1. 2016-2017 Algebra 2 Midterm Exam Review Solutions

2. Use the Distributive Property. Use the Distributive Property.
1. Simplify the expressions.
a. −2 3y+6 +5y b 𝑐−15𝑑 −11(4𝑐−2𝑑+7)
Use the Distributive Property.
−6y−12+5y
2𝑐−5𝑑−44𝑐+22𝑑−77
−y−12
−42𝑐+17𝑑−77
Collect like terms.
c 𝑤 (−3𝑤+7)
Use the Distributive Property.
16 5 𝑤+ 2 5 − 12 5 𝑤+ 28 5
Collect like terms.
4 5 𝑤+6

3. d. (4 𝑥 3 +2 𝑥 2 +3𝑥+4)÷(𝑥+4) f. ( 𝑥 4 +2 𝑥 3 +3 𝑥 2 +2𝑥+1)( 𝑥 2 +1)
Use synthetic division.
d. (4 𝑥 3 +2 𝑥 2 +3𝑥+4)÷(𝑥+4)
Use the coefficients from the dividend.
𝒙+𝟒=𝟎
𝒙=−𝟒 −4
−16 56
−236
These are the coefficients of the quotient. 59
−232
Bring down the first term.
Multiply.
Multiply.
Multiply. 4
−14
Add down the column.
Add down the column.
Add down the column.
Since we divided by a linear term, the degree of the quotient decreases by 1.
4 𝑥 2 −14𝑥+59− 232 𝑥+4
(9 𝑔 2 −100)(3𝑔−10) e.
Use the Distributive Property.
27 𝑔 3 −90 𝑔 2 −300𝑔+100
f. ( 𝑥 4 +2 𝑥 3 +3 𝑥 2 +2𝑥+1)( 𝑥 2 +1)
𝑥 𝑥 𝑥 𝑥
Use the Distributive Property in an area model.
𝑥 2 +1
𝑥 6
2𝑥 5
3𝑥 4
2𝑥 3
𝑥 2
𝑥 4
2𝑥 3
3𝑥 2 2𝑥 1
𝑥 6 +2 𝑥 5 +4 𝑥 4 +4 𝑥 3 +4 𝑥 2 +2𝑥+1
Combine like terms.

4. g. 𝑥 3 +2𝑥−3 𝑥 3 +2𝑥 −3 𝑥 3 +3 𝑥 6 +0 𝑥 5 +2𝑥 4 +0 𝑥 3 +0 𝑥 2 +6𝑥−9
Use long division since the divisor is NOT linear.
𝑥 3 +2𝑥−3
𝑥 6 +2 𝑥 4 +6𝑥−9 𝑥 3 +3
𝑥 3
+2𝑥 −3
Use 0 as a place holder for missing terms in the dividend.
𝑥 3 +3
𝑥 6 +0 𝑥 5 +2𝑥 4 +0 𝑥 3 +0 𝑥 2 +6𝑥−9
𝒙 𝟑 × 𝒙 𝟑 = 𝒙 𝟔
−(𝑥 𝑥 3 )
𝑥 𝑥 3
2𝑥 4 −3 𝑥 3 +0 𝑥 2 +6𝑥−9
𝒙 𝟑 ×𝟐𝒙=𝟐 𝒙 𝟒
−(2 𝑥 𝑥)
2 𝑥 𝑥
𝒙 𝟑 ×−𝟑=−𝟑 𝒙 𝟑
−3 𝑥 3 +0 𝑥 2 +0𝑥−9
−(−3 𝑥 −9)
−3 𝑥 −9
−2 4𝑓−3𝑝+1 −3(−3+𝑝−7𝑓) h.
Use the Distributive Property.
−8𝑓+6𝑝−2+9−3𝑝+21𝑓
Collect like terms.
13𝑓+3𝑝+7

5. Not Invertible because the inverse would not be a function.
Find the inverse of the relation if possible, otherwise write not invertible.
a. 𝑓 𝑥 = 3 𝑥 +6 b. 𝑓 𝑥 =5 𝑥 2 +8
Interchange x and y.
𝑦= 3 𝑥 +6
Not Invertible because the inverse would not be a function.
𝑥= 3 𝑦 +6
𝑥−6= 3 𝑦
Solve for y.
(𝑥−6) 3 = 3 𝑦 3
(𝑥−6) 3 =𝑦
𝑦= 2 3 𝑥+7
Interchange x and y. d.
𝑦= 5𝑥+1
Interchange x and y. c.
𝑥= 5𝑦+1
𝑥= 2 3 𝑦+7
𝑥 2 = 5𝑦+1 2
Solve for y.
Solve for y.
𝑥−7= 2 3 𝑦
𝑥 2 =5𝑦+1
𝑥 2 −1=5𝑦
3 2 (𝑥−7)=𝑦
𝑥 2 −1 5 =𝑦

6. Move all y’s to one side of the equal sign.
Find the inverse of the relation if possible, otherwise write not invertible.
e. 𝑓 𝑥 = 𝑥 2 −16𝑥+20 f. 𝑓 𝑥 = 𝑥+1 𝑥
𝑦= 𝑥+1 𝑥
Not Invertible because the inverse would not be a function.
Interchange x and y.
𝑥= 𝑦+1 𝑦
Solve for y.
Move all y’s to one side of the equal sign.
𝑥𝑦=𝑦+1
𝑥𝑦−𝑦=1
Factor.
𝑦(𝑥−1)=1
𝑦= 1 𝑥−1

7. 3. Designate the following as even, odd or neither. b. 𝑓 𝑥 =5 𝑥 2 +8
Even: symmetric about the y-axis.
a. 𝑓 𝑥 = 3 𝑥 +6
𝑓 −𝑥 =5 (−𝑥) 2 +8
𝑓 −𝑥 = 3 −𝑥 +6
𝑓 −𝑥 =5 𝑥 2 +8
Odd: symmetric about the origin.
𝑓 −𝑥 =− 3 𝑥 +6
Even
Neither
e. 𝑓 𝑥 = 2𝑥 3 −5𝑥 d.
c. 𝑓 𝑥 = 𝑥 4 −3 𝑥 2
𝑓 −𝑥 = 2(−𝑥) 3 −5(−𝑥)
𝑓 −𝑥 = (−𝑥) 4 −3 (−𝑥) 2
𝑓 −𝑥 = −2𝑥 3 +5𝑥
𝑓 −𝑥 = 𝑥 4 −3 𝑥 2
Odd
Even
Odd h.
f. 𝑓 𝑥 = 2 3 𝑥+7 g.
𝑓 −𝑥 = 2 3 (−𝑥)+7
𝑓 −𝑥 =− 2 3 𝑥+7
Algebraic test: Evaluate 𝒇(−𝒙) .
If 𝒇 −𝒙 =𝒇(𝒙), the function is even.
If 𝒇 −𝒙 =−𝒇(𝒙), the function is odd.
Otherwise, it is neither even nor odd.
Even
Neither
Neither

8. 4. Use the following graph to answer parts a – g.
a. (0, ∞) or 𝑥>0
b. (−∞, 0) or 𝑥<0
c. (−∞, −3) & (3, ∞) or 𝑥<−3 & 𝑥>3
d. (−∞, −3) & (3, ∞) or 𝑥<−3 & 𝑥>3
e. Minimum = −3 (no maximum)
f. Domain: all real numbers
g. Range: 𝑦≥−3

