Chapter 7: Rational Functions Section 7–1: Zeros, Vertical Asymptotes, and POEs 4–1 Free Response

1. ( )( )2

5 52 8 4 2

yx x x x

= =− − − +

Zeros: none VAs: x = 4 and x = –2 POEs: none

3. ( )( )( )

2 5 32 153 3

x xx xyx x

+ −+ −= =+ +

Zeros: ( ) ( )5,0 , 3,0− VAs: 3x = − POEs: none

5. 2

3 2

301

x xyx x x

+ −=+ − −

Numerator: ( )( )6 5x x+ − Denominator:

( ) ( )( )( )( )( )( )

3 2

2

2

11 1 1

1 1

1 1 1

x x xx x x

x x

x x x

+ − −+ − +

+ −

+ − +

Therefore, ( )( )( ) ( )

2

23 2

6 5301 1 1

x xx xyx x x x x

+ −+ −= =+ − − + −

Zeros: ( ) ( )6,0 , 5,0− VAs: 1, 1x x=− = POEs: none

7. 3 2

2

2 6 46

x x xyx x+ − −=

+ −

Using synthetic for numerator:

( )( )

( )( )2 22 4 2 4 23 2 3

x x x x xyx x x

− + + + += ≈+ − +

Zeros (Quadratic Formula): ( ) ( )0.586,0 , 3.414,0− − VAs: x = –3 POEs (plug x = 2 into simplified

fraction): 142,5

⎛ ⎞⎜ ⎟⎝ ⎠

9.

( )( )( )

23

4 2 22 2

225 6 33 2

x xx x xyx x xx x

−−= = ≈− + −− −

Zeros: (0, 0) VAs: 3x = ± POEs: ( )2, 2 , ( )2, 2− −

11. 2 12 4x xy

x x−= −

− +

Write as a single fraction with a common denominator:

2 4 1 22 4 4 2x x x xy

x x x x+ − +⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠

( )( )( )

( )( )

2 2

2

2 8 22 4

7 22 4

x x x xy

x x

x xyx x

+ − + −=

+ +

+ +=+ +

Zeros (Quadratic Formula): ( )0.298,0− , ( )6.702,0− VAs: x = –2, x = –4

POEs: none

13. 2

7 243 5 2 15xyx x x x

= − −− + + −

2

5 7 3 243 5 5 3 2 15x x xyx x x x x x

+ −⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟− + + − + −⎝ ⎠ ⎝ ⎠

( )( )

2 5 7 21 243 5

x x xyx x

+ − + −=− +

( )( )( )( )( )( )

2 3 12 3 13 5 3 5 5

x xx x xyx x x x x

− +− − += = ≈− + − + +

Zeros: ( )1,0− VAs: 5x = −

POEs: 13,2

⎛ ⎞⎜ ⎟⎝ ⎠

15. ( )( )2

3 312 4 3

x xyx x x x

+ += =+ − + −

Zeros: ( )3,0− VAs: 4x = − , 3x = POEs: none 17.

( )( )( )( )( ) ( )( )

2 3 22 3 1 2 1x x xy

x x x x x+ − += ≈

− − − − −

Zeros: ( )2,0− VAs: 2x = , 1x =

POEs: 53,2

⎛ ⎞⎜ ⎟⎝ ⎠

19. ( )( )( )( )

23 2

2

1 1130 6 5

x xx x xyx x x x

− ++ − −= =+ − + −

(see problem 5 for factoring process)

Zeros: ( )1,0 , ( )1,0− VAs: 6x = − , 5x = POEs: none 7–1 Multiple Choice 1. A zero at ( )3,0− means ( )3x + is a factor of the numerator only; a POE at

74,20

⎛ ⎞⎜ ⎟⎝ ⎠

means ( )4x − is a factor in both the

numerator and denominator. No vertical asymptotes means there are no remaining factors in the denominator following cancellation (or solutions to factors remaining in the denominator are imaginary). Therefore, eliminate answers A, B, and E. Check answers C and D:

