguyen is 115
3
km away from the White House. Mr. Nguyen seems to be catching up. (c) We can plug in 15 for ​
m​
to find ​
t​
: 15 = 50 − 5t ⇒ 5t = 35 ⇒ t = 7 . Now, we can plug in 7 for 25
65
t​
to find ​
n​
: n = 80 − 3 (7) = 3 , so when m = 15 , then n = 653 . To make this easier, we can express ​
n​
in terms of ​
m​
. We can multiply the overall equations for the mouse and Mr. Nguyen to get the respective equations: 5m = 250 − 25t, 3n = 240 − 25t . Subtracting, we get 3n − 5m =− 10 , or n = 53 m − 103 . This gives us a direct way to find ​
n​
given ​
m​
, and the other way around. Looking back to the original problem, we can plug in 15 for ​
m​
to get n = 53 (15) − 103 = 653 , as desired. (d) Plugging in 35 for Mr. Nguyen’s equation, we get 35 = 80 − 253 t ⇒ 253 t = 45 ⇒ t = 5.4 , so Mr. Nguyen has been flying for 5.4 minutes, or 5 minutes and 24 seconds. (e) 14 distance (km)
(f) The slopes of the lines are in terms of time (minutes)
, which represents speed. (g) We would first find their respective speeds. Because the slope of the line for the mouse is ­5, then the mouse moves at 5 km/min, or 300 km/h. Because Mr. Nguyen’s slope is ­25/3, then he moves at 25/3 km/min, or 500 km/h. That means that Mr. Nguyen is going 200 km/h faster than the mouse. (h) Mr. Nguyen will indeed catch up to save the world from imminent destruction. We can see this by plugging in 0 for ​
n​
: 0 = 53 m − 103 , or m = 2 . That means that when Mr. Nguyen is at the White House, the mouse still has two kilometers to go, so Mr. Nguyen would be able to stop the mouse. Section 4.5 Problem 1 (a) (3,­4,2) is the ordered triple, 4x ­ 3y + 6z = 12 (b) (3,­2,1) is the ordered triple, 6x ­ 9y + 18z = 18 15 (c) (4,1,­2) is the ordered triple, 4x + 16y ­ 8z = 16 (d) (2,­2,2) is the ordered triple, x ­ y + z = 2 Problem 2 16 (a) To find the four intercepts, you must set all values except the one you want the intercept of to equal 0. w­intercept: 6w = 36→ w = 6 x­intercept: ­3x = 36→ x = ­12 y­intercept: 9y = 36→ y = 4 z­intercept: 12z = 36→ z = 3 (b)The xyz trace equation is found when the w value equals 0. xyz trace: ­3x + 9y + 12z = 36 Section 5.3 Problem 1 Determine whether or not the following equations produce real solutions. If so, find the x­intercepts of the equation. 2​
1. y = 3x​
­ 10x + 6 2​
The discriminant is 10​
­ 4(3)(6), which is 28. Thus, the solutions are real numbers and we can solve for x to find the x­intercepts. x=
x=
−b±√b2−4ac
2a
10±√28
6
5±√7
3 x=
x ≈ 2.55, 0.785 2​
2. y = ­x​
+ 4x ­ 1 2​
The discriminant is 4​
­ 4(­1)(­1), or 12, so we know the solutions are real. Let’s solve for x: x=
−b±√b2−4ac
2a
−4±√12
8
x=
x ≈− 0.07, − 0.93 2​
3. y = 10x​
+ 19x +12 2​
The discriminant is 19​
­ 4(10)(12). This equals ­119, which means the solutions will not be real numbers! Problem 2 Determine the vertex, y­intercept, and point of symmetry of the above problems and use them to graph the equations. 2​
1. y = 3x​
­ 10x + 6 a. X­coordinate of vertex: 10/2(3) = 5/3 b. Y­coordinate of vertex: ­7/3 c. Y­intercept: (0, 6) 17 d. Point of symmetry: (10/3, 6) Graph: 2​
2. y = ­x​
+ 4x ­ 1 a. X­coordinate of vertex: ­4/2(­1) = 2 b. Y­coordinate of vertex: 3 c. Y­intercept: (0, ­1) d. Point of symmetry: (4, ­1) Graph: 18 2​
3. y = 10x​
+ 19x +12 a. X­coordinate of vertex: ­19/2(10) = ­19/20 = ­0.95 b. Y­coordinate of vertex: 119/40 = 2.975 c. Y­intercept: (0, 12) d. Point of symmetry: (­1.9, 12) Graph: 19 Problem 3 Solve for x: 2​
1. x​
­ 10x +25 = 0 x=
−b±√b2−4ac
2a
10±√100−100
2
10±0
2 x=
x=
x = 5 2​
2. x​
­ 2x + 2 = 2x Bring all terms to one side so you are left with 0 on the other. 2​
x​
­ 4x + 2 = 0 x=
x=
x=
−b±√b2−4ac
2a
4±√16−8
2
4±2√2
2 Problem 4 (Bonus) ax2 + bx + c = 0 x2 + ba x + ac = 0 x2 + ba x =− ac 2
2
x2 + ba x + 4ab 2 =− ac + 4ab 2 20 2
−4ac
(x + 2ab )2 = b 4a
2 2
√
b −4ac
b
x+ =
2a
2a
−b+√b2−4ac
=
2a
x
Challenge Problem First, we can expand the quadratic into (a12 + a22 + a32 + ... + an2)x2 + 2(a1b1 + a2b2 + a3b3 + ... + anbn)x + (b12 + b22 + b32 + ... + bn2) . Second, because x2 ≥ 0 for all ​
x​
, which should be intuitive, then that means the original quadratic must be nonnegative for all values of ​
x​
. That means that the discriminant must be nonpositive; if it were positive, then there would be two roots for the quadratic, which means that at one point the value of the quadratic must be negative, reaching a contradiction. Thus, 4(a1b1 + a2b2 + a3b3 + ... + anbn)2 − 4(a12 + a22 + a32 + ... + an2)(b12 + b22 + b32 + ... + bn2) ≤ 0 , or (a12 + a22 + a32 + ... + an2)(b12 + b22 + b32 + ... + bn2) ≥ (a1b1 + a2b2 + a3b3 + ... + anbn)2 , as desired. Section 6.10 Problem 1 3​
a. log 27 can be written as log 3​
and using the Logarithm of a power property we can write that as 3 log 3. It is given that log 3 ≈ 0.477, so log 27 is the same thing as 3 * 0.477 which is 1.431 b. log 15 is the same thing as log(10/2)*3 and using the Logarithm of a Quotient property and Logarithm of a Product property, it can also be written as log 10 ­ log 2 + log 3. We know that log 3 ≈ 0.477, log 2 ≈ 0.301, and log 10 = 1, so it is the same thing as 1 ­ 0.301 + 0.477 = 1.176 6​
c. log 729 = log 3​
which, using the Logarithm of a power property, can also be written as 6 log 3. Since we know log 3 ≈ 0.477, the answer is 6(0.477) = 2.862 d. log 30 can also be written as log 3*10 which, using the Logarithm of a product property, is the same thing as log 3 + log 10 and, knowing that log 3 ≈ 0.477 and log 10 = 1, we can say it is the same thing as 0.477 + 1 = 1.477 Problem 2 a. log​
7 + log​
3 is, because of the logarithm of a product property, the same thing as log​
7*3 4​
4​
4​
= log​
21 4​
2 ​
b. 2 log​
5 = log​
5​
(because of the logarithm of a power property) = log​
25 3​
3​
3​
3 ​
c. 3 log​
7 + log​
9 ­ log​
5 = log​
7​
+ log​
9/5 (because of the logarithm of a quotient property 8​
8​
8​
8​
8​
and logarithm of a power property) = log​
(343 * 9/5) (because of the logarithm of a 8​
product property) = log​
3087/5 8​
2 ​
d. log​
24 ­ 2 log​
3 = log​
24 ­ log​
3​
(because of the logarithm of a power property) = 11​
4​
11​
11​
log​
24/9 (because of the logarithm of a quotient property) = log​
8/3 11​
11​
21 Problem 3 9
a. Using the formula we can say R = log 1,000,000,000 which can be written as log 10​
which is the same thing as 9 log 10 because of the logarithm of a power property and since log 10 = 1, the answer is 9*1 = 9. b. We know that the magnitude of the Basel Earthquake has to equal the magnitude of the Yucatan Peninsula Earthquake / 2; to find the magnitude of the Yucatan Peninsula Earthquake, we can use the formula; the magnitude = (log 10,000,000,000,000) / 2 which 13 ​
can also be written as log 10​
/ 2 and, using the logarithm of a power property, that also equals (13 log 10) / 2; we know that log 10 = 1, so the final answer is 13 / 2 = 6.5 c. To find the magnitude of the Kamchatka Earthquake we find the magnitude of the other earthquake and subtract two from that magnitude; we use the formula to find the magnitude of the other earthquake; the magnitude of the Kamchatka Earthquake = (log 11​
100,000,000,000)­2 = (log 10​
) ­ 2 = (11 log 10) ­ 2 because of the logarithm of a power property = 11 ­ 2 since log 10 is 1 = 9. Problem 4 x​
x
a. 2.7​
= 55.1; the directions say to take the log of both sides so that is what we do: log 2.7​
= log 55.1; we then, using the logarithm of a power property, can write that as: x log 2.7 = log 55.1; finally, we can calculate x by dividing log 2.7 from both sides: x = log 55.1 / log 2.7 2x​
b. 6.38​
= 41.3; following a similar procedure as in the problem above, we take the log of 2x​
both sides: log 6.38​
= log 41.3; we then use the logarithm of a power property and also divide log 6.38 from both sides, like in the previous exercise: 2x = log 41.3 / log 6.38; we then isolate x by dividing 2 from both sides: x = log 41.3 / 2 log 6.38 ­0.2x​
­0.2x​
c. 9.67 * 10​
= 15.7; first, we divide 9.67 from both sides: 10​
= 15.7 / 9.67; then we ­0.2x​
take the log of both sides as said in the instructions: log 10​ = log 15.7/9.67; we then use the logarithm of a power property: ­0.2x log 10 = log 15.7/9.67; then we divide ­0.2 log 10 from both sides to isolate x and get the answer: x = ­(log 15.7/9.67)/0.2 log 10 5x​
5x​
d. 13.4 * 10​
= 21.5; we start by dividing 13.4 from both sides: 10​
= 21.5/13.4; we 5x​
continue by taking the log of both sides: log 10​ = log 21.5/13.4; then we use the logarithm of a power property to get: 5x log 10 = log 21.5/13.4; finally, we isolate x by dividing 5 log 10 from both sides: x = (log 21.5/13.4)/5 Section 7.3 Problem 1 x4 + 4x3 + 5x2 + 4x + 1 can be expressed as x4 + 4x3 + 6x2 + 4x + 1 − x2 , which is (x + 1)4 − x2 . This is a difference of squares, so we can factor this to be ((x + 1)2 + x)((x + 1)2 − x) , or (x2 + 3x + 1)(x2 + x + 1) , and we cannot factor this any more. 22 Problem 2 We can group the factors into (a − b − c − d)(a + b + c + d)(a2 + (b + c + d)2) . We can recognize the first two terms as a difference of squares, namely (a2 − (b + c + d)2)(a2 + (b + c + d)2) . This is another difference of squares, so we can multiply it out again. a4 − (b + c + d)4 . Multiplying this out, we get a4 − b4 − 4b3c − 4b3d − 6b2c2 − 12b2cd − 6b2d2 − 4bc3 − 12bc2d − 12bcd2 − 4bd3 − c4 − 4c3d − 6c2d2 − 4cd3 − d4 You
2
2
2
2
would obtain this by finding that (b + c + d) = b + c + d + 2bc + 2cd + 2db , and then multiplying that by itself. Finally, you would subtract that entire thing from a4 to get the final answer. Problem 3 We first attempt to factor out a two from the equation, but that doesn’t help us because that leads us further away from getting a sum of squares. Remember that part of this section is about multiply two special binomials together. Because the equation is symmetric in variables ​
a​
, ​
b​
, and 2
2
2
c​
, then we can try to add (a + b) + (b + c) + (c + a) , which is also symmetric, and is the sum of three perfect squares. Summing them, we get a2 + 2ab + b2 + b2 + 2bc + c2 + c2 + 2ca + a2 = 2a2 + 2b2 + 2c2 + 2ab + 2bc + 2ca , which works. For the sum of four perfect squares, from the previous problem, we recognize 2ab + 2bc + 2ca as the second part of the square of a trinomial, so we can separate a2 + b2 + c2 + 2ab + 2bc + 2ca from the equation to get a2 + b2 + c2 + (a2 + b2 + c2 + 2ab + 2bc + 2ca) , or a2 + b2 + c2 + (a + b + c)2 , which is the sum of four perfect squares, as desired. 23 1.2 ­ The Field Axioms 1. What is a Field? a. A field is any set of elements that satisfies the field axioms for both addition and multiplication. b. The four basic math operations are addition, subtraction, multiplication, and division. Because subtraction is addition of the opposite and division is multiplication by the reciprocal, the properties of real numbers focus on addition and multiplication. 2. Simplifying any expression or solving any equation requires the use of the following axioms: Closure­ {real numbers is closed under addition and under multiplication} i. Additive: x+y is a unique, real number if x and y are real numbers ii.
Multiplicative: xy is a unique, real number if x and y are real numbers Commutativity iii.
Additive: If a and b are any real numbers, then a + b = b + a iv. Multiplicative: If a and b are any real numbers, then ab = ba. Associativity v.
Additive: If a, b, and c are real numbers, then a + (b + c) = (a + b) + c vi. Multiplicative: If a, b, and c are any real numbers, then a(bc) = (ab)c. Distributivity vii.
Multiplication distributes over addition viii.
For all real numbers a, b, and c, a(b + c) = ab + ac Identity Elements ix. Additive Identity: There is a number 0 such that for all numbers a, a + 0 = a x.
Multiplicative: There is a number 1 such that for all numbers a, a*1 =a Inverses xi. Additive: For every real number a there is a real number −a such that a + (−a) = 0. xii.