9. 5. The following graph represents a family’s distance from home as they return from vacation.
a. 5 to 6 hours
b. 0 to 3 hours, 4 to 5 hours, 6 to 8 hours
c. Domain: 0≤𝑥≤8
d. Range: 0≤𝑦≤700

10. 6. Let 𝑓 𝑥 = 𝑥 2 . The graph of 𝑔 𝑥 is vertically compressed by a factor of 1 2 , translated 5 units to the right and 4 units down. Write the function rule for 𝑔 𝑥 .
𝑓(𝑥)= 1 2 𝑥 2
The vertical compression results in this equation:
𝑓(𝑥)= 1 2 𝑥−5 2
Adding a translation 5 units to the right results in this equation:
𝑓(𝑥)= 1 2 𝑥−5 2 −4
Adding a translation 4 units down results in this equation:
𝑔(𝑥)= 1 2 𝑥−5 2 −4
The function rule for 𝑔 𝑥 is:

11. 7. Describe the transformations applied to
that yields the following functions.
Then graph each function. a) b) c)
The function is translated 4 units up.
The function is reflected over the x axis…
AND
The function is translated 9 units to the right.
The function is translated 3 units left…
AND
The function is translated 2 units down.

12. Divide both sides of the equation by 𝟔𝒕
Solve each equation for the designated variable.
a. 𝑆=6 𝑟 2 𝑡 for t 𝑆=6 𝑟 2 𝑡 for r
Divide both sides of the equation by 𝟔𝒕
𝑆 6 𝑟 2 =𝑡
Divide both sides of the equation by 𝟔 𝒓 𝟐
𝑆 6𝑡 = 𝑟 2
Take the square root of both sides of the equation
± 𝑆 6𝑡 =𝑟
b. A= 1 2 ( 𝑏 1 + 𝑏 2 ) for b A= 1 2 ( 𝑏 1 + 𝑏 2 ) for b2
Multiply both sides of the equation by 𝟐.
2A= 𝑏 1 + 𝑏 2
2A= 𝑏 1 + 𝑏 2
2A− 𝑏 2 = 𝑏 1
2A− 𝑏 1 = 𝑏 2
Subtract.

13. 9. Describe the meaning of the slope and y-intercept in each situation.
a. The cost for prom can be represented by the function 𝑐=25𝑝+110 where c is the total cost for prom and p is the number of people attending.
slope: There is an additional $25 cost for each additional person attending prom.
y-intercept: There is a $110 flat fee in the cost of the prom. (Perhaps the venue, music, etc.)
b. Andrew has planted flowers in his yard and measures their height each week. The height of the flowers can be represented by the equation ℎ=12+4.2𝑤 where h is the height in cm and w is the number of weeks since he planted the flowers
slope: Plants grow an additional 4.2 cm each week.
y-intercept: The plants started at a height of 12 cm.

14. 10. Solve the following systems of equations. This is one method.
Substitute x = 2 into one of the equations to solve for y.
Subtract y from both sides of the second equation.
𝟒𝒙 = 𝒚+𝟓
−𝒚 −𝒚
𝟒𝒙−𝒚=𝟓
𝟑𝒙+𝒚=𝟗
𝟑(𝟐)+𝒚=𝟗
𝟑𝒙+𝒚=𝟗
+ 𝟒𝒙−𝒚=𝟓
Add the 2 equations together.
𝟔+𝒚=𝟗
−𝟔 −𝟔
Divide both sides of the equation by 7.
𝟕𝒙=𝟏𝟒
𝒚=𝟑
𝟕 𝟕
𝒙=𝟐
(𝟐, 𝟑)

15. This is one method. b.
(−𝟒, 𝟎)

16. 11. Identify the system of equations shown in the graph with the solution indicated.
This line has a y-intercept of –2 and a slope of 𝟐 𝟑 .
This line has a y-intercept of 3 and a slope of −𝟏.
𝒚= 𝟐 𝟑 𝒙−𝟐
𝒚=−𝒙+𝟑

17. Write a system of equations to model the situation. 𝒉+𝒑=36 𝟖𝒉+𝟐𝒑=𝟏𝟒𝟒
12. At a bookstore, used hardcover books sell for $8 each and used paperback books sell for $2 each. Sheila purchased 36 used books and spent $144. Write a system of equations that can be used to find how many hardcover and paperback books Sheila bought. Then state how many paperback books Sheila actually purchased
Write a system of equations to model the situation.
𝒉+𝒑=36
𝟖𝒉+𝟐𝒑=𝟏𝟒𝟒
Let h = # hardcover books and p = # of paperback books.
Let’s eliminate h since we are looking for p.
Mulitply the first equation by –8.
−𝟖 𝒉+𝒑 =−𝟖(𝟑𝟔)
−𝟖𝒉−𝟖𝒑=−𝟐𝟖𝟖
Add the 2 equations together.
−𝟖𝒉−𝟖𝒑=−𝟐𝟖𝟖
+ 𝟖𝒉+𝟐𝒑=𝟏𝟒𝟒
−𝟔𝐩=−𝟏𝟒𝟒
Divide both sides of the equation by –6.
−𝟔 −𝟔
𝒑=𝟐𝟒
24 paperbacks

18. Write a system of equations to model the situation. 𝒂+𝒄=𝟓𝟐 𝒄=𝟑𝒂
13. A group of 52 people attended a ball game. There were three times as many children as adults in the group. Set up a system of equations that represents the numbers of adults and children who attended the game and solve the system to find the number of children who were in the group
Write a system of equations to model the situation.
𝒂+𝒄=𝟓𝟐
𝒄=𝟑𝒂
Let a = # adults at the game and c = # of children at the game.
Since 𝒄=𝟑𝒂, substitute 𝟑𝒂 into the first equation for 𝒄.
𝒂+𝒄=𝟓𝟐
Combine like terms.
𝒂+𝟑𝒂=𝟓𝟐
𝟒𝒂=𝟓𝟐
Divide both sides of the equation by 4.
𝒂=13
𝒄=𝟑𝒂
Substitute a = 13 into one of the equations to find c.
39 children
𝒄=𝟑(𝟏𝟑)
𝒄=𝟑𝟗

19. 14.
Let g = # of pounds of gummy worms
c = # of pounds of candy corn
s = # of pounds of sourballs
𝒈+𝒄+𝒔=𝟏𝟑
𝟐𝒈+𝟏𝒄+𝟏𝒔=𝟏𝟗
𝒈=𝟑𝒄
𝒎+𝒐+𝒓=𝟏𝟏𝟔
Let m = # of mums
o = # of orchids
r = # of roses
𝟐.𝟐𝟓𝒎+𝟏𝟎.𝟕𝟓𝒐+𝟒.𝟓𝟎𝒓=𝟓𝟑𝟐
𝒓=𝒎+𝟐𝟎

20. Let h = # hamburgers and d = # of hot dogs.
15. Marco is buying hamburgers and hot dogs for a party and only has $40 to spend. Hamburgers cost $6.49 per pound. Hot dogs cost $4.99 per pound and Marco needs at least 5 pounds of hot dogs. Write a system of inequalities to represent this situation.
𝟔.𝟒𝟗𝒉+𝟒.𝟗𝟗𝒅≤𝟒𝟎
𝐡≥𝟓
Let h = # hamburgers and d = # of hot dogs.
16. The boys and girls golf teams are trying to raise $750 for an out-of-county tournament. The boys team is selling candy bars for $2 each. The girls team is selling candles for $4 each. The girls need to sell at least 100 candles to clear out already purchased inventory. Write a system of inequalities to represent this situation.
Let b = # of candy bars the boys sell and
g = # of candles the girls sell.
𝟐𝒃+𝟒𝒈≥𝟕𝟓𝟎
𝐠≥𝟏𝟎𝟎

21. 17.
𝒙<−𝟏
𝒙≥−𝟏
The line begins at 𝒙=−𝟏 and continues to ∞. The closed circle indicates – 1 IS included.
The quadratic is the parabola and its domain is from −∞ to x = – 1, but the open circle at – 1 means it is NOT included.