C. ( )( )( )( )

2

3 2 2

3 4124 4 16 4 4

x xx xyx x x x x

+ −− −= =− + − + −

D. ( )( )

( )( )( )2

3 2

3 4124 4 16 2 2 4

x xx xyx x x x x x

+ −− −= =− − + + − −

(C) is the correct response 3. A vertical asymptote at 0x = means a factor of x in the denominator only; a POE

at 131,7

⎛ ⎞−⎜ ⎟⎝ ⎠ means ( )1x + is a factor in both

the numerator and denominator. A zero at ( )12,0 means ( )12x − is a factor of the numerator only. Eliminate answers B, C, and D. Check answers A and E: A.

( )( )( )

( )2

2

12 1 1211 121

x x xx xyx x x x x

− + −− −= = =+ +

; POE at ( )1,13− E.

y = x2 −11x −12

7x2 + 7x=

x −12( ) x +1( )7x x +1( ) =

x −12( )7x

; POE at 131,7

⎛ ⎞−⎜ ⎟⎝ ⎠

(E) is the correct response 5. ( ) 3 24 3 60 88v t t t t= + − +

( ) 212 6 60a t t t= + −

( )( )( )

2

2

0 12 6 60

0 6 2 10

0 6 2 5 25 ,22

t t

t t

t t

t

= + −

= + −

= + −

= −

52

t ≠ − (not in the domain), so 2t =

( ) ( ) ( )( )2 4 8 3 4 120 88

2 12

v

v

= + − +

=

(B) is the correct response Section 7–2: End Behavior: Horizontal and Slant Asymptotes Section 7–2 Free Response

1. 2

3 2

7 2lim3 11 5 2x

x xx x x→∞

+ −− − +

n m< , so the limit = 0

3. ( )( )( )( )( )

4 1lim

5 4 4 1x

x xx x x→∞

− ++ − +

n m< , so the limit = 0

5. 2

2

6 2lim4x

x xx→∞

− +−

n m= , so the limit = 1 11=

7.

( )

( )

2

2

104 6 5

4

10 5 10 40

45

xx x x

x x

xx

+− + +

− −

+− −

Quotient: 45104

xx

+ +−

9.

( )

( )

2

2

1 22

2 4 6 10

2

4 10 4 8

2

x

x x x

x x

xx

−

− − +

− −

− +− − +

Quotient: 1 2 1 12 22 2 4 2 2x x

x x− + = − +

− −

11. n m= ;

HA at y = 1 SA: none

13. 1n m= +

HA; none SA at 1y x= +

( )

( )

2 3 2

3 2

2

2

14 4 3 25 21

4 4

29 21

4 4

25 25

xx x x x x

x x x

x x

x x

x

+− + − − −

− − +

− −

− − +

− −

15.

( )

( )( )

2 3

2 3

2 3

3

2

2

2

4 1 04 1

4 1

141 1

4

x y x yx y y x

y x x

xyxx x x

yx

− + + =+ = −

+ = −

−=+− + +

=+

Zeros: ( )1,0 VAs: None POEs: None HA: none SA: y x=

( )

2 3 2

3 2

0 4 0 0 1

0 4

4 1

xx x x x x

x x x

x

+ + + + −

− + +

− −

17. 2

22

yx

=−

Zeros: none VAs: 2x = , 2x = − POEs: none HA: 0y =

SA: none

19. 2

242 15

yx x

=+ −

( )( )

245 3

yx x

=+ −

Zeros: none VAs: 5x = − , 3x = POEs: none HA: 0y = SA: none

21. 2

312

xyx x

+=+ −

( )( )

34 3xy

x x+=

+ −

Zeros: ( )3,0− VAs: 4x = − , 3x = POEs: none HA: 0y = SA: none 23.

