Chapter 5 review solutions
1) 𝑓(𝑥) = (𝑥 − 5) 2 (𝑥 − 3) a) List each x-intercept (zero) and its multiplicity (round to 2 decimal places when needed)
(x-5) 2 (x-3) = 0
(x-5)(x-5)(x-3)=0 x – 5 = 0 or x – 3 = 0 x = 5 or x = 3
Answer: (5,0) even multiplicity, (3,0) odd multiplicity b) Determine whether the graph crosses or touches the x-axis at each x-intercept
Answer: graph touches x-axis at (5,0) graph crosses x-axis at (3,0) c) Determine the maximum number of turning points on the graph
Multiply the first in the expanded polynomial to get x*x*x = x 3 , this is a third degree polynomial.
Answer: will have at most 2 turning points d) Which function will the graph of f(x) behave like for large values of |𝑥|
Answer: graph will behave like f(x) = x 3 for large values of x. e) Describe the end behavior
Answer: falls to the left, rises to the right f) sketch a graph and approximate the turning points, also label the x-intercepts
) y
8
6
4
-9 -8 -7 -6 -5 -4 -3 -2 -1
2
(3.67,1.19)
1
(3,0) (5,0)
2 3 4 5 6 7 8 9 x
-2
-4
-6
-8 g) state the intervals where the function is increasing and decreasing answer: increasing (−∞, 𝟑. 𝟔𝟕) ∪ (𝟓, ∞) decreasing (3.67, 5)

2) 𝑓(𝑥) = (𝑥
2
− 16)(3𝑥 − 21) a) List each x-intercept (zero) and its multiplicity (round to 2 decimal places when needed)
(x 2 - 16)(3x – 21) = 0
(x+4)(x-4)(3x-21) = 0 x+ 4 = 0 x – 4 = 0 3x – 21 = 0 x = -4 x = 4 x = 7
Answer: (-4,0) odd mult (4,0) odd mult (7,0) odd mult b) Determine whether the graph crosses or touches the x-axis at each x-intercept
Answer: crosses at (-4,0) crosses at ( 4,0) crosses at ( 7,0) c) Determine the maximum number of turning points on the graph multiply the firsts in the expanded problem x 2 *3x = 3x 3 this is a 3rd degree polynomial
Answer: will have at most 2 turning points. d) Which function will the graph of f(x) behave like for large values of |𝑥|
Answer: will behave like f(x) = 3x 3 e) Describe the end behavior
Answer: fall left and rises to the right f) sketch a graph and approximate the turning points, also label the x-intercepts y
450
400
(-.95,360)
350
300
250
200
150
100
50
2
(4,0)
3 4 5
(7,0)
6 7 x
-7 -6
(-4,0)
-5 -4 -3 -2 -1 1 8 9
-50
(5.6,-64.5) g) state the intervals where the function is increasing and decreasing
Answer: increasing (−∞, −. 𝟗𝟓) ∪ (𝟓. 𝟔, ∞) decreasing (-.95, 5.6)

3) 𝑓(𝑥) = 𝑥
4
− 4𝑥
3
− 21𝑥
2 a) List each x-intercept (zero) and its multiplicity (round to 2 decimal places when needed) x 4 – 4x 3 – 21x 2 = 0 x 2 (x 2 – 4x – 21) = 0 x(x+3)(x-7) = 0 x 2 = 0 x+3 = 0 x – 7 = 0 𝑥 = ±√0 x = 0 x = -3 x = 7
Answer: (0,0) has even multiplicity and both (-3,0) (7,0) have odd multiplicity b) Determine whether the graph crosses or touches the x-axis at each x-intercept
Answer: graph will cross x-axis at (-3,0) and (7,0) and touch at (0,0) c) Determine the maximum number of turning points on the graph
Answer: this is a fourth degree polynomial and will have at most 3 turning points. d) Which function will the graph of f(x) behave like for large values of |𝑥|
Answer: for large values of x graph will behave like f(x) = x 4 e) Describe the end behavior
Answer: rises to the left, rises to the right f) sketch a graph and approximate the turning points, also label the x-intercepts y
400
300
200
100
-7
(-3,0)
-6 -5 -4 -3 (-2.07,-36.14) -2
(0,0)
-1
-100
1 2 3 4 5
(7,0)
6 7 8 9 x
-200
-300
-400
(5.07,-400.36) g) state the intervals where the function is increasing and decreasing
Answer: 𝒅𝒆𝒄𝒓𝒆𝒂𝒔𝒊𝒏𝒈 (−∞ − 𝟐. 𝟎𝟕) ∪ (𝟎, 𝟓. 𝟎𝟕) 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒊𝒏𝒈 (−𝟐. 𝟎𝟕, 𝟎) ∪ (𝟓. 𝟎𝟕, ∞)