Multiplicative: There is a number 1/a such that for all numbers a a*1/a = 1. By Samantha Kuo Visual Representation: Given if a,b, and c are real numbers. Property Addition Multiplication Associative (a + b) + c = a + (b + c) (ab)c = a(bc) Commutative a + b = b + a ab = ba Identity a + 0 = a = 0 + a a ∙ 1 = a = 1 ∙ a Inverse a + (−a) = 0 = (−a) + a a ∙ 1/a = 1 = a ∙ (1/ a), if a ≠ 0 Distributive a(b + c) = ab + ac ab + ac = a(b + c) By Samantha Kuo 1.2 Practice Problems 1. For each of the following, tell which of the axioms is used: a. x+(y+z) = (x+y)+z b. x(y+z) is a real number if x,y,z are real numbers c. x(y+z) = x(z+y) d. x(y+z) = (y+z)x e. x(y+z) = xy+xz f. x(y+z) = x(y+z)+0 g. x(y+z) = (­[x(y+z)]=0 h. x(y+z) = x(y+z)1 i. x(y+z)*1/x(y+z) = 1 2. What is the additive inverse of 2/3? 3. What is the multiplicative inverse of 2/3? 4. Write an example of the distribution property 5. Tell whether or not the following are fields under addition and multiplication. If the set is not satisfied as a field, show which of the field axioms do not apply. a. {rational numbers} b. {integers} c. {positive numbers} d. {non­negative numbers} 6. Explain why 0 has no multiplicative inverse 7. Write an example which shows that: a. Subtraction is not commutative b. {negative numbers} is not closed under multiplication c. {digits} is not closed under addition 8. What is meant by closure? By Samantha Kuo 1.2 Answer Key To Practice Problems 1. For each of the following, tell which of the axioms is used: a. x+(y+z) = (x+y)+z ​
Additive Associativity b. x(y+z) is a real number if x,y,z are real numbers ​
Additive & Multiplicative Closure c. x(y+z) = x(z+y) ​
Additive Commutativity d. x(y+z) = (y+z)x ​
Multiplicative​
​
Commutativity e. x(y+z) = xy+xz ​
Multiplicative Distributivity f. x(y+z) = x(y+z)+0 ​
Additive Identity g. x(y+z) + ­[x(y+z)]=0 ​
Additive Inverse h. x(y+z) = x(y+z)1 ​
Multiplicative Identity i. x(y+z)*1/x(y+z) = 1 ​
Multiplicative Inverse 2. What is the additive inverse of 2/3? ­2/3 3. What is the multiplicative inverse of 2/3? 3/2 4. Write an example of the distributive property a(b+c)=ab+ac 5. Tell whether or not if the set of numbers satisfy all the axioms. If the set is not satisfied as a field, show which of the axioms do not apply. a. {rational numbers} ​
Satisfied b. {integers} ​
No, multiplicative inverse: 2 x ½ = 1, but ½ is not an integer. c. {positive numbers} ​
No, additive identity and additive inverse: 0 is not a positive number, and a positive number plus another positive number never equals zero. d. {non­negative numbers} ​
No, additive inverse, a non­negative number plus another non­negative number other than zero never equals zero. 6. Explain why 0 has no multiplicative inverse Satisfied 0 does not have a multiplicative inverse because 0 has no reciprocal that will have the product as 1. 7. Write an example which shows that: j. Subtraction is not commutative ​
8­7 ≠ 7­8 (1 ≠ ­1) k. {negative numbers} is not closed under multiplication ​
(­8 x ­7=56) l. {digits} is not closed under addition ​
(8+7=15), and 15 is not a digit 8. What is meant by closure? The set of real numbers is closed under addition and multiplication if: 1. x+y is a unique, real number 2. xy is a unique, real number ​
By Samantha Kuo 3.3 ­ Other Forms of the Linear Function Equation 1. Slope intercept form:​
y = mx+b​
, where ​
m​
is the slope and ​
b​
is the y­intercept. This equation is called slope intercept because the y­intercept and the slope of the line can both be found in the equation. Example 1: In the equation y = 2x­6, 2 is the slope, and 6 is the y intercept. 2. Point slope form: ​
y­y​
​
, where ​
m​
is the slope and ​
(x​
is a point on 1 = m(x­x
​
1)
​​
1, y
​ ​
1)
​​
the graph. This equation is called point slope because the coordinates of a point and the slope of the line can both be found in the equation. Example 2: In the equation y­4 = 2(x­5), 2 is the slope, and (5,4) is a point on the graph. 3. “Ax+By = C” form: ​
Ax+By = C​
, where ​
A​
, ​
B​
, and ​
C​
stand for integer constants. This form will be used if both variables are on one side, and the constant is on the other. Example 3: If y­3 = ­5/2(x+1), transform the equation into: a. slope intercept form b. Ax+By = C form Slope intercept form: y­3 = ­2.5(x+1) y­3 = ­2.5x­2.5
Distribute the ­2.5 y = ­2.5x+0.5
Add 3 on both sides Ax+By = C form: y = ­2.5x+0.5
Use the slope intercept form of the equation 2.5x+y = 0.5
Add 2.5x on both sides 5x+2y = 1
Multiply by 2 to make the coefficients an integer By Kelly Tan 3.3 ­ Practice Problems 1. Let f (x) = 5x – 4 a. Determine the slope and y­intercept b. Graph f(x) c. Find the average­rate­of­change d. Determine whether f(x) is increasing, decreasing, or constant and give the interval. e. Find the x­intercept. By Samantha Kuo 3.3 ­ Answer Key 1. Let f (x) = 5x – 4 a. Determine the slope and y­intercept Since this is already written in the slope­intercept form we know that the slope is 5. We want to make sure we write the y­intercept in the proper form. We don’t just want to write ­4 as the answer. We write it as (0, ­4). b. Graph f(x) In order to graph we first will plot the y­intercept. We said the slope was 5. This is the same thing as 5/1 . The top number is the vertical change and the bottom number is the horizontal change. We will start at the y­intercept. From here we will go up 5 units and then go 1 unit to the right. c. Find the average­rate­of­change For any linear functions the average­rate­of­change is the slope. So the answer is 5. d. Determine whether f(x) is increasing, decreasing, or constant and give the interval. As we move from left to right the graph is always going uphill, so the graph is increasing on (­​
∞,∞​
). e. Find the x­intercept. Whenever you find the x­intercept this means you need to put in a value of 0 for y. We get 0 = 5x – 4. By solving you will get that x=4/5 . Then we write it in the proper form for x­intercepts: (⅘,0) By Samantha Kuo 4.2 ­ Solving Systems of Linear Equations 1. A system of equations is a system of two or more equations that have the same variables. a. Example: 5x + 2y = 11 x + y = 4 2. A solution set for a system of equations is an ordered pair that satisfies both equations by substituting the ordered pair for each corresponding variable. a. Example: The solution set for the system of equations above is ​
(1,3) b. Explanation: ​
(1,3)​
is the solution set because ​
5(1) + 2(3) = 11​
, and 1 + 3 = 4,​
therefore, the ordered pair satisfies both equations. 3. Now that you know what a system of linear equations and a solution set is, how do you find the solution set for a particular system of equations? There are actually three methods to find the solution set, and in this article, all three will be explained in­depth. The three methods are: a. Substitution b. Linear Combination c. Graphing 4. First, we will start with ​
substitution​
, and we can use the system of linear equations from the example above. a. The first step of substitution is to isolate one of the variables in one of the equations. b. Then, you plug in the isolated variable into the second equation. Now you should have a single equation with only one variable. c. The next step is to solve for the one variable in your new equation. d. Once you solve for the one variable, substitute that value back into one of the two original equations. Now you should have another equation with the one variable you have not solved for yet. e. The final step is to solve the equation you have just created in the last step to find the second variable. Now that you have found the two values of the variables, combine them in an ordered pair to create the solution set. i. Example: 5x + 2y = 11 x + y = 4 ii.
Solution: ​
1. x = 4 – y​
​
Isolate one variable (subtraction) ​
2. 5(4 – y) + 2y = 11​
​
Substitution By Evan Labuda ​
3. 20 – 5y + 2y = 11 ​
Distribution Property of Multiplication ​
4. 20 – 3y = 11 ​
Combining Like Terms ​
5. ­3y = ­9
​
Subtraction ​
6. ​
y = 3
​
Division ​
7. x + 3 = 4
​
Substitution ​
8. ​
x = 1
​
Subtraction ​
9. The solution set is (1,3) 5. The next way to solve a system of equations is ​
linear combination​
. This is probably the most useful, as well as the easiest and fastest way to solve a system of linear equations. a. The first step is to multiply either one or both equations by a constant so that the coefficients of one of the variables in each of the equations is the same. b. Next, subtract one equations from the other so that you will end up with one equation with one variable. c. Now solve the equation for the one variable. d. Once you have found the first variable, follow the same steps as the substitution method in order to find the second variable for the solution set. i. Example: 5x + 2y = 11 x + y = 4 ii.
Solution: ​
1. 5x + 2y = 11 ​
2x + 2y = 8 ​
Multiplication (x2) ​
2. 3x = 3 ​
Subtraction ​
3. ​
x = 1​
​
Division ​
4. 1 + y = 4
​
Substitution ​
5. ​
y = 3
​
Subtraction ​
6. The solution set is (1,3) 6. The last thing you will need to know about systems of linear equations is the different types of systems. If your answer consists of one ordered pair, the system is considered ​
independent. ​
If at one point, both equations are the same after being multiplied by a constant, the system is ​
dependant,​
meaning the equations are the same line. The solution set for a set of dependant equations is all real numbers. Lastly, if both variables cancel out, yet the constant on the other side of the equation does not, so one side of the equation is 0 and the other is a real number such as: ​
0 = 4,​
the system is inconsistent,​
meaning the equations are parallel lines. An inconsistent system of equations has no solutions set. By Evan Labuda 4.2 Real World Application The main way that systems of linear equations are applied in real life is to figure out a piece of information that you do not necessarily have. For example, if you are taking a survey at your school for how many adults and how many children attended a play, but you only know the total number of attendees, the prices of the different tickets, and the total profit made, you can use a system of linear equations to find out how many children and how many adults attended. Let’s say that children tickets cost $2, and adult tickets cost $4, 2000 tickets were sold, and $6400 was made. You could set up this system of linear equations using “x” as the number of children tickets sold, and “y” as the number of adult tickets sold: (x + y = 2000) ­2 2x + 4y = 6400 Add both equations 2y = 2400 y = 1200 x + 1200 = 2000 x = 800 Using this system of linear equations, you know know that 1200 adult tickets were sold, and 800 child tickets were sold. By Evan Labuda 4.2 Practice Problems: Solve each system using substitution: 1. x + 3y = 3 4x – 3y = 17 2. x = 1.2 (300 + y) y = 0.4 (300 + x) Solve each system using linear combination: 3. 3x + 10y = ­24 6x + 7y = ­9 4. 7x + 5y = 9 3x – 2y = 37 Solve each system using any method, or state that there is no solution: 5. (2/x) – (5/x) = 5 (3/x) + (10/y) = 18 6. 3x – y = 5 2x + 3y = 7 4x + 2y = 10 Determine whether the equations in the system are independant, dependant, or inconsistent: 1. 24x – 56y = 72 15x – 35y = ­45 2. 15x – 12y = 18 10x – 8y = 12 Turn to the next page for answers By Evan Labuda 4.2 Answer Sheet 1. x = 3 ­ 3y 4(3 ­ 3y) ­ 3y = 17 12 ­ 12y ­ 3y = 17 ­15y = 5 y = ­⅓ ​
Substitute x ­ 3(⅓) = 3 x ­ 1 = 3 x = 4 ​
SS = (4, ­⅓) 2. y = 0.4(300 + (360 + 1.2y)) y = 0.4(660 + 1.2y) y = 264 + 0.48y 0.52y = 264 y = 507.7 ​
Substitute ​
x = 1.2(300 + 507.7) x = 1.2(807.7) ​
x = 969.2 ​
SS = (969.2, 507.7) 3. (3x + 10y = ­24) 2 6x + 7y = ­9 ​
Subtract both equations 13y = ­39 ​
y = ­3 6x + 7(­3) = ­9 6x = 12 ​
x = 2 ​
SS = (2,­3) By Evan Labuda 4. (7x + 5y = 9) 2 (3x – 2y = 37) 5 ​
Add both equations 29x = 203 x = 7 ​
49 + 5y = 9 5y = ­40 ​
y = ­8 ​
SS = (7,­8) 5.( (2/x) – (5/y) = 5 ) 2 (3/x) + (10/y) = 18 ​
Add both equations 7/x = 28 x = 0.25 ​
8 – 5/y = 5 ­5/y = ­3 ​
y = 5/3 ​
SS = (¼, 5/3,) 6. (3x – y = 5) 3 2x + 3y = 7 ​
Add both equations 11x = 22 x = 4 ​
4 + 3y = 7 3y = 3 ​
y = 1 ​
SS = (4,1) By Evan Labuda 7. (24x – 56y = 72) 5 (15x – 35y = ­45) 8 ​
Subtract both equations ​
0 = 720 ​
Inconsistent 8. (15x – 12y = 18) 2 (10x – 8y = 12) 3 ​
Subtract both equations ​
0 = 0 ​
Dependant By Evan Labuda 6.8 ­ Exponential Equations Solved by Logarithms y​
1. Definition ­ Base 10 Logarithm: log x = y only if 10​
= x 2​
Example 1: log 100 = 2 because 10​
= 100 x​
2. Property ­ Logarithm of a power of 10: log 10​
= x x​
x​
x​
x​
x​
Log 10​
will only equal x if 10​
= 10​
. Since 10​
= 10​
is true because it is an identity, the first equation has to be true. ­0.5x​
Example 2: Solve for x: 16 x 10​