22. a. Define the piecewise-defined function.
18. Use the piecewise function shown in the graph below to answer parts a – d.
a. Define the piecewise-defined function.
Write the quadratic equation whose domain is from
x = 4 to ∞. The open circle at 4 means it is NOT included.
Write the absolute value equation whose domain is from
−∞ to x = 3, but the open circle at 3 means it is NOT included.
𝒇 𝒙 = 𝟏 𝟐 𝒙+𝟑 −𝟐
if 𝒙<𝟑
𝒇 𝒙 = (𝒙−𝟔) 𝟐 −𝟐
if 𝒙>𝟒
The vertex of 𝒚= 𝒙 shifted left 3 units and down 2 units. It has also been vertically compressed by a factor of 𝟏 𝟐 .
𝑓 𝑥 = 𝑥+3 −2 𝑖𝑓 𝑥<3 (𝑥−6) 2 − 𝑖𝑓 𝑥>4
The vertex of 𝒚= 𝒙 𝟐 shifted right 6 units and down 2 units.
d. 0, − 1 2
b. (−∞, −3) and (4, 6)
c. Relative min = −2

23. a. How much does it cost to groom a pet who weighs 47 pounds?
19. The local pet store charges for grooming according to your pet’s weight. If your pet is 10 pounds or less, they charge $30. If your pet is between 10 and 30 pounds, they charge $35. If your pet is 30 pounds or more, they charge $40, plus an additional $2 for each pound over 30.
a. How much does it cost to groom a pet who weighs 47 pounds?
𝟑𝟎 𝒊𝒇 𝒙≤𝟏𝟎 𝟑𝟓 𝒊𝒇 𝟏𝟎<𝒙<𝟑𝟎 𝟒𝟎+𝟐 𝒙−𝟑𝟎 𝒊𝒇 𝒙≥𝟑𝟎
Let’s define the function where x = # pounds a pet weighs.
Since 47 > 30, we use the last piece of the function.
𝟒𝟎+𝟐(𝟒𝟕−𝟑𝟎)
=𝟒𝟎+𝟐(𝟏𝟕)
=$𝟕𝟒
b. Over what interval is this piecewise-defined function increasing?
𝟑𝟎, ∞ 𝒐𝒓 𝒙≥𝟑𝟎
c. What does the y-intercept of this graph mean in context?
A pet that weighs 0 pounds (or no pet) costs $0 to groom.

24. 20. Graph the equations.

25. 21.
The vertex of 𝒚= 𝒙 shifted left 2 units and up 1 unit. It has also been vertically stretched by a factor of 3.
𝒚=𝟑 𝒙+𝟐 +𝟏

26. 22. Fill in the missing properties for the solution of the equation given below.
[5x was subtracted from both sides. This is an example of the…] Subtraction Property of Equality.
[The left side of the equation was factored. This is an example of …] Factor.
[Each factor is set equal to zero. This is an example of the…]
Zero Product Property.
[Each factor experiences the same process. A number was added to each side of both equations. This is an example of the…] Addition Property of Equality.
[The left equation is multiplied by ½ or is divided by 2. Either (but not both) are correct. This is an example of either the…]
Multiplication Property of Equality OR Division Property of Equality (but NOT both).

27. 23. Select all of the following justifications NOT used in the solution of the equation shown below.
75 was added to or –75 was subtracted from both sides of the equation. This is either the Addition Property of Equality OR the Subtraction Property of Equality. But it cannot be both.
Both sides of the equation are divided by 3 or multiplied by This is either the Division Property of Equality OR the Multiplication Property of Equality. But it cannot be both.
Square root both sides of the equation.
Simplify the radical.
Answers may vary:
a or b (but not both), c,
d or e (but not both), f, g, i

28. When the ball hits the ground, the height is 0.
24.
𝒉=− 𝟏 𝟐 𝒕 𝟐 +𝟐𝒕+ 𝟓 𝟐
Factor out − 𝟏 𝟐 .
𝒉=− 𝟏 𝟐 𝒕 𝟐 −𝟒𝒕−𝟓
Factor 𝒕 𝟐 −𝟒𝒕−𝟓.
𝒉=− 𝟏 𝟐 𝒕−𝟓)(𝒕+𝟏
𝟎=− 𝟏 𝟐 𝒕−𝟓)(𝒕+𝟏
When the ball hits the ground, the height is 0.
𝟎=𝒕−𝟓 𝒐𝒓 𝟎=𝒕+𝟏
Since time cannot be negative, the ball hit the ground after 5 seconds.
𝒕=𝟓
𝒐𝒓 𝒕=−𝟏

29. 𝑨 𝑳 = 𝑳 𝟐 −𝟏𝟔𝑳+𝟔𝟒 𝑨 𝑳 =(𝑳−𝟖)(𝑳−𝟖) 𝑨 𝑳 = 𝑳−𝟖 𝟐 𝑳−𝟖
25. Jesse has a square rug whose area can be represented with the function 𝐴 𝐿 = 𝐿 2 −16𝐿+64, where A(L) is the total area and L is the length of the rug in feet.
a. What expression can be used to find the length of one side?
𝑨 𝑳 = 𝑳 𝟐 −𝟏𝟔𝑳+𝟔𝟒
Factor.
𝑨 𝑳 =(𝑳−𝟖)(𝑳−𝟖)
𝑨 𝑳 = 𝑳−𝟖 𝟐
𝑳−𝟖
b. What is the length of the rug?
(Answers may vary) The length of the rug must be greater than 8 feet.

