FINAL EXAM - REVIEW
OVERVIEW
Use this as a REVIEW for the Final Exam
Period 1 will have theirs on Tuesday, January 22 nd
Period 2 will have theirs on Wednesday, January 23 rd
OVERVIEW (CONTINUED)
Remember … you are allowed a SINGLE side of ONE 8.5” x 11”
sheet of paper to write down anything you may find useful for
the exam
Think of the equations and rules that would be helpful to solve
the problems
LET’S GET STARTED!
First thing you need is the Final Exam Review that was handed
out in class.
If you have a graphing calculator, that would be helpful also.
And maybe some scratch paper to do some work on.
AND a pencil and an eraser.
OK!
PROBLEM #1
You are presented with a graph of a function.
A graph is used to VISUALLY represent math.
In this case the graph shows us a particular function.
Before we move on … can you describe to yourself what a
function is? Can you define the word function?! Take 45
seconds to think about that before moving on …
PROBLEM #1 (CONTINUED)
Here’s our graph that represents a function.
What do you think this function could model?
PROBLEM #1 (CONTINUED)
You are asked to define various characteristics of the graph of
the function. Let’s start with the domain. What is domain?
PROBLEM #1 (CONTINUED)
Domain tells us all of the “x” values where the function exists.
It’s easy to see which “x” values exist when we graph a function.
PROBLEM #1 (CONTINUED)
The least “x” value is 2 and the greatest “x” value is 13. And
you’ll notice that the function, which is a squiggly line, or curve,
contains ALL of the other possible x values.
PROBLEM #1 (CONTINUED)
So, our DOMAIN for this function defines ALL of the possible
values for “x” of the function. Therefore, our domain is every x
value between 2 and 13, including 2 and 13. We write this as:
PROBLEM #1 (CONTINUED)
Let’s move on to RANGE. If domain tells us all of the “x” values,
then what do you think range tells us? Right, all of the “y”
values for the function, from least to greatest.
PROBLEM #1 (CONTINUED)
So, our RANGE for this function defines ALL of the possible
values for “y” of the function. Therefore, our range is every y
value between 1 and 8, including 1 and 8. We write this as:
PROBLEM #1 (CONTINUED)
Where is the graph CONCAVE UP?
(think: “bowl facing up”)
PROBLEM #1 (CONTINUED)
Where is the graph CONCAVE DOWN?
(think: “bowl facing down”)
PROBLEM #1 (CONTINUED)
Where is the graph INCREASING? (“trace” from left to right)
PROBLEM #1 (CONTINUED)
Where is the graph DECREASING ? (“trace” from left to right)
PROBLEM #1 (CONTINUED)
What are the ABSOLUTE MINIMUM and ABSOLUTE MAXIMUM?
PROBLEM #2
For the function
𝑓 𝑥 =
1
𝑥−6 𝑥+2
where is there a vertical asymptote and WHY?
There exists a vertical asymptote for every value of x that
creates a “0” in the denominator
Therefore when x= 6 and x = -2 we have a vertical asymptote
GRAPH THIS FUNCTION ON YOUR CALCULATOR TO VERIFY!
PROBLEM #3
Evaluate the following …
Given:
𝑓 𝑥 = 3𝑥 + 2
g 𝑥 = −2𝑥 − 𝑥 2
Find:
a.
b.
g(𝑓
1
3
)
f(g −2 )
We are given T WO DIFFERENT FUNCTIONS: one named “f(x)” and
a second named “g(x)”. Then we have to find the result when
they are composed together.
CLICK TO THE NEXT SLIDE!
PROBLEM #3 (CONTINUED)
Let’s start with
a.
g(𝑓
1
3
)
Think of x as a new HTC phone being given as a present. You
have to put the phone safely in a box f(). This gives us f(x).
Then you have to take that box with the phone f(x) and wrap it in
wrapping paper g(). After wrapping the box with the phone, we
have g(f(x)). But we have to go STEP BY STEP!