= 28 ­0.5x​
10​ = 28/16 = 7/4 Divide both sides by 16 ­0.5x​
log 10​ = log (7/4)
Take the log of both sides ­0.5x = log (7/4)
Property: Log of a power of 10 x = (log (7/4))/­0.5
Divide both sides by ­0.5 x​
Example 3: Solve for x: 8​
= 138 k​
Let 10​
= 8, so k = log 8 k​x​
(10​
)​
= 138
Substitution (10^k as 8) kx​
10​= 138
Property: Power of a power kx​
log 10​ = log 138
Take the log of both sides kx = log 138
Property: Log of a power of 10 x = log 138/k
Divide both sides by k x = log 138/log 8
Substitution (k as log 8) By Kelly Tan 6.8 Practice Problems 1. Solve for x: x​
a. 10​
= 305 ­x​
b. 10​
= 0.008 x​
c. 10​
= ­5.6 x​
d. 10​
= 0 x​
e. 4.5×10​
= 90 0.4x​
f. 10​ = 101 ­0.3x​
g. 61.5×10​
= 123 2. Solve for x by first changing the base to 10: 3x​
a. 4.13​
= 940 5x​
b. 11×9​ = 121 8x​
c. 3.6×12​
= 72 ­0.8x​
d. 4×25​ = 100 5x​
e. 13​
= 1001 ­9x​
f. 29×13​
= 31 By Kelly Tan 6.8 Answer Key 1. Solve for x: x​
a. 10​
= 305 ​
log 305 = x ­x​
b. 10​
= 0.008 ​
log 0.008 = ­x ­(log 0.008) = x x​
c. 10​
= ­5.6 ​
No solution x​
d. 10​
= 1 ​
0 x​
x​
e. 4.5×10​
= 90 ​
10​
= 20 log 20 = x 0.4x​
f. 10​ = 101 ​
log 101 = 0.4x (log 101)/0.4 = x ­0.3x​
­0.3x​
g. 61.5×10​ = 123 ​
10​
= 2 log 2 = ­0.3x (log 2)/­0.3 = x 2. Solve for x by first changing the base to 10: 3x​
k​
a. 4.13​
= 940 ​
Let 10​
= 4.13 so k = log 4.13 k​3x​
(10​
)​ = 490 3kx​
10​ = 490 3kx = log 490 3(log4.13)x = log 490 x = (log 490)/(2 log 4.13) 5x​
b. 11×9​ = 121 ​
95x​
​ = 11 k​
Let 10​
= 9 so k = log 9 k​5x​
(10​
)​ = 11 5kx​
10​ = 11 log 11 = 5kx log 11 = 5(log 9)x x = (log 11)/(5 log 9) ­0.8x​
­0.8x​
c. 4×25​ = 100 ​
25​
= 25 x = ­1/0.8 5x​
k​
d. 13​ = 1001 ​
Let 10​
= 13 so k = log 13 k​5x​
(10​
)​ = 1001 5kx​
10​ = 1001 log 1001 = 5kx log 1001 = 5(log13)x (log 1001)/(5 log 13) = x ­9x​
­9x​
e. 29×13​ = 31 ​
13​
= 31/29
By Kelly Tan k​
​
Let 10​
= 13 so k = log 13 k​­9x​
​
(10​
)​ = 31/29 ­9kx​
10​ = 31/29 ​
­9kx = log (31/29) ­9(log 13)x = log (31/29) ​
x = (log (31/29))/(­9 log 13) By Kelly Tan 7.4 ­ More Factoring and Graphing Section Description: Section 7­4 covers the factoring of sums and differences of two cubes. These factoring 3​
3​
3​
3​
techniques are used to factor expressions like x​
+ y​
or x​
– y​
. Furthermore, you will have the opportunity to practice factoring. Factoring a Sum or Difference of Two Cubes: 3​
3​
2​
2​
Sum of two cubes: x​
+ y​
= (x + y)(x​
– xy + y​
) 3​
3​
2​
2​
Difference of two cubes: x​
– y​
= (x – y)(x​
+ xy + y​
) Prior Knowledge: 2​
2​
Difference of two squares: x​
– y​
= (x – y)(x + y) Example Problems: 2​
1. Factor: x​
– 4 3​
3
2. Factor: x​
– y​
9​
9
3. Factor: x​
– y​
6​
6
4. Factor: x​
– y​
3​
3
5. Factor: x​
+ y​
9​
9
6. Factor: x​
+ y​
6​
6
7. Factor: x​
+ y​
By Gunnar Omander Solutions and Steps: 2​
1. Factor: x​
– 4 2​
2
= x​
– 2​
= (x – 2)(x + 2) 3​
3 2. Factor: x​
– y​
2​
2​
= (x – y)(x​
+ xy + y​
) 9​
9
3. Factor: x​
– y​
3​
3​ 6​
3​3​
6​
= (x​
– y​
)(x​
+ x​
y​
+ y​
) 2​
2​ 6​
3​3​
6​
= (x – y)(x​
+ xy + y​
)(x​
+ x​
y​
+ y​
) 6​
6 4. Factor: x​
– y​
3​
3​ 3​
3​
= (x​
– y​
)(x​
+ y​
) 2​
2​ 3​
3​
= (x – y)(x​
+ xy + y​
)(x​
+ y​
) 2​
2​
2​
2​
= (x – y)(x​
+ xy + y​
)(x + y)(x​
– xy + y​
) or 2​
2​ 4​
2​2​
4​
= (x​
– y​
)(x​
+ x​
y​
+ y​
) 4​
2​2​
4​
= (x – y)(x + y)(x​
+ x​
y​
+ y​
) 3​
3
5. Factor: x​
+ y​
2​
2​
= (x + y)(x​
– xy + y​
) 9​
9
6. Factor: x​
+ y​
3​
3​ 6​
3​3​
6​
= (x​
+ y​
)(x​
– x​
y​
+ y​
) 2​
2​ 6​ 3​3​
6​
= (x + y)(x​
– xy + y​
)(x​
– x​
y​
+ y​
) 6​
6
7. Factor: x​
+ y​
2​
2​ 4​
2​2​
4​
= (x​
+ y​
)(x​
– x​
y​
+ y​
) By Gunnar Omander 7.4 Practice Problems 2​
1. Factor: x​
– 9 2​
2. Factor: x​
– 64 3​
3. Factor: x​
– 27 3​
4. Factor: x​
– 125 9​
5. Factor x​
– 512 3​
6. Factor: x​
+ 216 3​
7. Factor: x​
+ 8 9​
8. Factor: x​
+ 1 12​
12
9. Factor: x​
+ y​
2​4​
4​2
10. Factor: x​
y​
– x​
y​
2​
11. Factor: x​
– 2x – 15 2
12. Factor: 64 – (x – y)​
4​
2 13. Factor: 8x​
– 32x​
4​
2​
14. Factor: x​
– x​
+ 16 4​
15. Factor: x​
+ 4 By Gunnar Omander 7.4 Answer Key 2​
1. Factor: x​
– 9 = (x – 3)(x + 3) 2​
2. Factor: x​
– 64 = (x – 8)(x + 8) 3​
3. Factor: x​
– 27 2​
= (x – 3)(x​
+ 3x + 9) 3​
4. Factor: x​
– 125 2​
= (x – 5)(x​
+ 5x + 25) 9​
5. Factor x​
– 512 3​
6​
3​
= (x​
– 8)(x​
+ 8x​
+ 64) 2​
6​
3​
= (x – 2)(x​
+ 2x + 4)(x​
+ 8x​
+ 64) 3​
6. Factor: x​
+ 216 2​
= (x + 6)(x​
– 6x + 36) 3​
7. Factor: x​
+ 8 2​
= (x + 2)(x​
– 2x + 4) 9​
8. Factor: x​
+ 1 3​
6​
3​
= (x​
+ 1)(x​
– x​
+ 1) 2​
6​
3​
= (x + 1)(x​
– x + 1)(x​
– x​
+ 1) 12​
12
9. Factor: x​
+ y​
4​
4​ 8​
4​4​
8​
= ​
(x​
+ y​
) (x​
– x​
y​
+ y​
) 2​4​
4​2
10. Factor: x​
y​
– x​
y​
2​2​
2​2​
= ​
x​
y​
(y – x)(x + y) or –x​
y​
(x – y)(x + y) 2​
11. Factor: x​
– 2x – 15 = (x + 3)(x – 5) 2
12. Factor: 64 – (x – y)​
= (8 + x – y)(8 – x + y) ​
By Gunnar Omander 4​
2
13. Factor: 8x​
– 32x​
2​
= ​
8x​
(x – 2)(x + 2) 4​
2​
14. Factor: x​
– x​
+ 16 2​
2​
= ​
(x​
– 3x + 4)(x​
+ 3x + 4) 4​
15. Factor: x​
+ 4 2​
2​
= ​
(x​
– 2x + 2)(x​
+ 2x + 2) By Gunnar Omander The Calculator ​
1 HOW WILL SOLDIERS IN FUTURE WARS USE PARABOLAS TO THEIR ADVANTAGE? Harold Rogers is​
a former lieutenant of the United States army. Since he quit the army in 1964, he pursued his dreams of becoming a mathematician. Shortly after he became employed at the Einstein Physics Institute (EPI), he decided to use his math skills to further serve his country. “What I believe in most is serving my country, no matter what obstacles fall in my path,” says Rogers. “I have always stayed true to this piece of wisdom, and it has served me well.” At EPI, Rogers has worked tirelessly on how to quickly calculate angles for shooting instruments such as cannons or catapults, or even throwing grenades. “Yes, cannons and catapults may be a little outdated, but if used correctly, they can cause massive damage to enemy bases without spending huge sums of money on bombs and fighting jets.” The main mathematical concept used when researching these objects is called a parabola. A parabola is the graph of a quadratic function that appears as a U­shaped curve on the coordinate plane. It can curve up or down, but objects such as a cannon ball would curve down. This is because the cannonball is launched upwards and is then pulled down by gravity. Here is what a parabola looks like on a graph: ​
into the form of ​
y=ax2​
+bx+c​
. A, B, and C are constants. Let’s use our cannonball again. The cannon is located at point (0,0). Let’s say the wall that you need to shoot is 10 meters away and 9 meters high. To make sure the cannonball flies clear over the wall, we will add an extra meter to be safe. Now you have two points, (0,0) and (10, 10). The third point is where the enemy is. Let’s say they are 20 meters away, at ground level as well. So, the third point is (20, 0). To get the parabola’s equation, plug each of these points ​
into the equation ​
y=ax2​
+bx+c​
. You should get three equations: 0 = c , 10 = 100a + 10b + c , and 0 = 400a + 20b + c . “Imagine a catapult positioned at the point (0,0),” explains Rogers. “The ammunition has to hit the enemy 11 meters away, but has to pass over a wall that is 8 meters high. If we don’t do it within 3 seconds, the enemy will hit us. This is where the calculations come into the picture.” “Before you can calculate the trajectory of the cannonball, you need to find the equation of the parabola,” adds Harold’s coworker, Sandy Adams. “But you cannot calculate the equation unless you have three points.” Quadratic equations fall “Most people use substitution and elimination to solve systems of three equations, but here at EPI we find using augmented matrices to be much easier.” says Adams. To learn how to use an augmented matrix, see the​
Detailed Descriptions ​
on page 4​
. To practice, flip to page 8 ​
on the Practice Pages​
. The Calculator ​
2 “Once you’ve discovered your equation, graphing it really helps. However, you really need a good understanding of how a parabola works to graph them.” says Adams. Let’s look at the vertex. This point is the turning point in a parabola. It could be either the lowest or the highest point. The axis of symmetry is the vertical line that crosses the vertex. If you fold one half of the parabola over the axis of symmetry, the lines will fall upon each other.
“If you are looking to solve for the roots of the parabola, which is what ​
x​
equals when y = 0 , then completing the square is a great method.” adds Adams. “When the numbers are rational and factorable, this method works the best. This can be good if you want to check that your cannonball will actually hit your target, not go too far or too near. Remember, when put on a graph, you and your target make up the roots of the parabola.” Again, see the ​
Practice Pages ​
or the ​
Detailed Descriptions ​
for more information about completing the square. As a reader of this magazine, you know what the quadratic formula is. If you do, then you are familiar with b2 − 4ac , which is the discriminant of an equation of the form ax2 + bx + c = 0 . When determining a discriminant, the discriminant can ​
either be positive or negative. If ​
b2​
­4ac​
is negative, then there are no real roots, which means that your equation for your cannonball cannot hit your target at ground level. “That’s one way that we know that we’ve hit a dead end,” says Roberts. “If the roots are imaginary, then we look for alternate equations.” Of course, it’s possible to graph equations with imaginary roots. See how in “Another way to write the equation of a quadratic equation is in vertex form,” says Rogers. Vertex form is written in the form Detailed Descriptions​
on ​
page 5​
. ​
y­k=a(x­h)2​
. The vertex of the parabola falls at the point (h,k). “This is useful if you know the –Reported by Bennett Rosenberg, Karina Halevy, and Lucas Nguyen. vertex, or how high the wall is. You can simplify to find ​
a​
, and then you can transform it to standard form to have your quadratic equation.” If you want more information on vertex form, see ​
Detailed Descriptions ​
on page 5​
. You can also practice more on ​
page 8 with the ​
Practice Pages​
. The Calculator ​
3 EXPONENTIALLY GROWING POPULATIONS OF BACTERIA INVOLVED IN EXPERIMENT A recent study of a population of bacteria has proved mathematician’s methods of isolating and managing exponents. Scientists used exponentiation, which is the process of raising a number to a power, to study population growth of a species of bacteria. For simplicity’s sake, we will call the species bacteria X. When bacteria X was put in a petri dish, the population was just 1. This was the 0th hour. Over the course of the next hour, the population was two. The 2nd hour, the population was 4, then 8, then 16, and so on. Scientists realized that the population doubled with each hour. After discovering this, they decided to make an equation, with the hour represented as ​
x​
. This was the function of the growth of the population: f (x) = 2x . Every progressive hour, the population doubled, so as you increase x, the function multiplies. “The math behind this is called exponentiation,” says Dr. Joe Berg, who conducted this experiment. “Basically, it’s raising a number to a power. What defines an exponential function is that for every constant you add to the independent variable, the dependent variable is multiplied by a constant. In the above example, for every one you add to x, the population multiplies by 2.” But you have to be careful with exponentiation. If we have the equations g(x) = − 2x , and h(x) =− (2x) , they are NOT the same. “Many scientists make this mistake often, but if you graph the functions, they will come up with very different results,” says Berg. “But what happens when you know the number of bacteria and have lost track of time?” continues Berg. “Here, we use a powerful technique called logarithms. Let’s say there are 2048 bacteria. Using the derived equation, we substitute 2048 for f (x) and get the equation 2048 = 2x . From here, we take what is called the base 2 log of both sides. A log is defined as if logbx = y , then by = x . So we do log22048 = log22x . Using the definition of a logarithm, we can say that 11 = log22x . From here, we use the property that says logbbx = x to find that x = 11 . That means that it has been 11 hours since we put in the first bacteria. Of course, the exponents do not have to be integers. If there are, say, 2090 bacteria, then we know that the time is some number between 11 and 12.” For further exploration of logs and exponentiation, see the ​
Detailed Descriptions on ​
page 5​
, the ​
Practice Pages ​
on ​
page 8​
, and the ​
Solutions​
on ​
page 10​
. It is evident that seemingly otherworldly mathematics concepts may not be so otherworldly after all. If even logarithms and matrices can be applied, what is next?