30. 26. a. 𝒚=𝟏𝟎 𝒙 𝟐 −𝟏𝟏𝒙−𝟔 𝟎=𝟏𝟎 𝒙 𝟐 −𝟏𝟏𝒙−𝟔 𝟎=(𝟓𝒙+𝟐)(𝟐𝒙−𝟑) 𝟎=𝟓𝒙+𝟐 𝟎=𝟐𝒙−𝟑
To identify the zeros, set y equal to 0 and factor.
𝟎=𝟏𝟎 𝒙 𝟐 −𝟏𝟏𝒙−𝟔
Identify two numbers that multiply to –60 and add to –11.
Factor out the GCF up the columns and across the rows.
10 𝑥 2
−15𝑥 4𝑥 −6
−60 𝑥 2 −3
10 𝑥 2
−15𝑥 4𝑥 −6
−60 𝑥 2 2𝑥 −3 5𝑥
10 𝑥 2
−15𝑥 2 4𝑥 −6
−60 𝑥 2 2𝑥 −3
10 𝑥 2
−15𝑥 4𝑥 −6
−60 𝑥 2
10 𝑥 2 −6
10 𝑥 2 −6
−60 𝑥 2
Multiply
–15 and +4
𝟎=(𝟓𝒙+𝟐)(𝟐𝒙−𝟑)
𝟎=𝟓𝒙+𝟐
and
𝟎=𝟐𝒙−𝟑
−𝟐=𝟓𝒙
and
𝟑=𝟐𝒙
− 𝟐 𝟓 =𝒙
and
𝟑 𝟐 =𝒙

31. b. 𝒚= 𝒙 𝟐 +𝟐𝒙−𝟏𝟓 𝟎= 𝒙 𝟐 +𝟐𝒙−𝟏𝟓 𝟎=(𝒙+𝟓)(𝒙−𝟑) 𝟎=𝒙+𝟓 𝟎=𝒙−𝟑 −𝟓=𝒙 𝟑=𝒙
To identify the zeros, set y equal to 0 and factor.
𝟎= 𝒙 𝟐 +𝟐𝒙−𝟏𝟓
Identify two numbers that multiply to –15 and add to 2.
Factor the GCF up the columns and across the rows 𝑥 −3
𝑥 2
−3𝑥 5𝑥
−15
−15 𝑥 2
𝑥 2
−15
−15 𝑥 2
𝑥 2
−15 𝑥 −3
𝑥 2
−3𝑥 5 5𝑥
−15
−15 𝑥 2
𝑥 2
−3𝑥 5𝑥
−15
−15 𝑥 2
Multiply
+5 and –3
𝟎=(𝒙+𝟓)(𝒙−𝟑)
𝟎=𝒙+𝟓
and
𝟎=𝒙−𝟑
−𝟓=𝒙
and
𝟑=𝒙

32. c. 𝒚=𝟑 𝒙 𝟐 +𝟑𝟑𝒙+𝟗𝟎 𝟎= 𝟑𝒙 𝟐 +𝟑𝟑𝒙+𝟗𝟎 𝟎= 𝟑(𝒙 𝟐 +𝟏𝟏𝒙+𝟑𝟎) 𝟎=𝟑(𝒙+𝟓)(𝒙+𝟔)
To identify the zeros, set y equal to 0 and factor.
𝟎= 𝟑𝒙 𝟐 +𝟑𝟑𝒙+𝟗𝟎
Factor out the GCF.
𝟎= 𝟑(𝒙 𝟐 +𝟏𝟏𝒙+𝟑𝟎)
Identify two numbers that multiply to 30 and add to 11.
Factor the GCF up the columns and across the rows
𝑥 2 6𝑥 5𝑥 30
30 𝑥 2 𝑥 6
𝑥 2 6𝑥 5 5𝑥 30
30 𝑥 2 𝑥 6
𝑥 2 6𝑥 5𝑥 30
30 𝑥 2
𝑥 2 30
𝑥 2 30
30 𝑥 2
Multiply
+5 and +6
𝟎=𝟑(𝒙+𝟓)(𝒙+𝟔)
𝟎=𝒙+𝟓
and
𝟎=𝒙+𝟔
−𝟓=𝒙
and
−𝟔=𝒙

33. Distribute the negative.
27. Simplify the expressions.
a. −25
b. −54
𝒊= −𝟏
−25 = −1×25
−54 = −1×9×6
𝟗 =𝟑
𝟐𝟓 =𝟓
−25 =5𝑖
−54 =3𝑖 6
c. 3− −18 −(12+ −27 )
Simplify the radicals.
3− −1×9×2 −(12+ −1×9×3 )
3−3𝑖 2 −(12+3𝑖 3 )
Distribute the negative.
3−3𝑖 2 −12−3𝑖 3
Combine like terms.
−9−3𝑖 2 −3𝑖 3
d −20 +(−21− −45 )
Simplify the radicals.
8+ −1×4×5 +(−21− −1×9×5 )
8+2𝑖 5 +(−21−3𝑖 5 )
Combine like terms.
−13−𝑖 5

34. 28. Simplify the following.
a. −5−2𝑖 −(3−4𝑖)
b. 8−𝑖 +(2−7𝑖)
Distribute the negative.
−5−2𝑖−3+4𝑖
8−𝑖+2−7𝑖
−8+2𝑖
10−8𝑖
Use the Distributive Property
d. (−5−2𝑖)(3−4𝑖)
c. (−10+𝑖)(3−9𝑖)
−15+20𝑖−6𝑖+8 𝑖 2
−30+90𝑖+3𝑖−9 𝑖 2
𝒊 𝟐 =−𝟏
−15+14𝑖+8(−1)
−30+93𝑖−9(−1)
−23+14𝑖
−21+93𝑖
e. 3−4𝑖 −(−5−2𝑖)
Distribute the negative.
f. −10+𝑖 +(3−9𝑖)
3−4𝑖+5+2𝑖
−10+𝑖+3−9𝑖
8−2𝑖
−7−8𝑖

35. 29. Find the solutions to the following: a. 25 𝑥 2 +49=0
Subtract 49 from both sides of the equation.
− −49
25𝑥 2 =−49
Divide both sides of the equation by 25.
𝑥 2 =− 49 25
Use square roots.
𝑥 2 =± − 49 25
𝒊= −𝟏
𝟒𝟗 =𝟕
𝑥=± 7 5 𝑖
𝟐𝟓 =𝟓

36. Find the solutions to the following: b. 𝑥 2 +2𝑥−2=0
This equation is not factorable, so we can use the Quadratic Formula.
a = 1
b = 2
c = -2
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= −2± −4(1)(−2) 2(1)
Substitute the values for a, b, and c into the formula.
𝑥= −2± 4−(−8) 2
Simplify.
𝑥= −2±
𝟏𝟐 = 𝟒×𝟑
𝑥= −2±
Simplify the numerator and denominator using a factor of 2.
𝑥= −1±
𝑥=−1± 3

37. Find the solutions to the following: c. 𝑥 2 −6𝑥=1
Use completing the square and square roots to solve.
Take half of the x coefficient and square it.
𝟏 𝟐 −𝟔 =−𝟑
𝑥 2 −6𝑥+____=1
To keep the equation balanced, add 9 to both sides of the equation.
Add this number to create a perfect square trinomial.
(−𝟑) 𝟐 =𝟗
𝑥 2 −6𝑥+9=1
𝑥 2 −6𝑥+9=1+9
Factor the perfect square trinomial and combine like terms.
(𝑥−3) 2 =10
(𝑥−3) 2 =± 10
Square root both sides of the equation.
𝑥−3=± 10
Add 3 to both sides of the equation.
𝑥=3± 10

38. Find the solutions to the following: d. 3 𝑥 2 +3𝑥+2=0
This equation is not factorable, so we can use the Quadratic Formula.
a = 3
b = 3
c = 2
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= −3± −4(3)(2) 2(3)
Substitute the values for a, b, and c into the formula.
𝑥= −3± 9−24 6
Simplify.
𝑥= −3± −15 6
𝒊= −𝟏
𝑥= −3±𝑖