( )( )( )( )( ) ( )( )

2 3 22 3 1 2 1x x xy

x x x x x+ − += ≈

− − − − −

Zeros: ( )2,0− VAs: 2x = , 1x =

POEs: 53,2

⎛ ⎞⎜ ⎟⎝ ⎠

HA: 0y = SA: none

25. ( )( )( )( )

23 2

2

1 1130 6 5

x xx x xyx x x x

− ++ − −= =+ − + −

(see Section 4–1, FR5 for factoring process)

Zeros: ( )1,0 , ( )1,0− VAs: 6x = − , 5x = POEs: none HA: none SA: y x=

( )

2 3 2

3 2

30 1

30

29 1

xx x x x x

x x x

x

+ − + − −

− + −

−

27. 3 2

2

2 9 7 66

x x xyx x

− + +=− −

( )( )( )

( )( )( )( )2 2 1 3 2 1 3

2 3 3x x x x x

yx x x

− + − − + −= ≈

− + +

Zeros: 1 ,02

⎛ ⎞−⎜ ⎟⎝ ⎠, ( )3,0

VAs: 3x = − POEs: ( )2,1 HA: none SA: 2 11y x= − +

( )

( )

2 3 2

3 2

2

2

2 116 2 9 7 6

2 2 12

11 19 6

11 11 66

30 60

xx x x x x

x x x

x x

x x

x

− +− − + − + +

− + −

− + +

− − − +

−

Section 7–2 Multiple Choice 1. A horizontal asymptote of 2y =

means that 2axx= , so 2a = . A vertical

asymptote of 3x = − means the denominator of the fraction must be 3x + , so 3c = .

3 2 5a c+ = + = .

(E) is the correct response

3. 2

2

4lim10000n

nn n→∞ +

n m= , so the limit is 4 41=

(D) is the correct response

5. The given line is 5 6y x= + , so its slope (and the slopes of all parallel lines) is 5.

( ) ( )( )( )( )

2

1

2 1

p x x x k

p x x kx x k

p x x k

= − +

= + − −′ = + −

( )5 2 15 2 4 15 72

x kk

kk

= + −= + −= +

− =

(E) is the correct response Section 4–3: Derivatives: The Quotient Rule Section 4–3 Free Response

1. ( )( ) ( )( )

( )2 2

22

4 2 3 2

4

x x x xdydx x

− − −=

−

( )

( )

3 3

22

22

2 8 2 6

4

2

4

dy x x x xdx x

dy xdx x

− − +=−

−=−

3.

( ) ( )( ) ( )( )( )

2 2

22

3 2 2 2 8 2 1

3

x x x x x xf x

x x

− − + − + − −′ =

− −

( ) ( )( )

( )( )

3 2 2 3 2 2

22

2

22

2 2 6 2 2 6 2 4 16 2 8

3

3 10 14

3

x x x x x x x x x xf x

x x

x xf xx x

− − + − − − + − − − +′ =

− −

− + −′ =− −

5. 3

3 31

d xdx x

+⎡ ⎤⎢ ⎥+⎣ ⎦

( )( )( )2

2

3 11 1

31

xddx x x x

ddx x x

⎡ ⎤+= ⎢ ⎥

+ − +⎢ ⎥⎣ ⎦⎡ ⎤= ⎢ ⎥− +⎣ ⎦

( )( ) ( )( )( )

( )

2

22

2

1 0 3 2 1

1

6 31

x x x

x x

xx x

− + − −=

− +

− +=− +

7. 5 3

3

12 193

d c c cdc c

⎡ ⎤− −⎢ ⎥⎣ ⎦

4 2

2

12 193

d c cdc c

⎡ ⎤− −= ⎢ ⎥⎣ ⎦

( )( ) ( )( )

( )

( )

2 3 4 2

4

5 3 5 3

4

5

4

4

4

4

3

3 4 24 12 19 69

12 72 6 72 1149

6 1149

3 2 389

2 383

c c c c c cc

c c c c cc

c cc

c cc

cc

− − − −=

− − − −=

+=

+=

+=

9. 2 2 15

3x xyx+ −=−

( )( )5 3

53

1

x xy x

xy

+ −= ≈ +

−′ =

11. 2

2

31xyx

−=+

( )( ) ( )( )( )( )