­­Reported by Karina Halevy, Bennett Rosenberg, and Lucas Nguyen. The Calculator ​
4 Detailed Descriptions 4.8: Solutions of Higher Order Systems By Augmented Matrices Once you know how to use augmented matrices, you can solve equations with more than three variables using the same method. For a given three­variable equation, the first step is to write an augmented matrix. Then, perform row operations in the appropriate order so that there is exactly one variable in each row whose coefficient equals one and the rest of the variables have coefficients of zero, like the following matrix: This form is called reduced ​
row­echelon form​
. The row operations are: a) multiplying an equation by a constant b) adding or subtracting equations c) swapping rows of equations. 5.2: Graphs of Quadratic Functions Since parabolas are crucial in the world of math and science, learning about their traits and graphs is just as important. Finding the vertex, and the axis of symmetry of a parabola will make graphing these functions a breeze. To make it easier to find the equation, scientists and mathematicians use both standard form and vertex form. They also complete the square to find the roots of a parabola. 2​
Standard Form: y= ax​
+bx+c 2​
Vertex form: y­k=a(x­h)​
→ vertex falls at point (h,k) How to complete the square (example): 2​
ax​
+bx+c=0 Begin with standard form 2​
x​
­10x+16 Plug in appropriate numbers 2​
x​
­10x=­16 move ​
c​
(16) to the other side of equation 2​
2​
2
x​
­10x+​
(b/2)​
=­16+​
(b/2)​
2​
Soon, you will add (b/2)​
to both sides to make a perfect square trinomial 2​
x​
­10x+​
25​
=­16+​
25 2​
(10/2)​
=25. Add 25 to both sides. 2​
(x­5)​
=9 Done! You have completed the square. You can continue to solve for x, starting by taking the square root of both sides. The Calculator ​
5 5.4: Imaginary and Complex Numbers 2​
When faced with the equation x​
= ­1, mathematicians of the past concluded that there was no possible numeric solution. That is, until 1572, when Italian mathematician Rafael Bombelli described the rules of multiplying imaginary numbers in his book L’Algebra. Today, it is accepted that ​
i​
is used to represent the “unit” imaginary number. i = √− 1 . A number like √− 25 can be expressed in terms of i. √− 25 = √(− 1)(25) = √− 1 * √25 = i√25 = 5i Similar to the previous example, any square root of a negative number can be expressed as an imaginary number. Complex Numbers are numbers found in the form: a + bi where the real number a is added to the product of the real number b and the value i, which is √− 1 . a is considered the “real part”, while bi is considered the “imaginary part”. The complex numbers a+bi and a­bi are considered complex conjugates. Imaginary numbers, just like real numbers, have their own number line and can be plotted. Since both the imaginary number line and the real number line contain 0, they intersect at the origin and can be drawn as a Cartesian coordinate system. The expression 3+5i would be plotted as such: 6.2: Exponentiation for Positive Integer Exponents The operation of raising a number to a power is called ​
exponentiation​
. For example, x3 = x • x • x , 6
and x = x • x • x • x • x • x . This leads to the definition of exponentiation for positive integer ​
exponents: ​
xn​
is equal to the product of ​
n x​
’s. It is also beneficial to familiarize oneself with some ​
nomenclature regarding exponentiation. In the expression ​
xn​
, ​
x​
is called the ​
base​
, ​
n​
is called the exponent​
, and the entire expression is called a ​
power​
. When combining exponentiation with other operations, it is important to remember the order of operations, in which exponentiation comes before multiplication, which comes before addition. The Calculator ​
6 ​
Therefore, an expression of the form ​
nxm​
would be the product of ​
n​
and ​
m x​
’s. However, if we want ​
to raise ​
nx​
to the ​
m​
, the entire base must be in parentheses, as such: ​
(nx)m​
. Note that adding a negative sign before a power is also considered multiplication, and the parentheses must also be placed around the entire base (​
­x​
or some expression of such form) in order to raise the entire expression to an exponent. In sum, we can agree that if the base of a power has more than one symbol, the entire base must be in parentheses. If not, the exponent only applies to one symbol. After this lesson, you should be comfortable using exponentiation and its definition and agreement to evaluate expressions with powers that have positive integer exponents. 6.11: Proofs of Properties of Logarithms When using logarithms, it is important to understand how to derive the properties you use. In this section, these properties will be proven in the form of formal two­column proofs. To refresh, these proofs are written with the “Statements” column to the left that contains various equations and statements, and the “Reasons” column to the right containing axioms and definitions to be used as justification for the left column. Remember that when writing proofs, you can use previous theorems already proven in your proof. Example: log 2
Prove that log​
2 = log 7
7​
Statement Reason x = log​
2 7 ​
​
Given x​
7​
= ​
2 Definition of logarithm x​
log 7​
= log ​
2 Take the log​
of each member 10​
x log 7 = log ​
2 Log of a power x = log 2
log 7 Divide each side by log​
7 ​
log 2
log​
7 2​
​ = log 7 Substitute for ​
x The Calculator ​
7 Practice Pages 4.8: Solution of Higher Order Systems By Augmented Matrices Solve these equations using augmented matrices. 1. x + y + 2z = 10 2x + 3y + 3z = 17 4x + 9y + 2z = 29 2. w + x + y + z = 10 2w + 3x + 4y + 5z = 14 2w ­ 3x + y ­ 4z = ­17 3w + 6x + 2y + z = 25 3. a + b + c + d + e = 10 a ­ b + c ­ d + 5e = ­2 2a + 2b + 3c + 3d + 3e = 27 4a + 3b + 5c + 7d + e = 53 ­a + 2b + 3c + 4d + 5e = 28 5.2: Graphs of Quadratic Functions What number must be used to complete the square? 2​
1. x​
­10x+___ 2​
2. x​
+16x+___ Use the vertex and y­intercept to write the equation in both vertex and standard form. Then graph: 3. vertex: (­5,0) y­ intercept: 7 4. vertex: (3,5) y­intercept: 3 5.4: Imaginary and Complex Numbers Solve for ​
x​
: 1.
2.
x2​
​
+ 27 = 0 2x2​
​
­ 20​
x​
+ 52 = 0 Plot the following complex numbers on a complex­number plane: 3. 1+2​
i 4. 3+​
i 5. 3​
i​
­1 The Calculator ​
8 6. ­1­2​
i 7. 2­​
i 8. 4­2​
i 6.2: Exponentiation for Positive Integer Exponents For problems 1­5, evaluate the expression. 1. a. (3 • 5)2 b. 3 • 52 2. 3
a. (3 ­ 6)​
3
b. 3 ­ 6​
3 ​ 3
c. 3​
­ 6​
3. 4
a. ­4​
4
b. (­4)​
4. a. − 2 • 33 b. ( − 2 • 3)3 c. − (2 • 3)3 5. 4 • 63 + 2 • 36 6. Evaluate x4 • x9 . What property does this suggest (what do you notice about the exponents)? 23
7. Evaluate xx20 . What property does this suggest? 8. Evaluate (x7)8 . What property does this suggest?
x2y5
7
9. Evaluate xy • ( xy4 ) . 6.11: Proofs of Properties of Logarithms For problems 1­4, do a 2 column proof of the property log x
1. Change of Base: log​
x​
= logbba a ​
​
2. Product of Two Logs: (log​
a b)(log​
​
b c) = log​
​
a c ​
3. Log with a Power for its Base: log​
= 1n log​
( bn) x
​​
( b) x
​
​
4. Base and Argument are Like Powers: log​
(​
xn​
) = log​
x ( bn) ​
b ​
The Calculator ​
9 Answer Key 4.8: Solution of Higher Order Systems By Augmented Matrices Problem 1. x + y + 2z = 10 2x + 3y + 3z = 17 4x + 9y + 2z = 29 The original matrix can be written as a 3 by 4 augmented matrix. Therefore, x = 25, y = ­7, and z = ­4. Problem 2. w + x + y + z = 10 2w + 3x + 4y + 5z = 2w ­ 3x + y ­ 4z = ­17 3w + 6x + 2y + z = 25 This can be written with the following matrix: The Calculator ​
10 Therefore, w = 1, x = 2, y = 3, and z = 4. Problem 3. a + b + c + d + f = 10 a ­ b + c ­ d + 5f = ­2 2a + 2b + 3c + 3d + 3f = 27 4a + 3b + 5c + 7d + f = 53 ­a + 2b + 3c + 4d + 5f = 28 This can be written with the following matrix. The Calculator ​
11 5.2: Graphs of Quadratic Functions 2​
2
1. 25​
­­ x​
­ 10x + ​
25​
= (x ­ 5)​
2​
2 2. 64​
. x​
+ 16x + ​
64​
= (x + 8)​
3. The vertex form is y = 257 (x + 5)2 ​
. This can be derived from the formula y − k = a(x − h)2 . We can substitute ­5 for ​
h​
and 0 for ​
k​
, giving us y = a(x + 5)2 . Now, to find ​
a​
, we can 7
substitute 0 for ​
x​
and 7 for ​
y​
, giving us 7 = 25a , which means a = 25 , which means the equation is y = 257 (x + 5)2 . To convert this to standard form, we complete the square. First, we distribute (x + 5)2 , which gives us y = 257 (x2 + 10x + 25) . When we distribute that, we get y = 257 + 145 x + 7 . The Calculator ​
12 4. In the general vertex form, we can substitute 3 for ​
h​
, 5 for ​
k​
, 0 for ​
x​
, and 3 for ​
y​
. We know this because the y=intercept is 3, so y is 3 when x is 0. This gives us 3 = 9a , which means a = 13 . Therefore, the vertex form is y − 5 = 13 (x − 3)2 . To convert this to standard form, we first distribute (x − 3)2 , which gives us y − 5 = 13 (x2 − 6x + 9) . When we further distribute this, we get y − 5 = 13 x2 − 2x + 3 , which is y = 13 x2 − 2x + 8 . The Calculator ​
13 5.4: Imaginary and Complex Numbers x2​
​
+ 27 = 0 1.
In the case of an equation containing a variable squared and a constant like the one above, there is no need to factor. Simply move the constant to the other side, then take the square root of both sides. x2​
​
+ 27 = 0 2​
x​
= ­27 x = √− 27 x = 3i√3 2. ​
2x2​
​
­ 20​
x​
+ 52 = 0 For this problem, we will use the Quadratic Formula to find the solution. −b±√b −4(a)(c)
2(a)
2
= −20±√202(2)−4(2)(52) 2
= −20±4√−16 = −20±√−1 × 16
4
= −20±i4 √16 = −20±4i
4
= −5±i
1
= ­5 ± i The Calculator ​
14 Lesson 6.2 For problems 1­5, evaluate the expression. 1. a. (3 • 5)2 i.
This is equal to the quantity of three times five squared, which means we first do three times five, which is fifteen. Then, we square fifteen, and our final answer is 225. 2
b. 3 • 5 2​
i.
This is equal to 5​
multiplied by three. Therefore, we first square five, which is 25, and then we multiply it by three, which gets us a final answer of 75. 2. 3
a. (3­6)​
i.
We always do the expression in parentheses first and then raise it to the exponent. Therefore, we have ­3 cubed, which is ­27. 3
b. 3­6​
i.
First, we do 6 cubed, which is 216. Then, we subtract 216 from 3, which gets us ­213. 3​ 3
c. 3​
­6​
i.
First, we evaluate the exponents. This leaves us with 27 ­ 216, which is ­189. 3. 4
a. ­4​
i.
The exponent only applies to the 4. Therefore, we get ­256. 4
b. (­4)​
i.
Here, the exponent applies to the entire ­4. Therefore, we get 256. 4. a. − 2 • 33 i.
First, we evaluate the exponent, which gives us ­2 times 27, which is ­54. b. ( − 2 • 3)3 i.
Here, we evaluate ­2 times 3, which is ­6, and then cube it, giving us ­216. c. − (2 • 3)3 i.
Here, we first evaluate 2 times 3, which is 6. Then, we cube it, which gives us 216. Finally, we multiply 216 by ­1, which gives us ­216. 3
6
5. 4 • 6 + 2 • 3 a. Here, we first evaluate the exponents, which gives us 4 • 216 + 2 • 729 . Then, we evaluate the multiplication, which leaves us with 864 + 1458, which is 2,322. 6. Evaluate x4 • x9 . What property does this suggest (what do you notice about the exponents)? a. Here, we can write out all of the ​
x​
’s, and we find that there are a total of 13. Therefore, using the definition of exponentiation, the answer is ​
x13
​​
. We notice that the sum of the original exponents, 4 and 9, is 13. We can then conclude that xn • xm = xn+m . 23
7. Evaluate xx20 . What property does this suggest? The Calculator ​
15 a. We can write out all the ​
x​
’s and cancel them out, leaving us with only 3 ​
x​
’s in the ​
numerator. The answer is therefore ​
x3​
. We note that the difference between 23 and 20 n
is 3, which leads us to conclude that xxm = xn−m . 8. Evaluate (x7)8 . What property does this suggest? ​
a. Here, we can write out the eight ​
x7​
’s. This gets us ​
x56
​​
. We note that 56 is the product m
n
nm
of 7 and 8, which suggests that (x ) = x . 2 5
9. Evaluate xy7 • ( xxyy4 ) . a. We can start by simplifying what is in the parentheses. Using the properties we just discovered, the expression is simplified down to xy7 • xy . From here, we can continue using the properties to find that the final answer is x2y8 . Solutions for 6.11 Author’s Note: Many of these properties of logarithms can be proven in a multitude of ways. The following solutions are just 1 of the many ways to prove the properties. log x
1. log​
x​
= logbba a ​
​
Statement Reason Assume y = log​
x a ​
​
– y​
a​
= ​
x Definition of logarithm y​
log​
a​
= log​
x b​
b ​
​
Take the base ​
b​
log of each member y log​
a = log​
x b​
b ​
​
Log of a power log x
y = logbba log x
log​
x = logbba a ​
​
Divide each side by log​
a b​
Substitute for ​
y 2. (log​
b)(log​
c) = log​
c a​
b​
a​
The Calculator ​
16 Statement Reason log b
log c
(log​
a b) = ​
b c) = ​
log a , (log​
log b Change of base (log b)(log c)
(log​
a b)(log​
​
b c) = ​
(log a)(log b) Substitution (log b)(log c)
(log b)(log c)
(log a)(log b) = (log b)(log a) Commutative Property of Multiplication (log b)(log c)
(log c)
(log b)(log a) = (log a) Simplify (log c)
a c
​ (log a) = log​
Change of base (log​
a b)(log​
​
b c) = log​
​
a c ​
Substitution 3. log​
= 1n log​
( bn) x
​​
( b) x
​ Statement Reason Assume log​
x = ​
y ( bn) ​
– n​y​
(b​
)​
= ​
x Definition of logarithm n​y​
n*y (b​
)​
= b​
Power of a Power Property n*y​
b​
= ​
x Substitution n*y​
log​
b​
= log​
x b​
b ​
Take the log base ​
b​
of both sides ny log​
b = log​
​
x b​
b​
Logarithm of a Power log​
b = 1 b​
Logarithm of the base = 1 ny • 1 = log​
​
x b​
Substitution ny = log​
x b ​
Identity Property of Multiplication ( 1n ) ny = ( 1n ) log​
x b ​
Multiplication Property of Equality y = ( 1n ) log​
x b ​
Simplify log​
= 1n log​
( bn) x
​​
( b) x
​ Substitute log​
for y ( bn) x
​​
n​
4. log​
) = log​
( bn) (​
​x​
b x
​ The Calculator ​
17 Statement Reason n​
log​
) = n log​
( bn) (​
​x​
( bn) x
​ Logarithm of a Power n log​
x​
= n ( 1n log​
x​
) ( bn) ​
b ​
Log with a Power for its Base n ( 1n log​
x​
) = log​
x b ​
b ​
Property of Reciprocals ​
log​
(​
xn​
) = log​
x ( bn) ​
b ​
Transitive Property The Calculator ​
18 What are
sets of
numbers?