39. Divide both sides of the equation by 5.
Find the solutions to the following:
e. 5 𝑥 2 =−120
Divide both sides of the equation by 5.
𝑥 2 =−24
Use square roots.
𝑥 2 =± −24
𝒊= −𝟏
𝑥=± −1×4×6
𝟒 =𝟐
𝑥=±2𝑖 6

40. Find the solutions to the following: f. 3 𝑥 2 +10=4𝑥
Set the equation equal to zero by subtracting 4x from both sides of it.
−4𝑥 −4𝑥
This equation is not factorable, so we can use the Quadratic Formula.
3 𝑥 2 −4𝑥+10=0
a = 3
b = -4
c = 10
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= −(−4)± (−4) 2 −4(3)(10) 2(3)
Substitute the values for a, b, and c into the formula.
𝑥= 4± 16−120 6
Simplify.
𝑥= 4± −104 6
−𝟏𝟎𝟒 = −𝟏×𝟒×𝟐𝟔
𝒊= −𝟏
𝟒 =𝟐
𝑥= 4±2𝑖
Simplify the numerator and denominator using a factor of 2.
𝑥= 2±𝑖

41. 30. Identify the vertex and axis of symmetry for each of the following graphs.b. c.
Vertex (3, 1)
Vertex (0, 4)
Vertex (–2, –3)
Axis of symmetry 𝑥=3
Axis of symmetry 𝑥=0
Axis of symmetry 𝑥=−2

42. The parabola opens up and has a minimum value.
31.
The parabola opens up and has a minimum value.
Since a = +1, the parabola opens up and has a minimum value.
The minimum = 3 and occurs at the vertex (2, 3).
The minimum = 11 and occurs at the vertex (2, 11).
Part b has a smaller minimum.

43. 32. Which of the following functions has the largest maximum?b) c)
To find the vertex, solve for
𝒙= −𝒃 𝟐𝒂
Note the duplicated y values when x=0 and x=4 (and again at x=-2 & x=6). This shows the vertex of the parabola is at (2, -4).
𝒙= −𝟏 𝟐(−𝟏) = −𝟏 −𝟐 = 𝟏 𝟐
Plug in the x coordinate of ½ to find the y coordinate of the vertex.
Since the y values are decreasing on either side of the vertex, the parabola is opening down.
The above has a vertex at (-2,-1) and opens down.
𝒚=− 𝟏 𝟐 𝟐 + 𝟏 𝟐 −𝟑
Since it opens down, it has a maximum.
The maximum is at y = -4.
Since it opens down, it has a maximum.
𝒚=− 𝟏 𝟒 + 𝟏 𝟐 −𝟑
The maximum is at y = -1.
𝒚=− 𝟏𝟏 𝟒
The leading coefficient is negative, so this parabola opens down.
The vertex is (.5, -2.75).
The maximum is y= – 2.75.
Comparing the maximum values from (a), (b), and (c), the function with the largest maximum is (b).

44. 33. Give the equation for each of the following graphs.b. c.
Vertex (3, 1)
Vertex (0, 4)
Vertex (–2, –3)
𝑎=2
𝑎=− 1 2
𝑎= 2 3
𝑦=𝑎 𝑥−ℎ 2 +𝑘
𝑦=𝑎 𝑥−ℎ 2 +𝑘
𝑦=𝑎 𝑥−ℎ 2 +𝑘
𝑦=2 𝑥−
𝑦=− 1 2 𝑥 2 +4
𝑦=− 𝑥+2 2 −3

45. 34. What is the vertex form of the equation? a. y= 𝑥 2 −2𝑥+8
Vertex form of a quadratic equation is 𝒚=𝒂 (𝒙−𝒉) 𝟐 +𝒌.
Use completing the square.
y= (𝑥+ ________) __________
𝑏 2𝑎
𝑏 2𝑎 2
−2 2 =−1 1
y= (𝑥−1)
Combine like terms.
y= (𝑥−1) 2 + 7

46. 34. What is the vertex form of the equation? a. y= 𝑥 2 −2𝑥+8
Vertex form of a quadratic equation is 𝒚=𝒂 (𝒙−𝒉) 𝟐 +𝒌.
Add this number to create a perfect square trinomial.
Take half of the x coefficient and square it.
𝟏 𝟐 −𝟐 =−𝟏
Use completing the square.
To keep the equation balanced, subtract 1 as well. This way, a total of 0 has been added to the right side of the equation.
y= (𝑥 2 −2𝑥+____)+8
(−𝟏) 𝟐 =𝟏
y= (𝑥 2 −2𝑥+1)+8−1
y= (𝑥 2 −2𝑥+1)+8
y=(𝑥−1)(𝑥−1)+8−1
Factor the perfect square trinomial.
y= (𝑥−1) 2 +8−1
Combine like terms.
y= (𝑥−1) 2 +7
b. y= −2𝑥 2 +8𝑥−2
Factor out the GCF of the terms with variables.
y= −2(𝑥 2 −4𝑥+____)−2
Add this number to create a perfect square trinomial.
To keep the equation balanced, add 8 as well. This way, a total of 0 has been added to the right side of the equation.
Because we had a GCF, we did not just add 4. We really added −𝟐 𝟒 =−𝟖.
Take half of the x coefficient and square it.
𝟏 𝟐 −𝟒 =−2
Use completing the square.
y= −2(𝑥 2 −4𝑥+4)−2+8
y=−2( 𝑥 2 −4𝑥+4)−2
(−𝟐) 𝟐 =𝟒
y=−2(𝑥−2)(𝑥−2)+6
Factor the perfect square trinomial and combine like terms.
y= −2(𝑥−2) 2 +6

47. a. focus (6, 13) and directrix 𝑦=7
35. What is the equation of a parabola given the following information?
a. focus (6, 13) and directrix 𝑦=7
𝒑 is the distance from the focus to the vertex
Focus
Sometimes a graph can help to visualize the parabola.
p = 3
The vertex lies midway between the focus and directrix.
The focus lies inside the parabola, so this graph shows the parabola will open up. Therefore, p is positive and the equation is (𝒙−𝒉) 𝟐 =𝟒𝒑(𝒚−𝒌).
Directrix
𝒑 is also the distance from the directrix to the vertex
The vertex is represented by (h, k).
Substitute the corresponding values into the equation.
(𝒙−𝒉) 𝟐 =𝟒𝒑(𝒚−𝒌)
(𝒙−𝟔) 𝟐 =𝟒(𝟑)(𝒚−𝟏𝟎)
(𝒙−𝟔) 𝟐 =𝟏𝟐(𝒚−𝟏𝟎)