( )

( )

2 2

22

3 3

22

22

1 2 3 2

1

2 2 6 2

1

8

1

x x x xy

x

x x x xy

x

xyx

+ − − −′ =

+

− − − −′ =

+

−′ =+

(i) 8 0x− = 0x = (ii) ( )221 0x+ =

21 0x+ = None (iii) EoASD None CV: 0x =

13. 3

2

32 3xy

x x−=+ −

( )( ) ( )( )

( )( )

( )

( )( )

( )

2 2 3

22

4 3 2 4 3

22

4 3 2

22

2 2

22

2 3 9 3 2 2

2 3

9 18 27 6 6

2 3

3 12 27

2 3

3 4 9

2 3

x x x x xy

x x

x x x x xy

x x

x x xyx x

x x xy

x x

+ − − − − +′ =

+ −

− − + − − −′ =

+ −

− − +′ =+ −

− + −′ =

+ −

(i) ( )2 23 4 9 0x x x− + − = 0x = , 1.606x = , 5.606x = − (ii) ( )22 2 3 0x x+ − =

( )( )( )

2 2 3 0

3 1 0

x x

x x

+ − =

+ − =

3x = − , 1x = (these are not in the domain) (iii) EoASD None CVs: 0x = , 1.606x = , 5.606x = −

15. 2

3 2

3 5 26 11 3 2

x xyx x x

+ −=− − +

Using synthetic:

( )( )( )( )( ) ( )( )

2

3 1 2 22 2 1 3 1 2 2 12

2 3 2

x x xyx x x x xxy

x x

− + += ≈− + − − +

+≈− −

( )( ) ( )( )

( )( )

( )

( )

2

22

2 2

22

2

22

2 3 2 1 2 4 3

2 3 2

2 3 2 4 5 6

2 3 2

2 8 4

2 3 2

x x x xy

x x

x x x xy

x x

x xyx x

− − − + −′ =

− −

− − − + −′ =

− −

− − +′ =− −

(i) 22 8 4 0x x− − + =

( )2

2

2 4 2 0

4 2 0

x x

x x

− + − =

+ − =

0.449x = , 4.449x = − (ii) ( )222 3 2 0x x− − =

( )( )

22 3 2 02 1 2 0x xx x− − =+ − =

12

x = − , 2x = (these are not in

the domain) (iii) EoASD None EP: ( )0.449, 0.832− , ( )4.449, 0.048− −

17. 2

31xy

x−=+

At 1x = , the point on the function is 31,2

⎛ ⎞−⎜ ⎟⎝ ⎠

( )( ) ( )( )( )( )

( )

( )

2

22

2 2

22

2

22

1 3 3 2

1

3 3 6

1

3 3

1

x x xy

x

x xy

x

xyx

+ − − −′ =

+

− − − −′ =

+

−′ =+

At 1x = , 0y′ = . This is the slope of the tangent line (horizontal). The equation of the tangent line at

1x = is the horizontal line 32

y = −

The normal line at 1x = is the vertical line 1x = Section 7-3 Multiple Choice 1. 3u = , 0uD = , 24v x= + , 2vD x=

( )( ) ( )( )( )

( )

2

22

22

4 0 3 2

4

6

4

x xdydx x

dy xdx x

+ −=

+

−=+

(A) is the correct response 3. 1u = , 0uD = , 1v x= + , 1vD =

( )( ) ( )( )( )

( )

2

2

1 0 1 11

11

11 4

xdydx xdydx xdydxx

+ −=

+−=+

==

(D) is the correct response 5.

( ) ( )( ) ( )

3

3 31.1 0.70.4

f g

f g

′−′ ′= −

= −=

(B) is the correct response 7.