Next issue:
Matrices and
Linear
Programming!
Discover graphs
of functions!
Math Monthly!
Sets of numbers, properties of
exponentiation, systems as models, and
quadratic functions
By Iris Mang, Erik Li, and Brynn
Walther Period 1
Learn f(x)
terminology
How do we
use these
concepts in
our lives?
1.1: Sets of Numbers
There are two major sets of
numbers, real and
imaginary. Real numbers are
used for “real” things like
counting and measuring,
and imaginary numbers are
the square roots of negative
numbers and are used less.
Real numbers can be plotted
on a number line, and are
broken into subsets. The
more specific subsets, like
natural and whole numbers,
were invented first, to count
things.
2.4: Graphs of Functions and Relations
Most of the graphs so far
have only had one value for y
every value of x. However,
sometimes there can be
multiple values of y for each
value of x, like in the
equation y2=x for x=4, y can
=-2, or 2. Relations like this
with multiple values of y for
each value of x, are relations
and NOT functions. A
RELATION is a set of
ordered pairs. A FUNCTION
is a relation for which there
is exactly one value of the
dependent variable for each
value of the independent
variable.
RELATION
FUNCTION
4.4: f(x) Terminology, and Systems as Models
For two equations with the
same independent but
different dependent
variables, use the f(x)
terminology to differentiate
between the functions. f(x)
means f of x or f at x, and it
gives the value of the
dependent variable f, when
the independent variable is
x. f(x) does not mean f times
x, f is not a variable, it is just
the name of the function.
With this new concept, we
can model certain
situations with the f(x)
function, and compare
them to other functions
using the same independent
variable.
5.5: Evaluating
Quadratic
Equations
6.3: Properties of
Exponentiation
When a function is
used as a model, you
must be able to
calculate y when x is
known, and x when y is
known. If x is known, y
can be found by
plugging in the x value
into the equation. If y is
known, x can be found
through the quadratic
formula.
It is possible to
simplify
expressions with
powers by
operating with
their exponents.
There are five
special cases,
called the
properties of
exponentiation,
that would make
simplifying
expressions
much easier.
1.1 Sets of Numbers Practice
1.
2.
3.
Write the definition for each
set of numbers:
a. {integers}
b. {even numbers}
c. {negative numbers}
d. {irrational numbers}
e. {real numbers}
f. {counting numbers}
g. {digits}
h. {positive numbers}
i. {rational numbers}
j. {imaginary numbers}
k. {natural numbers}
l. {transcendental numbers}
Do decimals like 4.536
represent rational numbers or
irrational numbers? Why?
What real number is neither
positive nor negative?
2.4 Graphs of Functions and Relations Practice
1. Using the given equation, determine if the graph is a
function.
a. y² = x
b. |y| = |x|
2. Tell whether or not the given relation is a function, and
explain why or why not.
4.4 f(x) Terminology and Systems as Models
Practice
1.
a.
b.
c.
d.
e.
A particular brand of car with the normal engine costs $13,000
to purchase, and 25 cents a mile to drive. The same car with a
fuel-injection engine costs $14,700 to purchase, but only 21 cents
to operate.
Let d be a variable equal to the number of miles you have driven
the car, and f(d) be the total number of dollars it costs to own
the $16,000 car. Write the particular equation for function f.
Calculate f(2,500), f(25,000), and f(250,000).
Let g(d) be the total number of dollars it costs to drive the
$14,700 car for d miles. Write the particular equation for
function g.
Calculate g(2,500), g(25,000), and g(250,000).
When does the total cost of owning the car with the normal
engine equal the cost of owning the car with the fuel-injected
engine?
2. Let f(x)=7x+19
g(x)= 2x2+x+5
Evaluate:
g(f(g(2)))
5.5 Evaluating Quadratic Equations Practice
1. If f(x) = 3x2+2x-10, find
a. f(100)
b. x, if f(x) = -9
c. the x-intercepts
2. If f(x) = -2x2+6x-4, find
a. f(-15)
b. x, if f(x) = -5
c. the x-intercepts
3. If f(x) = 4x2-7x+1, does f(x) ever equal:
a. 15?
b. 1?
c. -10?
d. Write c. in form a+bi
6.3 Properties of Exponentiation Practice
1. Evaluate the following five properties of exponentiation
a. xaxb=
b. xa/xb=
c. (xa)b=
d. (xy)a=
e. (x/y)a=
2. Evaluate the expressions
a. -(-3)5=
b. -10(-10)3=
c. ((-32)2)3=
3. Simplify the expressions
a. (15x)(2x)2=
b. 6k11m2/72k8
c. (2x1066)10/(4x1184)9
4. Solve for x
a. 43x+1=2x-1
b. 16x=(23)(84)
How do we use this in real life?
In 4.4, we learned how to use systems as models. In the real world,
this can be used for business owners to calculate how much money
they either make or lose in a year. If a restaurant needs to pay a
property tax per year, and needs to buy products for each item in
their meals, they can set up the total money as a function, f(x)=ax+b.
f(x) being the total amount of money they owe, a being the the
number of food items they purchase, x being the price for each food
item, and b being the amount for property tax. If a restaurant pays a
property tax of 10,000 a year, and only has one item on their menu
purchased at $5, you can set up the function f(x)=5x+10,000. Using
this function, business owners can simply plug in how many times
they purchase that food, and they can find the total amount of
money they lose in a year. However, most restaurants buy more than
just one item from others stores, and so their function would be
extended to include more separate amounts and costs for food.
Appendix/Solutions 1.1
1.
2.
3.
Definitions:
a. {integers} - whole numbers and their opposites
b. {even numbers} - integers divisible by two
c. {negative numbers} - numbers less than 0
d. {irrational numbers} - cannot be expressed exactly as a ratio of two
integers, but are real numbers
e. {real numbers} - have points on the number line
f. {counting numbers} - positive integers or natural numbers
g. {digits} - numbers from which the numerals are made
h. {positive numbers} - numbers greater than 0
i. {rational numbers} - can be expressed exactly as a ratio of two integers
j. {imaginary numbers} - square roots of negative numbers have no points
on the number line
k. {natural numbers} - positive integers or counting numbers
l. {transcendental numbers} - irrational numbers that cannot be
expressed as roots of integers
Do decimals like 4.536 represent rational numbers or irrational numbers?
Why?
a. Numbers like 4.536 are rational because it can be represented as a ratio
of two integers, ie 4536/100.
What real number is neither positive nor negative?
a. 0 is neither positive or negative, and it is a real number.
Appendix/Solutions 2.4
1.
Using the given equation, determine if the graph is a function.
y² = x
- no, because if you substitute 9 for x, y can be either 3 or -3,
so it does not fit the definition of a function
|y| = 2x - no, because if you substitute 2 for x, then y can be either 4 or
-4, so it does not fit the definition of a function
2. Tell whether or not the given relation is a function.
a. This is a function because every value of x has only one value of y
b. This is a function because every value of x has only one value of y
c. This is not a function because a value of x has infinite values of y
d. This is not a function because some values of x have two values of y
Appendix/Solutions 4.4
1.
A particular brand of car with the normal engine costs $13,000 to
purchase, and 25 cents a mile to drive. The same car with a fuelinjection engine costs $14,700 to purchase, but only 21 cents to operate.
a. f(d)=0.25x+13000 Set the total cost for owning the car equal to the
amount per mile in dollars, .25, times
d, the number of miles, plus the constant 13,000 for purchasing the car.
b. f(2500)=13625 f(25000)=19250 f(250000)=75500 Substitute d for 2500,
25000, and 250000 to get the total cost.
c. g(d)=0.21x+14,700 Similar to part a, set the amount per mile in dollars
multiplied by the number of miles d, and add the constant value of
14700.
d. g(2500)=0.21(2500)+14700=15225 g(25000)=0.21(25000)+14700=19950 g
(250000)=0.21(250000)+14700=67200 Substitute 2500, 25000, and 250000
for the number of miles to get the total cost.
e. 0.25x+13000=0.21x+14700
0.04x=1700
x=42500 miles
Since f(d)=.25x+13000 and g(d)=.21x+14700 and they are asking for how many
miles with each engine for the cost to be equal. you can set the two equations
equal to each other by transitive property. .25d+13000=.21d+14700. Separate
the like terms on each side, 0.04d=1700, and divide 0.04 by both sides to get
42500.
Appendix/Solutions 4.4 Continued
2. f(x)=7x+19
g(x)= 2x2+x+5
g(f(g(2)))
g(f(2(22)+2+5)
g(f(15))
g(7(15)+19)
g(124)
g(2(1242)+124+5
g(124)=30, 881
Use the equations given and solve the problem one step at a
time. First do evaluate g(2), and plug 2 in for every x in the g(x)
function. g(2)=15. You now have g(f(15)). Next evaluate f(15) and
similarly plug in 15 for x in the f(x) equation. f(15)=124. You
know have g(124). Lastly, plug in 124 for every x in the g(x)
once again to find g(f(g(2)))=30,881.
Appendix/Solutions 5.5
1.
2.
If f(x) = 3x2+2x-10, find
a. f(100)
=3(100)2+2(100)-10
=1030190
b. x, if f(x) = -9
3x2+2x-1=0
x=⅓, -1
c. the x-intercepts
3x2+2x-10=0
x=1.522, -2.189
If f(x) = -2x2+6x-4, find
a. f(-15)
=-2(-15)2+6(-15)-4
=-544
b. x, if f(x) = -5
-2x2+6x+1=0
x=-0.158, 3.158
c. the x-intercepts
-2x2+6x-4=0
x=1, 2
3. If f(x) = 4x2-7x+1, does f(x)
ever equal:
a. 15?
4x2-7x-14=0
(-7)2-4*4*-14=-224
Yes.
b. 1?
4x2-7x=0
(-7)2-4*4*0=49
Yes.
c. -10?
4x2-7x+11
(-7)2-4*4*11=-127
No.
d. Write c. in form a+bi
4x2-7x+11=0
-7+-127/ 8, -7- -127/ 8
-⅞ + 127i/8, -⅞ -127
i/8
Appendix/Solutions 6.3
1. Evaluate
a. xa+b
b. xa-b
c. xab
d. xaya
e. xa/ya
2. Evaluate
a. -(-3)5 = -(-243)
= 243
b. -10(-10)3= -10
(-1000) =
10000
c. ((-32)2)3= (-3)12
= 531441
3. Simplify
a. (15x)(2x)2=(15x)
(4x2)=60x3
b. 6k11m2/72k8=
(k11-8m2)/12
c. (2x1066)10/(4x1184)
9
=x10/(2x118)9
=1/x1052
4. Solve for x
d. 43x+1=2x-1
26x+2=2x-1
6x+2=x-1
x=-(⅗)
e. 16x=(23)(84)
16x=(23)(212)
24x=215
x=15/4
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1.4 – Polynomials
Explanation:
Practice Problems:
A polynomial is an expression that only uses the
operations of addition, subtraction, and
multiplication of variables. But, any mathematical
operation may be done to the constants. Note that
this means that variable exponents can only be
non-negative integers.
If the following expressions are polynomials, name
them. If they are not, explain why.
1.
2.
Here is how to name a polynomial by the number 3.
of terms:
• If it has 1 term, it is called a monomial.
4.
• If it has 2 terms, it is called a binomial.
• If it has 3 terms, it is called a trinomial.
• If it has 4 or more terms, there is no special
name for it.
The degree of a polynomial is the same as the
degree of the monomial with the highest degree.
This is how to name a polynomial based on the
degree of the polynomial
• If the degree is 0, it is constant.
• If the degree is 1, it is linear.
• If the degree is 2, it is quadratic.
• If the degree is 3, it is cubic.
• If the degree is 4, it is quartic.
• If the degree is 5, it is quintic.
• If the degree is higher, say nth degree.
Then, combine the 2 names with the degree
coming first and the number of terms coming
second. So,
would be a cubic binomial.
If 0 is a coefficient of a term, then the term has no
degree.
x3 is the same as x*x*x, so the variable is
undergoing multiplication, a permitted operation.
x/3 is the same as x * ⅓, so this is another
permitted operation.
Answers are on the next page.
Answers:
1. No because it is taking the root of a variable.
2. No because it is dividing by a variable.
3. Constant monomial. 1 term, 0th power.
4. Quadratic trinomial. 3 terms, 2nd power
3.4 – Equations of Linear
Functions From Their Graphs
Explanation:
Practice Problems:
You can find the equation of almost any line by
substituting the given information into the various
linear function forms you already know.
Write the described equation. Form only matters if
specified.
1. Contains (3, 4) and has a slope of ⅔
• Point slope form:
• Slope intercept form:
• Intercept form:
2. Contains (2,1) and is parallel to the graph of
, where a is the x-
intercept and b is the y-intercept
3. Contains (6, 8) and is perpendicular to the graph of
• Standard form:
If you are not given the slope (which nearly all of
these forms require) you can use the formula
4. Has a y-intercept of 4 and an x-intercept of 1
to find the slope before substitution. If you are
5. Is vertical, and contains (12,16)
missing one piece of information from any form,
simply substitute one of the given points into x and y
6. Transform
to standard form and slopeand solve for the remaining piece of information.