48. What is the equation of a parabola given the following information?
b. focus (–3, 1) and directrix 𝑥=−1
𝒑 is the distance from the focus to the vertex
Sometimes a graph can help to visualize the parabola.
The vertex lies midway between the focus and directrix.
p = –1
The focus lies inside the parabola, so this graph shows the parabola will open left. Therefore, p is negative and the equation is (𝒚−𝒌) 𝟐 =𝟒𝒑(𝒙−𝒉).
Focus
𝒑 is also the distance from the directrix to the vertex
The vertex is represented by (h, k).
Directrix
Substitute the corresponding values into the equation.
(𝒚−𝒌) 𝟐 =𝟒𝒑(𝒙−𝒉)
(𝒚−𝟏) 𝟐 =𝟒(−𝟏)(𝒙−(−𝟐))
(𝒚−𝟏) 𝟐 =−𝟒(𝒙+𝟐)

49. Since 𝒚=𝟐𝒙−𝟑, substitute 𝟐𝒙−𝟑 in for y in the first equation.
36. Solve the system.
a. 𝑦=− 𝑥 2 +4𝑥+5 𝑦=2𝑥−3
Since 𝒚=𝟐𝒙−𝟑, substitute 𝟐𝒙−𝟑 in for y in the first equation.
𝒚 =− 𝒙 𝟐 +𝟒𝒙+𝟓
𝟐𝒙−𝟑 =− 𝒙 𝟐 +𝟒𝒙+𝟓
Set the equation equal to 0.
−2𝑥 −2𝑥+3
Divide both sides of the equation by –1 to create a positive leading coefficient.
𝟎=− 𝒙 𝟐 +𝟐𝒙+8
− −1
𝟎= 𝒙 𝟐 −𝟐𝒙−8
Factor and solve.
𝟎=(𝒙−𝟒)(𝒙+𝟐)
Substitute the x-values into one of the equations to solve for the corresponding y-values.
𝟎=𝒙−𝟒
and
𝟎=𝒙+𝟐
𝒙=𝟒
and
𝒙=−𝟐
𝒚=𝟐𝒙−𝟑
𝒚=𝟐𝒙−𝟑
𝒚=𝟐(𝟒)−𝟑
𝒚=𝟐(−𝟐)−𝟑
𝒚=𝟖−𝟑
𝒚=−𝟒−𝟑
𝒚=𝟓
𝒚=−𝟕
(𝟒, 𝟓)
(−𝟐, −𝟕)

50. Set the equation equal to 0.
Solve the system.
b. 𝑦= 𝑥 2 +16𝑥+32 𝑥+𝑦=2
Since 𝒚= 𝒙 𝟐 +𝟏𝟔𝒙+𝟑𝟐, substitute 𝒙 𝟐 +𝟏𝟔𝒙+𝟑𝟐 in for y in the second equation.
𝒙+𝒚 =𝟐
𝒙+ 𝒙 𝟐 +𝟏𝟔𝒙+𝟑𝟐=𝟐
Collect like terms.
𝒙 𝟐 +𝟏𝟕𝒙+𝟑𝟐=𝟐
Set the equation equal to 0.
−2 −2
𝒙 𝟐 +𝟏𝟕𝒙+𝟑𝟎=𝟎
Factor and solve.
𝟎=(𝒙+𝟏𝟓)(𝒙+𝟐)
𝒙+𝟏𝟓=𝟎
and
𝒙+𝟐=𝟎
𝒙=−𝟏𝟓
and
𝒙=−𝟐
Substitute the x-values into one of the equations to solve for the corresponding y-values.
𝒙+𝒚=𝟐
𝒙+𝒚=𝟐
−𝟏𝟓+𝒚=𝟐
−𝟐+𝒚=𝟐
𝒚=𝟏𝟕
𝒚=𝟒
(−𝟏𝟓, 𝟏𝟕)
(−𝟐, 𝟒)

51. 37. Find the inverse of the relation if possible. (Refer to Question #2.) For those that were previously not invertible, give a restriction on the domain of the function in order to make it invertible.
Only Parts b and e were not invertible.
b. 𝑓 𝑥 =5 𝑥 2 +8
e. 𝑓 𝑥 = 𝑥 2 −16𝑥+20
This graph will pass the horizontal line test if the domain is restricted to either 𝒙≤𝟎 or 𝒙≥𝟎.
With an axis of symmetry at 𝒙=𝟖, this graph will pass the horizontal line test if the domain is restricted to either 𝒙≤𝟖 or 𝒙≥𝟖.

52. Remember whole numbers are 0, 1, 2, 3, ….
38. Name all of the operations under which polynomials are closed.
Addition, Subtraction and Multiplication
Is the set of whole numbers closed under
a. Addition?
Remember whole numbers are 0, 1, 2, 3, ….
Adding whole numbers will result in equivalent or larger whole numbers, so YES
b. Subtraction?
Subtracting whole numbers may result in integers, so NO. Example: 3−7=−4
c. Multiplication?
Multiplying whole numbers will result in equivalent or larger whole numbers, so YES
d. Division?
Dividing whole numbers may result in rational numbers, so NO. Example: 3÷7= 3 7

54. 42. Describe the end behavior of the following polynomials.
a. y=2 𝑥 4 −4 𝑥 3 −6 𝑥 2 +8𝑥+8 b. 𝑦= 1 4 (𝑥+3) 3 −2
Make sure the equation is written in standard form. Then look to the leading term to determine end behavior.
The polynomial is odd (degree of 3), so the end behavior will either be down on the right and up on the left or up on the right and down on the left.
The polynomial is even (degree of 4), so the end behavior will either be up on the left and right, or down on the left and right.
Since the leading coefficient is positive, the end behavior goes down on the left and up on the right.
Since the leading coefficient is positive, the end behavior goes up on the left and right.
This can also be described by stating as 𝒙→−∞, 𝒇 𝒙 →+∞ and as 𝒙→+∞, 𝒇 𝒙 →+∞.
This can also be described by stating as 𝒙→−∞, 𝒇 𝒙 →−∞ and as 𝒙→+∞, 𝒇 𝒙 →+∞.

55. Describe the end behavior of the following polynomials
Describe the end behavior of the following polynomials. Then sketch a graph of each.
a. y=2 𝑥 4 −4 𝑥 3 −6 𝑥 2 +8𝑥+8
Identify the zeros of the polynomial.
Factor out the GCF of 2.
Use the Rational Root Theorem or other tools to test possible zeros.
y=2( 𝑥 4 −2 𝑥 3 −3 𝑥 2 +4𝑥+4) 2
1 −2 −
Let’s test +2 using synthetic division. 2 −6 −4
2 is a zero! Let’s test +2 again using synthetic division to see if it is a double root. −3 −2 1
𝐲=𝟐(𝒙−𝟐)( 𝒙 𝟑 −𝟑𝒙−𝟐) 2
−3 −2 2 4 2 1 2 1
Factor.
𝐲=𝟐(𝒙−𝟐)(𝒙−𝟐)( 𝒙 𝟐 +𝟐𝒙+𝟏)
𝐲=𝟐(𝐱−𝟐)(𝐱−𝟐)(𝒙+𝟏)(𝒙+𝟏)
Zeros = –1 and 2, each with a multiplicity of 2.
𝐲=𝟐 (𝒙−𝟐) 𝟐 (𝒙+𝟏) 𝟐

56. Sketch the graph using the end behavior and the plotted points.
Describe the end behavior of the following polynomials. Then sketch a graph of each.
a. y=2 𝑥 4 −4 𝑥 3 −6 𝑥 2 +8𝑥+8
𝐲=𝟐 (𝒙−𝟐) 𝟐 (𝒙+𝟏) 𝟐
Plot the zeros first. Even multiplicity means the graph will touch (not cross) the x-axis.
Test points between the zeros to determine possible local maximums and minimums.
𝒙= 𝟏 𝟐 is half way between the zeros.
𝐲=𝟐 ( 𝟏 𝟐 −𝟐) 𝟐 ( 𝟏 𝟐 +𝟏) 𝟐
𝐲=𝟏𝟎 𝟏 𝟖
Sketch the graph using the end behavior and the plotted points.