PROBLEM #3 (CONTINUED)
Let’s start with
a.
g(𝑓
1
3
)
Now, we know 𝑓 𝑥 = 3𝑥 + 2
We also know for this problem our “x” is
1
3
So let’s pack it up as 𝑓
𝑓
and put the
1
3
1
3
in for all “x” values:
1
1
=3∗
+2
3
3
3
=
+2
3
=1+2
=3
PROBLEM #3 (CONTINUED)
We just found 𝑓
present now
1
3
= 3 so all we have to do is wrap up the
Remember, we are solving for g 𝑓
So we just substitute in our 𝑓
1
3
1
3
= 3 as g 𝑓
And and since 𝑔 𝑥 = −2 − 𝑥 2
we can find 𝑔 3 = −2 − 3 2
= −2 − 9
= −11
1
3
= 𝑔(3)
PROBLEM #3 (CONTINUED)
Our answer:
g 𝑓
Because: STEP 1:
𝑓
1
3
1
3
= −11
=3∗
1
3
+2
3
+2
3
=1+2
=3
=
STEP 2:
g 𝑓
1
3
= 𝑔 3
= −2 − 3
= −2 − 9
= −𝟏𝟏
2
PROBLEM #3 (CONTINUED)
Your turn for
b. f g −2
Start with g(-2) and then use that result to put in for f( )
The answer will come out as -16
PROBLEM #4
Solve for x using any method:
𝑥 2 + 2𝑥 − 8 = 0
Step 1: Factor (𝑥 + 4)(𝑥 − 2) = 0
Step 2: Find the roots – the values for x
that make the equation true
𝑥 = −4 𝑎𝑛𝑑 𝑥 = 2
Step 3: Verify by multiplying factors and graphing
PROBLEM #5
𝑥 2 + 6𝑥 + 10 = 0
Solve for x by completing the square:
Step 1: Find the “square” by subtracting “1” from both sides
𝑥 2 + 6𝑥 + 10 = 0
_____
− 1
−1
2
𝑥 + 6𝑥 + 9 = −1
Step 2: Factor the lef t side, revealing the “square”
𝑥+3
2
= −1
Step 3: We have to get “x” by itself. First take the square
root of both sides … then subtract 3
𝑥 + 3 2 = ± −1
𝑥 + 3 = ±𝑖
𝑥 = ±𝑖 − 3
PROBLEM #6
Simplify:
3 + 2𝑖 + (−2 + 4𝑖 )
Step 1: Group the REAL and the IMAGINARY
3 + −2 + 2𝑖 + 4𝑖
Step 2: Combine
1 + 6𝑖
PROBLEM #7
Graph the point:
1 + 5𝑖 on the complex plane
Imaginary
------
Real
|
|
|
|
PROBLEM #8
Solve for the magnitude
|(2 − 𝑖 )(1 + 3𝑖 )|
Step 1: Multiply the inside quantity
2 1 + 2 3𝑖 + −𝑖 1 + −𝑖 3𝑖
|2 + 6𝑖 − 𝑖 − 3𝑖 2 )|
2 + 5𝑖 − 3 −1
2 + 5𝑖 + 3
5 + 5𝑖
Step 2: Find the magnitude
5
2
+ 5
2
=
25 + 25 =
50 =
2 25 = 5 2
PROBLEM #9
Simplify
4−𝑖
2+𝑖
Step 1: Find the complex conjugate
For
4−𝑖
2+𝑖
the complex conjugate is 2 − 𝑖
Step 2: Multiply by the complex conjugate
4−𝑖 (2−𝑖)
2+𝑖 (2−𝑖)
=
8−4𝑖 −2𝑖 +𝑖 2
4−2𝑖 +2𝑖 −𝑖 2
=
8−6𝑖 +(−1)
4−(−1)
=
7−6𝑖
5
PROBLEM #10
How do we represent SOH -CAH-TOA as fractions for the following
triangle?
4
𝑠𝑖𝑛𝜃 =
𝑜𝑝𝑝
ℎ𝑦𝑝
=
3
5
𝑐𝑜𝑠𝜃 =
𝑎𝑑𝑗
ℎ𝑦𝑝
=
4
5
𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝
𝑎𝑑𝑗
=
3
4
3
ө
5
Pay attention to
where 𝜃 is!
PROBLEM #11 & #12
Convert 25° to radians
25 ×
Convert
5𝜋
18
5𝜋
18
×
𝜋
180
=
25𝜋
180
=
5𝜋
36
radians
radians to degrees
180
𝜋
=
50
1
= 50 degrees
PROBLEM #13 & #14
Draw a 280° angle
Draw a -135° angle
90°
90°
0°
360°
180°
270°
0°
360°
180°
270°
PROBLEM #15
The amplitude is 0.5
The period is 120
The equation for the graph is constructed using:
A = amplitude = 0.5
B=
360
𝑝𝑒𝑟𝑖𝑜𝑑
=
360
120
=3
𝑦 = 0.5𝑠𝑖𝑛3𝜃
*You can check with your calculator!