Vertical lines have the equation x = c. Horizontal lines intercept form.
have the equation y = c. In these cases, c is the x or
y coordinate (with respect with what c is equal to) of a
Answers are on the next page.
point that is on the line.
Recall that lines are parallel when their slopes are
equal, and perpendicular when their slopes are the
negative reciprocals of each other. (ex:
)
3.4 – Equations of Linear
Functions From Their Graphs
Answers:
1. Use point-slope form with substitution:
2. Using slope-intercept form,
has a
slope of 10. Then use point-slope form with same
slope for parallelism.
3. Using slope-intercpet form,
has a
slope of -2. Then find that the slope is ½ by taking the
negative reciprocal. Use point-slope form.
4. Use intercept form.
5. The line is vertical, so set the value of x to the x
coordinate in the point.
6. Multiply both sides by 4.
subtracting 2x from both sides.
. Isolate y by
4.10 – Systems of Linear Equalities
Explanation:
Practice Problems:
Here's how to graph an inequality:
Graph the following systems of inequalities:
First, graph the line as if there was an equals sign
instead of an inequality symbol.
1.
Remember that the line must be dotted unless the
2.
inequality sign used has a line under it, the line is not
actually included in the solution. (≤, ≥)
To find the side of the line that the inequality is true
for, pick an easy test point ((0,0),(0,1), etc.) and
3.
substitute it into the inequality. If it comes out true,
shade the side the point is on. If it’s false, shade the
other side.
If you have the y term with
no coefficient isolated (y =
_____) then you can simply
use the sign of the
inequality to determine
the shading. > or ≥ means
above the line. < or ≤ mean
below the line.
Remember to double shade the intersection of
multiple inequalities, or shade only that section.
Tip: Instead of shading for every line, draw arrows
from them to indicate which side is true, and shade
only the answer at the end of the problem.
4.
Answers are on the next two pages.
4.10 – Systems of Linear Equalities
Answers:
1.
2.
3.
4.
5.6 – Equations of Quadratic
Functions from Their Graphs
Explanation:
Practice Problems:
Thanks to the work of Galileo in the seventeenth
century, it is now possible to measure and track the
change in the height of a projectile in a quadratic
relation with time or distance, where the height of the
projectile is determined by the distance it has
traveled, on account of the forces of gravity acting on
the projectile’s location.
1. Find the particular equation of the parabolic
function which passes through the given points of:
(3, 50), (5, 83), and (10, 8).
2. Find the particular equation of a parabola
containing (9, 40) as its vertex, and a point at (18,22).
Be Clever!
A parabola can be determined by the providing of any
three non-linear points on the path it travels through,
but only two points are needed to find the parabola’s 3. Find the particular equation of the parabolic
equation if one of the points given is the parabola’s function that passes through the given points
(5,108),(-4,-45),(4,35)
vertex.
One is able to plug a given value of x and y into the
formula
, where a, b, and c are Answers are on the next page.
real numbers, and a is not equal to zero. Soon after
Galileo proved this idea, he falsified Aristotle’s
ancient explanation of forces on parabolic motion,
and people from baseball players up at bat to circus
performers about to fire a person out of a cannon
have continued to use his principle to improve the
success and accuracy of their individual world.
Mathematical concepts such as Galileo’s
investigation involving the determining the equation
and path of a parabola are not only able to be
examined and discussed in the classroom, but are
also able to determine the behavior of real-world
entities.
To find the equation of a parabola given information
about its graph, simply substitute the given
information into one of the quadratic forms.
•
where a, b, c are
constants.
•
where k is the y
coordinate of the vertex and h is the x
coordinate.
Then solve for the remaining information.
5.6 – Equations of Quadratic
Functions from Their Graphs
Answers:
1. First substitute the points to get 3 equations.
25a + 5b + c = 108
16a + -4b + c = -45
16a + 4b + c = 35
Then solve the system using any of the methods
taught in previous chapters.
This yields that c = -117, b = 10, a = 7.
Now put these back into the general form, yielding:
y = 7x2 + 10x - 117
2. Substitute the points into vertex form.
22 - 40 = a(9 - 18)
-18 = 81a
a=
y - 40 =
(x – 18)2
3. First substitute the points to get 3 equations.
25a + 5b + c = 108
16a - 4b + c = -45
16a + 4b + c = 35
Then solve the system using any of the methods
taught in previous chapters.
This yields that c = -117, b = 10, a = 7.
Now put these back into the general form, yielding:
y = 7x2 + 10x – 117
6.14 – Exponential and other
Functions as Mathematical Models
Explanation through an example:
Take the root-14 of both sides
In 1630 The Massachusetts Bay Company, a large b = 1.203241
trading company consisting mainly of English
f(x) = 1.2 x 1.203241x
Puritans, had arrived in present-day North America in
order to establish a new religious community. One of
the first ever cities established by this group was the b. 1865 is 22.5 decades since 1640. Plug in 22.5 and
we get 7711. So the population would be 77,100
city of Boston in the northeastern region of the
present-day United States. By 1640, the population of
the city totaled approximately 1,200 residents, and by c. 500 = 1.2 x 1.203241x
1780 the population totaled approximately 16,000
416.66 = 1.203241x Then take the log base
residents. Assume that the population of the city
1.203241 of both sides.
varies exponentially with time and has continued to
x = 32.6036, which means about 326 years are
increase by that rate.
needed to surpass 500,000, which puts the date at
a. Derive the particular equation expressing the
1966.
population of Boston in terms of time (decades since
1640). When solving, express your answers in terms
d. 1622 is -1.8 decades. Plug this in. 1.2 x 1.20341-1.8
of thousands (i.e. 1200 would equal 1.2).
b. Toward the beginning of 1865, the Civil War of the This gives 0.8598 thousand people, which is 860,
rounded.
United States had ceased. However, much of
Boston’s population remained intact, as no battles
were fought in the entire state of Massachusetts
during this war. Calculate the predicted population
according to the mathematical model. Round to the
nearest hundred.
c. In what year will the model show the population
reaching 500,000 residents for the first time?
d. 1640 was the date of the first time the population of
Boston was recorded, but 1622 was the first year
when colonists began to travel over individually to
New England. Predict what the population at that time
would have been. Round to the nearest whole
number.
a. Use the exponential form f(x) = abx
Substitute the two given points of information.
f(0) = 1.2 = ab0
a= 1.2
f(14) = 16 = 1.2b14
16/1.2 = b14
6.14 – Exponential and other
Functions as Mathematical Models
were able to retrieve a sample of 40 grams of
holmium-166 to help aid future patients on March
1. The Richter Scale is an archaic device that was
22nd, 2015 at 2:30 PM. One half life (meaning after
used frequently during the twentieth century to
every half-life interval, exactly one-half of the
measure the intensities and sizes of earthquakes.
remaining substance is decomposed into more stable
The device analyzes the greatest distance that the
seismic (earthquake) waves vibrate (also known as elements) of this isotope is about 27 hours. The final
amount remaining of a radioisotope such as
amplitude). Because the amplitude of earthquakes
can be so large, Charles Richter created the Richter holmium-166 can be modeled after the equation:
Scale to read from the base 10 log of the given
amplitude of the waves.
Practice Problems:
Given:
The 1904 Maine earthquake was recorded as being
5.1 on the Richter Scale.
The Nepal earthquake disaster of April 2015
registered a 7.8 on the Richter Scale.
The Chilean earthquake of 1960, considered to be
one of the most deadly and destructive natural
disasters of all time, registered a 9.5 on the Richter
Scale.
where F equals the final amount remaining of the
initial amount of the chemical, A0 = the initial amount
of the substance, t = the time elapsed since the
retrieval of the chemical in hours, and h is the span of
one half-life in hours.
a. Find how much of the original 40 grams of
holmium-166 would be left after three days since the
receiving of the tracer (medicinal aid). Round to the
nearest tenth of a gram.
a. How many more times intense was the Chilean
earthquake from the Nepalese earthquake of this
year? Than the Maine earthquake? Remember, a
b. How much time would have elapsed if the final
logarithm is an exponent. Round your answers to the amount of holmium-166 had a mass of 3.75 g?
nearest whole number.
Round to the nearest minute.
b. In 1964 a powerful earthquake hit Alaska that
c. If the initial amount given to the hospital had been
caused roads and buildings to become pieces of
remnants from a larger original supply that had a
rubble. It is calculated to be approximately 25 times mass of 220g, when was the mass of the original
more intense than the earthquake of Nepal this past larger supply first recorded? Round to the nearest
year. What was the earthquake recorded as on the minute of the day, and be specific.
Richter Scale? Round to the nearest tenth.
Answers are on the next page.
2. Holmium-166, an unstable isotope, is used very
commonly in hospitals to detect the presence and
development of cancer in the human liver. At a
certain hospital in Metropolitan Town, a few doctors
6.14 – Exponential and other
Functions as Mathematical Models
Answers:
1.
a. log(amplitude) = 9.5 for the Chilean, meaning the
amplitude was 109.5 and the Nepal was 107.8 and the
Maine was 105.1. So, 109.5/107.8 = 101.7 = 50 times
more massive, and 109.5/105.1 = 104.4 = 25119 times
more massive.
b. Nepal earthquake was 107.8. Times 25 puts the
Alaskan one at 1577393361.2 amplitude. Take the log Then do some simple evaluating again and get x =
base 10 of that and we get a 9.2 on the Richter
-66.4047 hours. Converting to days and minutes
Scale.
gives 2 days, 18 hours, and 24 minutes. Counting
2.
backwards from March 22nd at 2:30 PM then gives us
March 19th at 8:06 PM.
a. First plug in information we have into the given
equation.
Evaluating gives F = 6.299. round to the nearest
gram and we get 6.3 grams.
b. First plug in information we have into the given
equation.
Then solve it:
Finally do some simple evaluating and get x = 92.206
hours. 0.206 * 60 is about 12 minutes. So 92 hours
and 12 minutes, or 3 days, 20 hours, and 12 minutes.
c. Plug the stuff into the equation again, but this time
we will get negative time since we are predicting
backwards.
By Josh Hejna, Jack Lilygren, and Henry Ying.
December 7, 2015
DIVISION TODAY
The Final Review Issue
Read about how to solve simple equations and inequalities and how to complete linear systems
with two variables with augmented matrices and about significant digits and scientific notation
inside! This includes detailed explanations of each topic, practice problems, and even answers and
solutions! Don't miss your chance to study for your final with this issue of Division Today. If you
would like to receive monthly issues like this, just go to divisiontoday.com to subscribe!
By: Romella Sagatelian, Ella Milliken, Kevin Gao
1
1.5­ Equations Expression vs. Equation! Expressions and equations are not the same. Similar, but not the same. An expression is when the value of the variable is know, and the overall value is being found. However, an equation is finding out the value of the variable when the value of the overall expression is already known. For example, an expression is 3x + 7, and x=5 However, an equation is 3x + 7= 22, where the variable value is unknown Justification All Multiplication, Division, Addition and Subtraction in equations is justified by the Property of Addition and Multiplication respectively. Also included, Square roots and variables to powers. How to solve them! To solve an equation is to find the value of the variable in said equation, and writing its solution set. The best way to solve an equation is by manipulating the terms to equal “x= a constant”. Before finalizing the solution set, the values should be substituted back into the equation in order to make sure you didn’t make any mistakes. The solution is not complete unless the answer is written as S={answer}. Also, in order for a number to be a solution, it ​must be in the domain of the variable. Example 1: 3x ­ 5 = 16 1. Add the 5 to both sides in order to cancel it 3x = 21 2. Divide both sides by 3 in order to get the x alone, transferring it into the “x= a constant” form. x = 7 3. However, we are not finished! We must put it into S={ }! S= {7} Side note: this solution would only be valid if it was included in the domain of x. For example, if x ϵ {positive integers} this solution would be valid, but if x ϵ {negative irrational numbers}, this solution set would be empty. Example 2: x ­1 = x 1. This is an impossible equation. When placed into the “x= a constant” form, it is: x = x+1, which is impossible, The solution set is ​empty​, so it would be written as: S= ø or S= { } Example 3: (x + 4)(2x ­ 5) = 0 1. Because these two terms are multiplied, only one of them must equal zero. Therefore, we can solve them separately, each equaling zero. x + 4 = 0 and 2x ­ 5 = 0 2. First, add/ subtract the terms, keeping in mind order of operations. x = 4 and 2x = 5 3. The first equation is finished, as the form is complete. However, the second one is not. In order to finish it, just divide both sides by 2 to get the x alone. x = 4 and x = 5/2 4. Finally, put the answer into S= { }. S = {4, 5/2} Example 4: |2x + 2| = 8 1. This is an absolute value equation, which means the value inside of the bars is turned into a positive number, no matter if it was negative or positive. Therefore, x can be negative or positive, as long as the equation is equal to x. So, create two equations, one equal to 8 and ­8. 2x + 2 = 8 and 2x + 2 = ­8 2. Then, use order of operations to solve the equations. 2x + 2 = 8 2x + 2 = ­8 2x = 6 2x = ­10 x = 3 x = ­5 3. Finally, put the answer into S= { }. S = {3, 5) Transformed equations Some equations have an irreversible step, where a previously valid answer in no longer true. This new solution set is called the extraneous solution, as it only satisfies the transformed equation, not the original one. Take this for example! (x ­ 3)(x ­ 7) = 0 1. When both sides are divided by (x ­ 3), a transformed equation is formed. While it is still valid, a solution is lost. (x ­ 7) = 0 2. The extraneous solution would be​ S = {7}​, but the original would be ​S = {3, 7} Practice Problems 1. x​2​ = 7 {irrational numbers} 2. x​2​ = 64 {negative numbers} 3. |9x ­ 17| = 10 4. (2x ­ 20)(3x+15)= 0 {real numbers} 5. Given the equation (x + 20)(x ­ 9)= 10 a. Write the solution set. b. Divide by ( x + 20) and write the solution of the transformed equation. c. Is the transformed equation equal to the original? Why? 1.6 ­Inequalities What is an inequality? An inequality is formed when the “=” sign in an equation is replaced by one of the order signs: <, >, ≤, or ≥. The solution set of an inequality is all the values with make it true. A solution set can contain an infinite number of solutions, so a number line is normally drawn. In order to reach a solution set, one must transform it into a simpler inequality. Solving and Graphing Take the inequality 3x ­ 5 > 16. 1. First add 5 to each side, to get: 3x > 21 2. Next, multiple by ⅓ , getting: x > 7 3. Finally, plot this inequality on the number line. Knowing that the”greater than” sign means an empty dot on 7, draw this, which a line to the right of it. Negative Numbers and Inequalities When dealing with negative numbers in inequalities, it is important to remember when to flip the inequality and when not to. When subtracting number, the inequality stays the same. However, when multiplying both sides by a negative number, the inequality switches. Take the inequality ­2x + 5 ​>​ 19 1. Add ­5 to both sides, creating the inequality: ­2x ​>​ 14 2. Next, divide both sides by ­2, but don’t forget to ​reverse the inequality. ​ The final inequality is: x ​<​ ­7 3. Last but not least, graph this on a number line. Don’t forget to use the solid dot, as it is “less than or equal to”. Some problems will have a given domain. For example, in the above equation, the domains of {real numbers} and {negative integers} would apply, while {positive numbers} would not. Multiplication Property of Order If x < y, then: xz < yz, if z is positive, xz > yz, if z is negative, and xz = yz, if z is zero. There are transformations pertaining to ​Absolute Values: If c is a non­negative constant, then |expression| > c is equivalent to: expression > c ​or e
​ xpression < ­c |expression| < c is equivalent to: expression < c or expression is > ­c This is justified, as an absolute value can be thought of as the distance between c and 0. Therefore, there could be a negative and positive distance from zero, so there must be two equalities formed. Example 1: Graph: ­3x ​>​ 27 1. First, multiply by ­⅓, then switch the order sign. x ​<​ ­9 2. Next, graph, keeping in mind to use a solid dot. Example 2: Graph the solution of: 7 ​<​ 2x + 5 < 11 1. Add ­5 to all three parts of the equation. 2 ​<​ 2x < 6 2. Next, divide all three numbers by 2. 1 ​<​ x < 3 3. Finally, graph. Because x is between 1 and 3, the line would be there as well. Use a filled point on the 1, and a non­filled point on the 3. Example 3: Graph: |c| > 5 1. Using the Absolute Value transformations given, c > 5 or c < ­5 2. Next, graph the two equations, both with non­filled dots. Practice Problems 1. 7 ­ 3x ​>​ 14 2. x + 3 < 5 {Positive numbers} 3. 5 ­ 4x < ­3 {Negative Numbers} 4. 6 < |2x + 4| ​<​ 8 Complete with the domains of: a. {Real Numbers} b. {Integers} c. {Positive Numbers} Ch. 4-7: Solution of Second-Order Systems by Augmented Matrices
Augmented matrices are another way to solve linear equations. These are
advantageous because in the real world you can use computers to solve them, instead
of just writing out the equations and solving them by hand. Matrices are also used in
many other places in computer science. Also, when you make a data table, the
information is represented as a matrix.