4) f(x) = x 3 + x 2 – 9x– 9 a) use your graphing calculator, or the rational root theorem to find a zero of the polynomial i) you need to find one zero for a third degree polynomial ii) you need to find two zeros for a fourth degree polynomial
Answer: I graphed and found the graph crossed the x-axis at x = 3. I will do my synthetic division with x =3, but you could also use -1 or -3 b) use synthetic division to completely factor the polynomial (use “double” synthetic division for fourth degree polynomials)
3
1
1
1
3
4
Synthetic division gives me: 𝑥
3
+𝑥
2
−9𝑥−9 𝑥−3
= 𝑥
2
+ 4𝑥 + 3 using the result of my synthetic division I get: x 3 + x 2 – 9x– 9 = (x-3)(x 2 + 4x+ 3)
Answer: x 3 + x 2 – 9x– 9 =(x-3)(x+3)(x+1)
-9
12
3
-9
9
0

1
5) f(x) = x 4 +3x 3 -8x 2 -12x +16 a) use your graphing calculator, or the rational root theorem to find a zero of the polynomial i) you need to find one zero for a third degree polynomial ii) you need to find two zeros for a fourth degree polynomial
Answer: Graph crosses x-axis at x = 1 and x = 2. I will use double synthetic division b) use synthetic division to completely factor the polynomial (use “double” synthetic division for fourth degree polynomials)
1
1
3
1
4
-8
4
-4
-12
-4
-16
16
-16
0
2
1
1
4
2
6
Synthetic division tells me: 𝑥
4
+3𝑥
3
−8𝑥
2
−12𝑥+16
(𝑥−1)(𝑥−2)
= 𝑥
2
+ 6𝑥 + 8
-4
12
8 x 4 + 3x 3 – 8x 2 -12x + 16 = (x-1)(x-2)(x 2 + 6x+ 8)
Answer: x 4 +3x 3 -8x 2 -12x +16 = (x-1)(x-2)(x+2)(x+4)
-16
16
0

2
#6-7: Solve
6) x 4 – 13x 2 + 36 = 0
I graphed and found the graph has x-intercepts of x = 2 and x = -2, I will do double synthetic division with 2 then -2
The fact that x = 2 and x = -2 are zeros I know (x-2) and (x+2) are factors
I need to think of the problem as x 4 + 0x 3 -13x 2 + 0x +36 = 0
1
1
0
2
2
-13
4
-9
0
-18
-18
36
-36
0
-2
1
1
The double synthetic division tells me : x 4 – 13x 2 + 36 = (x+2)(x-2)(x 2 – 9) now I can solve the problem x 4 – 13x 2 + 36 = 0
(x+2)(x-2)(x 2 – 9) = 0
(x+2)(x-2)(x+3)(x-3) = 0 x+2 = 0 x – 2 = 0
Answer: x = 2,-2,3,-3 x + 3 = 0
2
-2
0 x – 3 = 0
-9
0
-9
-18
18
0

-1
7) x 3 + x 2 + 2x + 2= 0
The graph of x 3 + x 2 + 2x + 2 crosses the x-axis at -1. I know that x+1 is a factor of x 3 + x 2 + 2x + 2. I will use synthetic division to factor completely.
1
-1
0
2
0
2
2
-2
0
1
1
Synthetic division tells me x 3 + x 2 + 2x + 2 = (x+1)(x 2 + 2)
I can now solve the problem x 3 + x 2 + 2x + 2= 0
(x+1)(x 2 + 2) = 0 x+1 = 0 or x 2 + 2 = 0 x = -1 or x 2 = -2 𝑥 = ±√−2
Answer: x = -1, ±𝒊√𝟐

#8-9: Create a function with lead coefficient 1 that satisfies the conditions.
8) degree 3; zeros -2 and 3 + i f(x) = (x- (3+i))(x-(3-i))(x-(-2))
= (x – 3 – i)(x – 3 + i)(x+2)
= (x 2 -3x+ix – 3x + 9 – 3i – ix + 3i – i 2 )(x+2)
= (x 2 -6x + 9 – (-1))(x+2)
=(x 2 – 6x + 10)(x+2)
= x 3 + 2x 2 – 6x 2 -12x + 10x + 20
Answer: f(x) = x 3 – 4x 2 – 2x + 20
9) degree 3: zeros – 5 and 4i f(x) = (x – (-5))(x-4i)(x – (-4i))
= (x+5)(x-4i)(x+4i)
=(x+5)(x 2 – 16i 2 )
=(x+5)(x 2 + 16)
= x 3 + 16x + 5x 2 +80
Answer: f(x) = x 3 + 5x 2 + 16x + 80