PROBLEM #16
Degree:
End Behavior:
Roots:
Factors:
Intercept:
EVEN
LEFT(+) RIGHT (+)
x = -4,-1 ,2,2
𝑥 + 4 , 𝑥 + 1 , 𝑥 − 2 , (𝑥 − 2)
𝑓 𝑥 = 𝑥+4 𝑥+1 𝑥−2 2
Standard form:
𝑓 𝑥 = 𝑥 4 + 𝑥 3 − 12𝑥 2 + 4𝑥 + 16
PROBLEM #17
Given:
𝑓 (𝑥 ) = 𝑥 4 +2𝑥 3 − 9𝑥 2 − 2𝑥 + 8
Step 1: Test the factors of 8
Test 1
1
1
Test -1
1
2
1
3
-1
2
Step 2: Quadratic Formula using
-9
3
6
-2
4
-2
-6
-8
8
0
a
b
8
-8
0
c
Step 3: Determine factors using the test values and quadratic form.
x = 1, -1, 2, -4
therefore the factors are
(x-1),(x+1),(x-2),(x+4)
PROBLEM #17 (CONTINUED)
Your graph should look like the one below.
Are the intercepts correct?
PROBLEM #18
Solve for x:
Step 1:
4 3𝑥 = 0.45
Change to logarithm in order to find
the exponent
log 4 0.45 = 3𝑥
Step 2:
Change of base to use your calculator
lo g 0 .45
= 3𝑥
log 4
−.576 = 3𝑥
Step 3:
Divide both sides by 3 to get “x” by itself
−.192 = 𝑥
PROBLEM #19
Solve for x:
Step 1:
log 𝑥 243 = 5
Convert to exponential form to solve for “x”
𝑥 5 = 243
Step 2:
Determine what value of “x” multiplied
five times is equal to 243
𝑥 5 = 𝑥 ∗ 𝑥 ∗ 𝑥 ∗ 𝑥 ∗ 𝑥 = 243
or
𝑥=
Therefore
x = 3
1
243 5
PROBLEM #20
You've got the flu that’s been going around, so you decide to stay home and
take 500 mg of Nyquil to help relieve congestion. Each hour, the amount of
Nyquil in your system decreases by about 15% . How much Nyquil is lef t in
your system af ter 8 hour s ?
Step 1: Use the equation 𝑃 𝑡 = 𝑃0 1 + 𝑟
Step 2: Determine what you know
𝑡
𝑃 t = ?
P0 = 500mg
r
= − 15% = −0 .15
t
= 8 hour s
Step 3: Substitute in the exponential growth/decay equation
𝑃 𝑡 = 500 1 − 0.15
𝑃 𝑡 = 500 0 .85
8
𝑃 𝑡 ≈ 221.85 𝑚𝑔
8
PROBLEM #20
O a t m e a l c o o k i e s n e e d to r e a c h a te m p er a t ur e o f 37 5 d e g r e e s b e f o r e b e i n g e d i b l e !
T h e d o u g h i s a t a n i n i t ia l r o o m t e m p e r a t ur e o f 7 2 d e g r e e s b e f o r e b e i n g p u t i n to t h e
o v e n . I f t h e c o o ki e s g et 3 6 % h o t te r e v e r y m i n ute , h o w l o n g b e f o r e w e c a n t a ke t h e
c o o ki e s o u t o f t h e o v e n ?
S te p 1 :
S te p 2 :
U s e t h e e q u a t i o n 𝑃 𝑡 = 𝑃0 1 + 𝑟
D ete r m in e w h a t yo u k n o w
𝑡
𝑃 t = 3 7 5 d e g re e s
P0 = 7 2 d e gre e s
𝑟
= 36% = 0.36
t
= ?
S te p 3 :
S u b s t it ute i n t h e ex p o n e n t i a l g r o w t h / d e c ay e q u a t i o n
375 = 72 1 + 0.36
𝑡
375 = 72 1.36
5.208 = 1.36
𝑡
𝑡
PROBLEM #20 (CONTINUED)
Oatmeal cookies need to reach a temperature of 375 degrees
before being edible! The dough is at an initial room temperature
of 72 degrees before being put into the oven. If the cookies get
36% hotter ever y minute , how long before we can take the cookies
out of the oven?
Step 4:
Convert to a logarithm to solve for time
log 1.36 5.208 = 𝑡
log 5.208
log 1.36
=𝑡
5.37 minutes = t