Let's say that your given system is
3x + 4y = 20
5x + 6y = 34
The coefficient of x can be made zero by multiplying the second equation by 3 and
adding -5 times the first equation.
0x - 2y = 2
The 0x term is left like that on purpose, so next all you have to do is divide the equation
by -2 and you get the value of y.
0x + y = -1
This means that y equals -1
You can write this system using matrices, like this
This is called an augmented matrix. The coefficients and constants are written in the
exact order that they are found in the equation. The x coefficients are in the first column,
the y coefficients are in the second column, and the constants are in the third column.
Operations can be performed on the rows of this augmented matrix just like you would
do it if you were solving it normally.
In this case, you would multiply the first row by -5 and the second row by 3 and then
adding them together. After you do this, substitute the sum of the two rows for the
second row. The first row will stay the same.
The big arrow shows the transition of the first matrix to the one after you have applied
arithmetic. The -5R1 + 4R2 R2 shows that you're multiplying the first row by -5,
1
Ch. 4-7: Solution of Second-Order Systems by Augmented Matrices
multiplying the second row by 3 and then adding them and then substituting the answer
for row two.
The second row can now be divided by -2. You should continue writing
R2/-2
R2
Now you know that y = -1 because the second row says 0x + 1y = -1
To find x you can multiply the second row by -4 and add it back to the first row. This will
make the y coefficient equal 0 because you're doing 4y + (-4y) and it will make the x
coefficient stay the same because you're doing 3x + 0x. After you have isolated the x
and its coefficient, just divide the whole first row by the coefficient of x.
R2/-2
R1/3
R2
R1
Now you know that x = 8 and y = -1 which can be written as (8, -1). You have
successfully solved a system of equations using augmented matrices.
Practice Problems: Solve these with augmented matrices
1. 7x + 3y = -5
2. 8x + 7y = 4
5x + 11y = 23
3x + 5y = -27
3. 8x - 3y = -19
-14x + 5x = 32
4. 16x + y = 11
-4x - 5y = 2
2
-4R2 + R1
R2
6.5 Fractional Exponents Raising numbers to a power with a rational exponent is a key skill in solving logs. Simple numbers that are small and give a good answer should be done mentally and quickly. Generally, an approach to these type of problems are recognizing a large number is smaller number to a power, such as 8 being equal to 2​3​. An example of a problem using this method is shown below: 729​1/3​ = (9​3​)​1/3 ​, which is then equal to ​9​. Radical expressions can also be expressed with roots. The key to solving it is to change the roots to fractional exponents. This is shown below: 4​
√128/​6​√32 = 128​1/4​/32​1/6 ​ = (2​7​)​1/4​/(2​5​)​1/6​ = 2​7/4​/2​5/6​ = ​211/12 ​
Whenever opposite powers are equivalent, such as √(2)​2​, then the answer is equal to itself, which is 2 in this case. However, this does not apply to negative numbers: √(­2)​2​ = √4 = 2, which is not equal to ­2. Example Problems Example 1: 64​x​ = 8 Since 8 is the square root of 64, and the square root symbol is equivalent to a power of ½, the answer is ​½​. Example 2: 625​x​=1 This would be a very hard problem to solve. However, we know that any number to the zero is equal to one. Therefore, the answer is ​0​. Example 3: 36​x​=0 There is no number that will raise 36 to a power such that the answer is 0. If x were to be 0, the answer would be 1. Therefore, the answer is ​undefined​, or no solution. Example 4: In this case, we would first solve the inside of the equation, giving us (x​5/2​y​7/3​) to the ­1/7 power. Then we remove the negative from the ­1/7, giving us the reciprocal of (x​5/2​y​7/3​). After applying 1/7, we find the answer is ​x5/7​
​ y​1/3​. Practice Problems 1. 216​2/3 2. (­64)​3/2 3. 243​­4/5 4.
5. 6 x*x
​ ​36​2 =
6 3x
√
7
72x 49 ×343 −2x
−3x
6.6 Scientific Notation Scientific notation is a special way of writing large numbers. It simplifies a big value by making it smaller. For example, instead of the large number 3,456,721, we can use 3.456721 × 10​6​, subsequently making it easier to work with this large value. Estimating Scientific notation lets you approximate large answers quickly. For example, if someone were to ask you to multiply two large numbers like 4983 and 9984, you would use scientific notation to round it to 50 million. Characteristics Numbers in scientific notation always contain two parts: a number between 1 and 10, not including ten but including one, a multiplying ten to a certain power. For example, 1056 would be written as 1.056 * 10​3 0.98 would be written as 9.8 * 10​­1 The two parts of a number in scientific notation have special names. 7.6 × 10​5 In the example above, the factor on the left, 7.6, is called the mantissa. The factor on the right is a power of 10, and the power, in this case, 5, is called the characteristic. From this number, it is easy to tell that the mantissa is much less significant than the power of 10 that “characterizes” the number. Significant Digits The number of significant figures in a number are the number of digits that are “important”, or carry meaning to the number. For example, 2.574 has four significant digits, since all the numbers are important. The number 1000 has 1 significant digit, since the other 0’s are placeholders, but 1000.0 has 5 significant digits, since the extra 0 after the decimal shows exact accuracy to the tenths. 0.00035 has two significant digits, while 0.000350 has three. Example 1: How many significant digits are in the number 3007? Since 3 and 7 are “interesting” numbers, and the two zeroes are in between, we must count the zeroes as significant digits. Therefore, there are ​4​ significant digits. Example 2: How many significant digits are in the number 560? The zero is a placeholder, so therefore there are ​2​ significant digits. Example 3: How many significant digits are in the number 560.0? The 0 and the end shows how precise the number is, meaning the other 0 is also important. Therefore, this number has ​4​ significant digits. Exponent Sign in Scientific Notation If you make one factor smaller by moving a decimal, you must make the other side larger, so that the number still has the same value as when it started. For example, the number 162 × 10​5​ would be equivalent to 1.62 × 10​7​. For each decimal place moved, the characteristic of the power of 10 must be changed by 1 or ­1. Example 1: (6 × 10​24​)(4 × 10​­15​) = 24 × 10​9​, which is equal to 2.4 × 10​10​. Precise vs. Accurate 8
Say we have a solution of 6.2×10 = 1.77142857*10​4​, which is around 1.8*10​4​. 3.5×10 4
When we round off a number, we want to round it to the number of significant digits in the least accurate factor of the numerator and denominator. In this case, we would round it to 1.77*10​4​. Accurate does not mean precise, however. The two words have completely different meanings. For the number above, 1.77142857 is much more precise than 1.78 because it fits more closely in the comparison of the two numbers. However, accuracy refers to how closely an answer fits a real world quantity being measured, so 1.8 may be more accurate than 1.77142857. Scientific Notation in Real Life Scientists often use scientific notation to simplify very large or small numbers, such as dealing with small numbers in cells or large numbers in outer space. In this case, large numbers such as 5.97*10​24​ are simplified. Aside from scientists, engineers and managers use scientific notation to manage numbers and save up space. Practice Problems 1. Light travels around 3.0 * 10​8​ miles per second. How far does light travel in a week? 2. (7.76*10​13​)/(3.2*10​­5​) 2
­7​
3. Electric power is modeled by the equation VR , where v is equal to 5.66*10​
, and R is equal to ­27​
800.89*10​ . Find the solution in terms of scientific notation. 4. 1000*1993*0.1993*10 Feature story ­ Nigerian Fraud caught by Financial Genius A Nigerian banker was recently caught by an Indian finance advisor by the name of Div E. Dand for tricking customers into switching compound and simple interest in his deals. Div had been helping some of his African friends with some calculations when he realized that their IRA, which was supposed to be compounded, was not rising at an increasing rate. While most would assume that this was a mistake, Mr. Dand investigated the case further. His finding shocked the nation. The Hawaiian­born Akela Kaiholo had moved to Nigeria after getting a Ph. D. in Finance and had been scamming people for an estimated seventeen years. When questioned, he confessed to all of his crimes and tricks, claiming that his parents needed money and his bank was not successful enough. Akela had scammed over thirty thousand people over the course of his career, and his bank was severely endangered. “All I wanted to do was to help my sister. She lost two legs and an arm to a vicious coral attack while surfing,” said the traumatized Akela. But how did the marvelous finance aid discover this? When approached by thankful natives and news reporters, he said, “All you have to do is keep a cool head and think. I noticed the pattern that resembled simple interest, and I got to work. I guess the natives were never properly educated about this.” After some time discussing this amazing breakthrough, Div agreed to show people his work. His math expressed his fluency in the arts of logarithmic operations, understanding of linear functions, and ability to find an exponential function. Throughout his career, Div E. Dand had saved up quite a large sum of money. Back in 2010, he deposited $20,000 into Akela’s bank. With a “compounded” interest of 5% annually, Div E. Dand realized that by 2012, he should have $22,050 in his bank account. However, he saw that in 2011, he had $21,000 which was normal. However, in 2012, he only had $22,000 in his account! Because of his math genius, he immediately knew that his interest was not compound. Using the data points he had from his account balance at different years, he found that his bank account grew ​
linearly​
, indicating the interest was non­compounded. Akela Kaiholo’s trial’s jury selection will begin early next year. The judge assigned to the case says that his punishment, very ironically, will be exponentially related to the amount of people he cheated. 1.7 ­ Properties Provable by Axioms Using the axioms, a multitude of ​
properties​
, or rules in math, can be proved. Unlike axioms, the vast majority of them can be proved. Axioms are the most basic “rules” of math. They are what everything is deduced from, including properties and theorems that are used in everyday algebra and math. For example, the following property may seem a bit basic: Similar to how protons and electrons make up atoms and molecules, axioms make up most properties and theorems in math, it just may take a while to show exactly how it unfolds. Furthermore, axioms cannot be broken down any further, so feel free to use them at leisure in any proofs. There are three commonly used methods to do proofs of theorems and properties: 1. Start with one member of the equation and try to achieve the other member. 2. Start with the given equation and transform it to the desired equation. 3. Start somewhere else and use multiple clever arguments and statements. The example above was an example of transforming the given equation to achieve the desired result. To start on one side and deduce the other side, you undergo a similar process: Prove: (− x)(− y) = xy (− x)(− y)
Given = (− 1 * x)(− 1 * y)
Definition of Negative = (− 1)(x * [− 1])(y)
Associativity = (− 1)(− 1 * x)(y)
Commutativity = (− 1 * [− 1])(xy)
Associativity = (1)(xy)
Arithmetic = xy
Identity Axiom of Multiplication ∴ (− x)(− y) = xy , Q.E.D.