10. f ( x )

2 x x


3
6 a) Domain x- 3 = 0 x = 3
Answer: all real numbers except 3 b) Vertical Asymptote (if any)
Answer: x = 3 c) Horizontal asymptote, or slant asymptote 𝑦 =
2𝑥
= 2 𝑥
Answer: y = 2 d) x- intercept(s) if any
2x + 6 = 0
2x = -6 x = -3
Answer (-3,0) e) y-intercept(s) if any 𝑓(0) =
2∗0+6
0−3
=
6
−3
= −2
Answer: (0,-2) f) Sketch a graph of the function : show all the features found in parts b – e y
8
6
4
2
-18 -16 -14 -12 -10 -8 -6 -4 -2 (-3,0)
-2
(0,-2)
-4
2 4 6 8 10 12 14 16 18 x
-6
-8

11. f ( x )
 x
2  x

4 x
1

21 a) Domain x 2 + 4x – 21 = 0
(x+7)(x-3) = 0 x + 7 = 0 x – 3 = 0 x = -7 x = 3
Answer: all real numbers except -7,3 b) Vertical Asymptote (if any)
Answer: x = -7 and x = 3 c) Horizontal asymptote, or slant asymptote 𝑦 = 𝑥 𝑥
2
=
1 𝑥
Answer: y = 0 (the x-axis) e) y-intercept(s) if any d) x- intercept(s) if any x – 1 = 0 x = 1
Answer: (1,0) 𝑓(0) =
0−1
0 2 +4∗0−21
Answer: (0, 1/21)
=
−1
−21
=
1
21 f) Sketch a graph of the function : show all the features found in parts b - e y
4
3
2
1
-14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 x
-2
-3
-4

12. f ( x )

2 x
2 
7 x x

5

5 a) Domain x – 5 = 0 x = 5
Answer: all real numbers except 5 b) Vertical Asymptote (if any)
Answer: x = 5 c) Horizontal asymptote, or slant asymptote 𝑦 =
2𝑥
2
= 2𝑥 𝑥
Answer: y = 2x slant asymptote d) x- intercept(s) if any
2x 2 + 7x+5 = 0
(2x+5)(x+1) = 0
2x + 5 = 0 x + 1 = 0 x = -5/2 x = -1 answer: (-5/2,0) (-1.0) e) y-intercept(s) if any 𝑓(0) =
2∗0
2
+7∗0+5
0−5
Answer: (-1,0)
=
5
−5
= −1 f) Sketch a graph of the function : label all the features found in parts b - e (I marked the x and yintercepts but didn’t label the points as they were on top of each other with this window)

#13 – 16: Solve
13) x 2 + 6x – 16 < 0
(x + 8)(x-2) < 0
This is true when x + 8 > 0 and x – 2 < 0 x > -8 and x < 2 this asks for numbers that are both less than 2 and greater than -8
Answer to this part -8 < x < 2
Answer: - 8 < x < 2
14) x 2 + 5x – 24 > 0
(x+8)(x-3) > 0
This is true when
x+8 < 0 and x – 2 > 0 x < -8 and x > 2 there is no number that is both less than -8 and greater than 2 so this gives no solution. x+8 > 0 and x – 3 > 0 x> -8 and x > 3
I need to find all numbers that are greater than both -8 and 3,
The answer to this part x > 3
Answer: x > 3 or x<-8 x+8 < 0 and x – 3 < 0 x < -8 and x < 3
I need to find all numbers that are less than 3 and less than -8 the answer to this part x < -8
15) 𝑥−4 𝑥−6
< 0
(x = 6 is not in the domain of the fraction and can’t be part of my answer) this is true when x-4 > 0 and x – 6 <0 x > 4 and x < 6 this is true for all numbers bigger than 4 and less than 6 the answer to this part is 4 < x < 6
x- 4 < 0 and x – 6 > 0 x < 4 and x > 6 this is true for all numbers that are both less than
4 and bigger than 6. There are no such numbers.
This part has no solution.
Answer: 4 < x < 6

16) 𝑥+4 𝑥−2
> 0 x = 2 can’t be part of my answer as it is not in the domain of the fraction the problem is true when x + 4 > 0 and x – 2 > 0 x > -4 and x > 2 this is looking for numbers that are greater than both -4 and 2. The answer to this part x > 2
Answer: x > 2 or x < -4 x+ 4 < 0 and x – 2 < 0
x < -4 and x < 2 this wants all numbers that are both less than -4 and less than 2. The answer to this part is x < -4