57. Describe the end behavior of the following polynomials
Describe the end behavior of the following polynomials. Then sketch a graph of each.
b. 𝑦= 1 4 (𝑥+3) 3 −2
Graph the parent function 𝒚= 𝒙 𝟑 . 𝒙 𝒚 −2 −8 −1 1 2 8
Translate the vertex 3 units to the left and 2 units down.
Vertically compress the function by a factor of 𝟏 𝟒 .

58. 43. What are the solutions (zeros) to the following equations?
Factor the difference of cubes.
𝒂 𝟑 − 𝒃 𝟑 =(𝒂−𝒃)( 𝒂 𝟐 +𝒂𝒃+ 𝒃 𝟐 )
(2𝑥−1)(4 𝑥 2 +2𝑥+1)=0
Set each factor equal to 0.
Use the quadratic formula to solve.
2𝑥−1=0
4 𝑥 2 +2𝑥+1=0
and
2𝑥=1
a = 4
b = 2
c = 1
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= 1 2
Substitute the values for a, b, and c into the formula.
𝑥= −2± (2) 2 −4(4)(1) 2(4)
𝑥= −2± 4−16 8
Simplify.
𝑥= −2± −12 8
−𝟏𝟐 = −𝟏×𝟒×𝟑
𝑥= −2±2𝑖 3 8
Simplify using a factor of 2.
𝑥= −1±𝑖 3 4

59. What are the solutions (zeros) to the following equations?
b. 𝑥 3 =−125
Factor the sum of cubes.
𝒂 𝟑 + 𝒃 𝟑 =(𝒂+𝒃)( 𝒂 𝟐 −𝒂𝒃+ 𝒃 𝟐 )
Set the equation equal to 0.
𝑥 =0
(𝑥+5)( 𝑥 2 −5𝑥+25)=0
Set each factor equal to 0.
𝑥+5=0
𝑥 2 −5𝑥+25=0
Use the quadratic formula to solve.
and
𝑥=−5
a = 1
b = -5
c = 25
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Substitute the values for a, b, and c into the formula.
𝑥= −(−5)± (−5) 2 −4(1)(25) 2(1)
𝑥= 5± 25−100 2
Simplify.
𝑥= 5± −75 2
−𝟕𝟓 = −𝟏×𝟐𝟓×𝟑
𝑥= 5±5𝑖 3 2

60. What are the solutions (zeros) to the following equations?
c. 𝑥 4 −40 𝑥 =0
Factor.
( 𝑥 2 −4)( 𝑥 2 −36)=0
Factor again – difference of squares.
(𝑥+2)(𝑥−2)(𝑥+6)(𝑥−6)=0
Set each factor equal to zero.
𝑥+2=0
𝑥−2=0
𝑥+6=0
𝑥−6=0
𝑥=−2
𝑥=2
𝑥=−6
𝑥=6
d. 2𝑥 3 +3 𝑥 2 −18𝑥=27
Set the equation equal to zero.
2𝑥 3 +3 𝑥 2 −18𝑥−27=0
Factor by grouping.
𝑥 2 (2𝑥+3)−9(2𝑥+3)=0
(𝑥 2 −9)(2𝑥+3)=0
Factor again – difference of squares.
(𝑥+3)(𝑥−3)(2𝑥+3)=0
Set each factor equal to zero.
𝑥+3=0
𝑥−3=0
2𝑥+3=0
𝑥=−3
𝑥=3
𝑥=− 3 2

61. Factor the difference of cubes.
44. Factor completely.
a. 27 𝑥
Factor the sum of cubes.
𝒂 𝟑 + 𝒃 𝟑 =(𝒂+𝒃)( 𝒂 𝟐 −𝒂𝒃+ 𝒃 𝟐 )
(3𝑥)
(3𝑥+5)( 3𝑥 2 − 3𝑥 )
(3𝑥+5)( 9𝑥 2 −15𝑥+25)
b. 𝑥 3 −64
𝑥 3 − 4 3
Factor the difference of cubes.
𝒂 𝟑 − 𝒃 𝟑 =(𝒂−𝒃)( 𝒂 𝟐 +𝒂𝒃+ 𝒃 𝟐 )
(𝑥−4)( 𝑥 2 + 𝑥 )
(𝑥−4)( 𝑥 2 +4𝑥+16)

62. −𝟑 is a zero of the polynomial. 𝟎 is a zero of the polynomial.
45.
−𝟑 is a zero of the polynomial.
𝟎 is a zero of the polynomial.
Since the graph touches the x-axis here and turns around, this zero must have an even multiplicity.
Since the graph touches the x-axis here and turns around, this zero must have an even multiplicity.
(𝑥+3) 2
(𝑥−0) 2
=𝑥 2
−𝟏 is a zero of the polynomial.
𝑦= 𝑥 2 𝑥+3 2 (𝑥+1)
The end behavior tells us that the leading coefficient is negative.
Since the graph crosses the x-axis here, this zero must have an odd multiplicity.
𝑦= −𝑥 2 𝑥+3 2 (𝑥+1)
(𝑥+1)

63. 𝟑 is a zero of the polynomial.b.
𝟑 is a zero of the polynomial.
𝟐 is a zero of the polynomial.
Since the graph crosses the x-axis here, this zero must have an odd multiplicity.
Since the graph touches the x-axis here and turns around, this zero must have an even multiplicity.
(𝑥−3)
(𝑥−2) 2
𝟎 is a zero of the polynomial.
The end behavior tells us that the leading coefficient is positive.
Since the graph crosses the x-axis here, this zero must have an odd multiplicity.
𝑦=𝑥 𝑥−2 2 (𝑥−3)
(𝑥+0) =𝑥

64. 46.