Transitivity As you may have noticed, the past two methods are only slightly different, as they undergo similar changes and steps. The last is a little different from both: Prove: ​
If x = y , then x + z = y + z x+z = x+z
Reflexive Property x=y
Given ∴x + z = y + z
Substitution We started by immediately achieving one side of the desired equation, then used the given information to transform the right side into the final product. The practice problems will help you use the axioms with the three methods described above while proving different new properties. Do try to remember the properties and ​
where you can use them, as they could be useful later on in your life. Prove the following properties. Show all steps and justify your reasoning. 1. Prove that multiplication distributes over subtraction: Prove: x( y − z ) = xy − xz Answers found on pg. x 2. Prove that Division distributes over addition: Prove: (x + y)
= xz + yz z
Justify each step of the following proof: 3. 3.2: Properties of Linear Function Graphs The graphs of linear functions are straight lines, modeled by the equation, ​
y​
=​
mx​
+​
b​
. Slope­Intercept Form If ​
y­mx+b​
, then ​
m​
equals the slope of the graph, and ​
b​
equals the y­intercept The slope of a line is the line’s “ rise
run ”, where the run is the horizontal distance between two points on the line, and the rise is the vertical distance between those two points. y −y
This can be modeled by the equation, ​
​
m​
= x2−x1 , ​
with coordinates (​
x​
,​
y​
) and (​
x​
,​
y​
). 1​
1​
2​
2​
2
1
Ex 1: Find the slope of a line that passes through the points (1,­1) and (­5, ­5) Step 1. Because we are given two points of the line, we can use the slope equation, plugging in the coordinates for their corresponding variable Step 2. Plugging the coordinates in, the equation states m = −5−(–1)
−5−1 −4
2
Step 3. Simplified, the equation comes out to m = −6 = 3 Step 4. The slope is 23 If ​
m>0​
, then the slope is positive, with the line sloping up as ​
x​
increases. If ​
m<0​
, then the slope is negative, with the line sloping down as ​
x​
increases. If ​
m=0​
, then the graph is horizontal. This type of graph is also known as a constant function. Ex: Positive Slope Negative Slope Slope of Zero The value of ​
b​
shows the ​
y­intercept​
, or where the graph crosses the y­axis. Although not represented in the equation, the ​
x­intercept​
is where the graph crosses the x­axis. Intercepts The ​
y­intercept​
of a function can be found when ​
x=​
0 The ​
x­intercept​
of a function can be found when ​
y=​
0 Ex 2: Plot the graph of a line with its x­intercept at ­4, and its y­intercept at 6. Step 1. From the given information, we already know two points of the line, so plot the points Step 2. Because two points make a line, connect the points, and the graph is plotted. Ex 3: Find the equation of the line that passes through the points (3,5) and (­1,2) Step 1. First, look at the points. Neither ​
b​
nor ​
m​
is given. Step 2. Solve for ​
m​
, as you can use ​
m​
to find ​
b​
. Use the equation, and plugging in the values 3
gives you m = 4 Step 3. Now, for the equation ​
y=mx+b​
, we can plug in three variables, leaving only ​
b​
. For ​
x​
and y​
, it does not matter which coordinates are used. Step 5. Taking the first point, we are left with the equation 5 = 34 (3) + b . Solving algebraically, b = 114 . Step 5.5: Just for learning’s sake, let’s use the second point. Using the second point, the equation is 2 = 34 (− 1) + b , and solving algebraically, b = 114 . Step 6. Therefore, the equation of the line is y = 34 x + b Practice Problems: 1. Plot the graph of the given equation a. y=4x+6 b. y= 47 x+2 c. y=5 2. Find the equation of the line a. x­intercept=5, y­intercept=7 b. y­intercept=3, passes through the point (5,1) c. Passes through the points (3,4) and (­1, ­1) 4.11­Linear Programing Linear programming is a method used to find the maximum or minimum value within a given set of restrictions. This section is most easily explainable using examples. However, the basic format of solving a linear programming problem is: 1.
2.
3.
4.
5.
Define variables Find all restrictions, and convert them to inequalities Graph the inequalities to find a feasibility region* Find the equation for the total profit/cost Use the corner points** of the feasibility region to find the maximum or minimum value *feasibility region is the region in which all inequalities are true, the “shaded” region **corner points are the corner points of the feasibility region, found by setting the two equations that form the point as a system of equations Example 1: Craig Browning is baking cookies for his school’s bake sale. His chocolate chip cookies sell for $1.25 a dozen, and his brownies sell for $2 a dozen. He will bake up to 25 dozens of chocolate chip cookies, and up to 40 dozen brownies. However, he will not bake more than 50 dozens of total desserts. Also, the number of brownies will be no more than three times the number of chocolate chip cookies. How many of each type of dessert should Craig make to maximize the profits of the bake sale? How much money will this be? Solution: 1. Set variables: let x=number of chocolate chip cookies, let y=number of brownies, and let p=the total profits 2. Given the restrictions, we can find the inequalities that we will graph. The inequalities are: x < 25 , as he will bake no more than 25 dozen cookies y < 40 , as he will bake no more than 40 dozen brownies x + y < 50 , as he will bake no more than 50 dozen of total desserts y < 3x , as the number of brownies will never succeed three times the number of cookies x > 0, y > 0 , as he cannot make a negative amount of cookies 3. Graph the inequalities to find the feasibility region 4. Find the equation for the profit: Because each dozen of chocolate chip cookies sells for $1.25, and each dozen brownies sells for $2, p=1.25x+2y 5. Use the corner points to find the maximum profit. By using systems of equations, the corner points are, from the bottom most left one and then counterclockwise, (0,0), (25, 0), (25,25), and ( 12.5, 37.5) . To find the profit that each value of x and y would make, plug them into the equation for profits, p=1.25x+2y. For (0,0), p=$0. For (25,0), p=$31.25. For (25,25) p=$81.25. For (12.5, 37.5), p=$90.63. Therefore, by baking 12.5 dozen cookies, and 37.5 dozen brownies, Craig can maximize his profits to $90.63 Practice Problem: You work at a small radio manufacturing company. They produce two types of radios, “Fixers” and “Bad Buddies”. The boss has assigned you to figure out how many of each radio should be produced per day to maximize profits. However, there are some restrictions. No more than 30 Fixers and 15 Bad Buddies can be produced per day. Also, no more than 35 total radios can be produced per day, but there must be at least 10. They can spend no more than a total of 189 man hours per day assembling radios, with a Fixer taking 7 hours to assemble, and a Bad Buddy taking 3. Finally, they can spend at most 68 man­hours per day testing the radios. It takes 1 hour to test a Fixer, and 4 hours to test a Bad Buddy. Using this information, answer the following: a. Plot the graph and feasibility region of this problem. Use inequalities b. The company makes a profit of $40 on each Fixer, and $20 on each Bad Buddy. Write an equation for the total profits the company can make each day, in terms of the number of Fixers and Bad Buddies. c. Shade the part of the feasibility region where the profit is at least $1000 d. Find the optimum point at which the daily profit is a maximum, and find this daily profit. Remember, you cannot produce half of a radio. 6.9 ­ Logarithms with other bases Remember logarithms? They were used to solve exponential equations like 7x = 12 . But the logarithms you know, written as ​
log x​
, actually have 3 parts, only that the base is hidden: log x = log​
10 x ​
Obviously, this means that ​
x​
to the ​
log x ​
is 10, and the small 10 is normally hidden. The ten is known as the ​
base ​
of the logarithm, and can be any value greater than 0, but cannot be 1. ​
logb ​
x = y if and only if by ​
= x ​
5​
For example, 5 = log​
32 because 2​
= 32. 2​
Each of the variables showed above has a specific name: y ​
is the logarithm x ​
is the argument b​
is the base. Using the shown statements, each one of those parts can be found as long as the other two are given: Example 1 Find x if ​
log​
x = 4. 3​
First, transform this into a more understandable equation by using the definition of a logarithm: 34 = x Now, just evaluate the new equation: x = 81 Example 2 Find x if log​
x 16 = 8 ​
Do the same as Example 1: x8 = 16 Cancel out the exponent to isolate x: (x8)1/8 = 161/8 x = √2 Example 3 Find x if log​
5 125 = x ​
5x = 125 5x = 53 x = 3 In the upcoming exercises, you will be able to practice your new knowledge of the definition of logarithms. Furthermore, you will experience how they may be used in a real life situation. Practice Problems Solve for x 1. log​
x = 3 7​
2. log​
144 = 2 x​
3. log​
128 = x 16​
4. log​
36 = x 6​
5. Bang Krupp is a financial manager who wants to find how long his client has been in debt. The client owes $50178.10 at the moment and the interest rate is 2.2%, compounded annually, but used to owe only $37000. How long has the client been in debt? Assume that the client never paid off or added to the debt. Put the exact answer. 6.13: Exponential and Logarithmic Functions x General equation of an exponential function: f(x) = a x b​
­ a and b are constants ­ b > 0 ­ b ≠ 1 ­ a ≠ 0 Theorem: x​
f(x) = a x b​
, then f(x + c) is a constant multiplied by f(x). Property: x​
c ­ If f(x) = a x b​
, then f(x + c) = f(x) x b​
­ If f(x) = mx + b, then f(x + c) = f(x) + mc Example Problem Find an exponential function that can model the following data: f(4) = 24 f(10)= 81 First make a table with the data you are given: f(x) x 24 4 81 10 x​
Then use the general equation for an exponential function, f(x) = a x b​
, and plug in the values from your data table. 4 25 = a x b​
10
81= a x b​
Now use systems of two equations to solve for a and b 10 81= a x b​
4
25 = a x b​
6
3.24 = b​
b = 6√3.24 b= 10.8 Now plug b into one of your original equation to find a 4
25 = a x 10.8​
25 = a x 13604.8896 a = .00183757 Now use a and b to create an exponential function x
f(x) = .0018375746 x 10.8​
You can now round to four significant figures x
f(x) = .0018 x 10.8​
Practice Problems Find the exponential function based on the given data: 1. f(3) = 20 ; f(7)= 24 2. f(­3) = 4.3 ; f(­½) = 11.35 Say whether or not the given points form an exponential functions: 1. x f(x) 1 15 5 25 10 35 15 45 2. The amount of money in the bank in 2003 was estimated to be $35,000 with an annual rate of increase of about 2.4%. a. Write an equation to model future growth. b. Use the equation you got to estimate the amount of money in 2007 to the nearest hundred dollar. Answer Appendix Answers to 1.7: 1. Prove: x( y − z ) = xy − xz
x( y − z ) = x( y + (− z))
x( y + (− z)) = xy + x(− z)
xy + x(− z) = xy − xz
∴ x( y − z ) = xy − xz , Q.E.D.
2. Prove: (x + y)
= xz + yz
z
(x +y)
1
z = z (x + y)
1
x y
z (x + y) = z + z
∴ (x + y)
= xz + yz
z
Given Definition of Subtraction Distributive Property Multiplication Transitive Property Given Definition of Division Distributivity Transitive Property Answers to 3.2: 1. y=4x+6 First, use the y­intercept to plot the first point at (0,6). Then, using the slope, plug in either x or y to find the second point, or just count rise over run in your head. 2. y= 47 x+2 Use the same method as #1. 3. y=5 Realize that the slope of the line is zero; as long as y=5, then x can equal any real number. Therefore the graph is horizontal. Find the equation of the line: 4. With an x­intercept at (5,0), and a y­intercept at (0,7) Because there are two points given, find the slope using the slope of equation. Also, we are given b​
, as ​
b​
=the y­intercept. Therefore, the equation is ​
y= − 75 x+7. 5. With a y­intercept of 3, and passes through (5,1) Again, use the slope formula to find the slope, as you are given two points. You are also given the y­intercept. Therefore, the equation is y =− 25 x+3. 6. That passes through (3,4) and (­1,­1) First, find the slope of the line using the slope formula. In this problem, you are not given the y­intercept, so you will need to solve for ​
b​
. After finding the slope (m=5/4), you can plug in any one of the points, along with the slope, to find ​
b​
(just as in Example 3). The equation will come 5
out to y = 4 x + 0.25 Answers to 4.11: 1. a. Let x=the number of Bad Buddies, Let y=the number of Fixers x ≤ 15 , y ≤ 30 , x + y ≤ 35 , x + y ≥ 10 , 3x + 7y ≤ 189 , 4x + y ≤ 68 , x ≥ 0, y ≥ 0 You can find all of these inequalities from the information given, in the same order as the info given in the paragraph. b. Let ​
p​
=profit ($), ​
p= ​
20​
x+​
40​
y c. To find the region, set up an inequality for finding the profit; 1000 ≤ 20x + 40y . To graph is, use the intercepts of the line. The two intercepts are (50,0) and (0,25). d. Just as in the example problem, set the inequalities as systems of equations to find the corner points. Because the bottom two corner points will obviously not produce the maximum profit, only concern yourself with the top two. The top right corner point, once solved for the equations 189=3x+7y and 68=4x+y, is (11.48, 22.08). The top left corner point, is at (0,27). Now, using the profit equation, for the point (11.48, 22.08), we need to convert it to whole radios. The most sensible assumption would be the production 11 Bad Buddies and 22 Fixers, bringing a profit of $1100. If there is production of only 27 fixers, the profit would be $1080. Therefore, you should produce 11 Bad Buddy radios and 22 Fixer radios per day to make a maximum profit of $1100. Answers to 6.9: 1. log7​
x = 3 ​
3​
7​
= x x = 343 2. log​
x 144 = 2 ​
2​
x​
= 144 x = 12 3. log​
16 128 = x ​
x​
16​
= 128 24x
​​
= 27​
4x = 7 x = 7/4 4. log​
6 36 = x ​
x ​
6​
= 36 x = 2 5. A = P (1 + nr )nt ; A = $50178.10 (current debt), P = $37000 (starting debt), r = 0.022 (interest rate), n = 1 (compounds per year), t = # of years passed 50178.1 = 37000 * (1 + 0.22)t 1.22t = 50178.1
37000 50178.1
log​
( 37000 ) = t 1.22​
The client has been in debt for log​
( 50178.1
1.22 ​
37000 ) years (~14). Answers to 6.13: 1. f(3) = 20 ; f(7)= 24 a. f(x) x 20 3 23 7 7 b. 23 = a x b​
3
20 = a x b​
4
c. 1.15 = b​
d. b = 1.035558076 3
e. 20 = a x 1.035558076​
a = 18.0097057 x
f. ​
f(x) = 18.009 x 1.036​
2. f(­3) = 4.3 ; f(­½) = 11.35 a. f(x) x 4.3 ­3 11.35 ­1/2 ­3
b. 4.3 = a x b ​
­½
11.35 = a x b​
2.5
c. 2.639534884 = b​
d. b = 1.474385199 ­3
e. 4.3 = a x 1.474385199​
a = 13.78165402 x f. ​
f(x) = 13.781 x 1.474​
2. The amount of money in the bank in 2003 was estimated to be $35,000 with an annual rate of increase of about 2.4%. c. Write an equation to model future growth. x
i. y = ab​
35000 + .024(35000) 35000 (1 + 0.024) = 35000(1.024) growth factor = 1.024 x
y = a(1.024)​
x y= 35,000(1.024)​
y = amount of money ($) x = number of years since 2003 d. Use the equation you got to estimate the amount of money in 2007 to the nearest hundred dollar. i. 2007 ­ 2003 = 4 years 4​
​
y = 35,000(1.024)​
≈ 38,482.91 ≈ ​
$38,500