65. Find the value of 𝑷(𝟐) to determine if the remainder is 0.
47. Use the Remainder Theorem to determine if the given value is a zero of the polynomial.
a. Is 2 a zero of 𝑃 𝑥 = 𝑥 3 +2 𝑥 2 −6𝑥−4?
𝑃 𝑥 = 𝑥 3 +2 𝑥 2 −6𝑥−4
Find the value of 𝑷(𝟐) to determine if the remainder is 0.
𝑃 2 = (2) 2 −6(2)−4
𝑃 2 =8+2(4)−12−4
𝑃 2 =8+8−12−4
Since the remainder is 0, 2 is a zero of the polynomial.
𝑃 2 =0
b. Is −1 a zero of 𝑃 𝑥 = 𝑥 3 + 𝑥 2 −3𝑥+8?
𝑃 𝑥 = 𝑥 3 + 𝑥 2 −3𝑥+8
𝑃 −1 = (−1) 3 + (−1) 2 −3(−1)+8
𝑃 −1 =−
Since the remainder is NOT 0, −𝟏 is NOT a zero of the polynomial.
𝑃 −1 =11

66. 𝑥−4 𝑥−4 × 𝑥 𝑥−2 + 𝑥−2 𝑥−2 × 1 𝑥−4 = 2 (𝑥−4)(𝑥−2)
48. Find the solutions to the following equations and name any extraneous solutions that arise.
Factor the denominator to find all of the factors needed for a common denominator. a.
𝑥 𝑥−2 + 1 𝑥−4 = 2 𝑥 2 −6𝑥+8
𝑥 𝑥−2 + 1 𝑥−4 = 2 (𝑥−4)(𝑥−2)
𝒙=𝟒 would have 0 in the denominator which is undefined!
𝑥−4 𝑥−4 × 𝑥 𝑥−2 + 𝑥−2 𝑥−2 × 1 𝑥−4 = 2 (𝑥−4)(𝑥−2)
Set the equation = 0 by subtracting 2 (𝑥−4)(4−2) from both sides of the equation and combine numerators with a common denominator.
𝑥 2 −4𝑥 (𝑥−4)(𝑥−2) + 𝑥−2 (𝑥−4)(𝑥−2) = 2 (𝑥−4)(𝑥−2)
𝑥 2 −4𝑥+𝑥−2−2 (𝑥−4)(𝑥−2) =0
𝑥 2 −3𝑥−4 (𝑥−4)(𝑥−2) =0
Clear the denominator.
𝑥 2 −3𝑥−4=0
Factor and solve.
(𝑥−4)(𝑥+1)=0
Since 𝒙≠𝟒, it is extraneous.
𝑥=4 𝑜𝑟 𝑥=−1

67. 5 𝑥 2 +8𝑥−4 (3𝑥+2)(𝑥+3) − 1 𝑥+3 × 3𝑥+2 3𝑥+2 = 𝑥 3𝑥+2 × 𝑥+3 𝑥+3
Factor the denominator to find all of the factors needed for a common denominator. b.
5 𝑥 2 +8𝑥−4 3 𝑥 2 +11𝑥+6 − 1 𝑥+3 = 𝑥 3𝑥+2
5 𝑥 2 +8𝑥−4 (3𝑥+2)(𝑥+3) − 1 𝑥+3 = 𝑥 3𝑥+2
Set the equation = 0 and combine numerators with a common denominator.
5 𝑥 2 +8𝑥−4 (3𝑥+2)(𝑥+3) − 1 𝑥+3 × 3𝑥+2 3𝑥+2 = 𝑥 3𝑥+2 × 𝑥+3 𝑥+3
5 𝑥 2 +8𝑥−4 (3𝑥+2)(𝑥+3) − 3𝑥+2 (3𝑥+2)(𝑥+3) = 𝑥 2 +3𝑥 (3𝑥+2)(𝑥+3)
5 𝑥 2 +8𝑥−4− 3𝑥+2 −( 𝑥 2 +3𝑥) (3𝑥+2)(𝑥+3) =0
Clear the denominator and simplify.
5 𝑥 2 +8𝑥−4−3𝑥−2− 𝑥 2 −3𝑥=0
4 𝑥 2 +2𝑥−6=0
2 𝑥 2 +𝑥−3=0
Factor and solve.
(2𝑥+3)(𝑥−1)=0
𝑥=− 𝑜𝑟 𝑥=1

68. The common denominator for the left side of the equation is 𝒙(𝒙+𝟐). a.
49. Find the value(s) of x for which f(x) = g(x). Name any extraneous solutions that arise.
The common denominator for the left side of the equation is 𝒙(𝒙+𝟐). a.
3 𝑥+2 − 1 𝑥 = 1 5𝑥
Cross multiply
𝑥 𝑥 × 3 𝑥+2 − 𝑥+2 𝑥+2 × 1 𝑥 = 1 5𝑥
𝒙=𝟎 would have 0 in the denominator which is undefined! b.
4 𝑥 = 3 𝑥−2
3𝑥−(𝑥+2) 𝑥(𝑥+2) = 1 5𝑥
Simplify
4 𝑥−2 =3𝑥
2𝑥−2 𝑥 2 +2𝑥 = 1 5𝑥
4𝑥−8=3𝑥
Cross multiply
−8=−𝑥
5𝑥(2𝑥−2)=1( x 2 +2x)
𝑥=8
Set the quadratic equal to 0 and solve.
10 𝑥 2 −10𝑥= 𝑥 2 +2𝑥
9 𝑥 2 −12𝑥=0
3𝑥(3𝑥−4)=0
𝑥=0 𝑜𝑟 𝑥= 4 3
Since 𝒙≠𝟎, it is extraneous.

69. 50. Name any vertical and horizontal asymptotes of the graphs of the following equations.
𝑦= 𝑥+7 𝑥 2 +3𝑥+2
VA:
Set the denominator = to 0 and solve.
𝑥 2 +3x+2=0
Vertical Asymptotes are
𝒙=−𝟐 and 𝒙=−𝟏
𝑥+2 𝑥+1 =0
𝑥=−2 𝑎𝑛𝑑 𝑥=−1
HA:
BOB0 BOTN EATS DC
BOB0
Horizontal Asymptote is
𝒚=𝟎
𝑦=0

70. The vertical asymptote is 𝒙=−𝟏. The horizontal asymptote is 𝒚=𝟏.
50. Name any vertical and horizontal asymptotes of the graphs of the following equations. b.
𝑦= (𝑥−2)(𝑥−3) (𝑥−3)(𝑥+1)
Simplify the fraction
𝑦= 𝑥−2 𝑥+1
VA:
𝑥−2=0
The vertical asymptote is 𝒙=−𝟏.
𝑥=2
HA:
EATS DC
The horizontal asymptote is 𝒚=𝟏.
𝑦= =1

71. Factor to see if the fraction can be simplified.
50. Name any vertical and horizontal asymptotes of the graphs of the following equations. c.
𝑦= 2 𝑥 2 +𝑥+3 3 𝑥 2 −8𝑥+5
Factor to see if the fraction can be simplified.
The numerator is not factorable.
𝑦= 2 𝑥 2 +𝑥+3 (3𝑥−5)(𝑥−1)
3𝑥−5=0
𝑥−1=0
The vertical asymptotes are 𝒙=𝟏 and 𝒙= 𝟓 𝟑 .
3𝑥=5
𝑥=1
𝑥= 5 3
Degree = 2
Look to the degree of the numerator and the degree of the denominator for the horizontal asymptote.
𝑦= 2 𝑥 2 +𝑥+3 3 𝑥 2 −8𝑥+5
Since the degree of the numerator = the degree of the denominator, divide the leading coefficients.
Degree = 2
The horizontal asymptote is 𝒚= 𝟐 𝟑 .

72. The denominator is not factorable. There are no vertical asymptotes.
50. Name any vertical and horizontal asymptotes of the graphs of the following equations.
Degree = 1 d.
𝑦= 𝑥−12 𝑥 2 +23𝑥−1
The denominator is not factorable.
There are no vertical asymptotes.
Degree = 2
Since the degree of the numerator < the degree of the denominator, the horizontal asymptote is 𝒚=𝟎.