600600

lw

lw

=

=

1

3 3 56006 5

30006

6 3000

Cost C w w l

C ww

C ww

C w w−

= = + +

⎛ ⎞= + ⎜ ⎟⎝ ⎠

= +

= +

2

2

6 300030006

C w

Cw

−′ = −

′ = −

2

2

2

2

30000 6

3000 6

3000 6500

10 5

w

ww

w

w

= −

=

==

± =

10 5 represents both the east and west walls

l

w w

North Wall

600

60010 560 55

12 5

lw

l

l

l

=

=

=

=

(A) is the correct response Section 7–4: Sign Patterns and Rational Functions Section 7–4 Free Response

1. 2 2 3 0

1x xx+ − ≤+

( )( )3 10

1x xx

+ −≤

+

0 03 1 1

DNEyx

− + − +− −←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

0≤ means that solutions that are either 0 or negative are included. DNEs are not included. ( ] ( ], 3 1,1x∈ −∞ − ∪ −

3. 2

3

4 01

xx− ≥−

( )( )( )( )2

2 20

1 1x xx x x

+ −≥

− + +

0 02 1 2

DNEyx

− + − +−←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

0≥ means that solutions that are either 0 or positive are included. DNEs are not included. [ ) [ )2,1 2,x∈ − ∪ ∞

5. 2 2

7 82 6 5

x xx x x x

− <+ − − +

( )( ) ( )( )( )( ) ( )

( )( )( )

( )( )( )

( )( )( )( )

( )( )( )( )( )

( )( )( )( )

( )( )

2 2

2

2

7 8 02 1 5 1

7 5 8 20

2 1 5

12 35 8 16 02 1 5

7 28 35 02 1 5

7 4 50

2 1 5

7 5 10

2 1 5

7 50

2 5

x xx x x x

x x x xx x x

x x x xx x x

x xx x x

x xx x x

x xx x x

xx x

− − <+ − − −

− − − +<

+ − −

− + − − <+ − −

− − + <+ − −

− + −<

+ − −

− + −<

+ − −

− +<

+ −

+ 0 5 2 1 5

DNE DNE DNEyx

− + + −− −←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

<0 means that negative solutions are included. DNEs and zeros are not included. ( ) ( )5, 2 5,x∈ − − ∪ ∞

7. 2

59

xyx−=−

( )( ) ( )( )( )

( )

( )( )( )

( ) ( )

2

22

2 2

22

2

22

2 2

9 1 5 2

9

9 2 10

9

10 9

9

9 13 3

x x xy

x

x x xyx

x xyx

x xy

x x

− − −′ =

−

− − +′ =−

− + −′ =−

− − −′ =− +

(i) ( )( )9 1 0x x− − − = 9, 1x = (ii) ( ) ( )2 23 3 0x x− + = 3x = ± (iii) EoASD: none CVs: 9, 1, 3, 3x = −

DNE 0 0 3 1 3 9

DNEyx

− − + + −−

′←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

Min at 1x = , max at 9x = , 3x = ± are DNEs of the function as well as of the derivative, so neither max nor min.

9. 2 11 3xy

x+=

−

( )( ) ( )( )( )

( )

( )

2

2

2 2

2

2

2

1 3 1 3 2

1 3

3 3 6 21 3

3 2 31 3

x x xy

x

x x xyx

x xyx

+ − − −′ =

−

− − + −′ =−

− −′ =−

(i) 23 2 3 0x x− − = 1.387, 0.721x = −

(ii) ( )21 3 0x− =

13

x =

(iii) EoASD: none

CVs: 1 1.387, 0.721, 3

x = −

0 + 01 0.721 1.3873

DNEyx

− + −

−

′←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

Min at 0.721x = − , max at 1.387x = .

13

x = is a DNE both of the derivative and

the function, so neither max nor min.

11. ( )2 11

tx tt+=+

( ) ( )( ) ( )( )( )

( )( )

( )( )

2

2

2 2

2

2

2

1 2 1 1

1

2 2 11

2 11

t t tv t

t

t t tv tt

t tv tt

+ − +=

+

+ − −=+

+ −=+

“When it stops:” ( ) 0v t = 2 2 1 0t t+ − = 0.414, 2.414t = − ; 0t > , so 0.414t =

“Where is it:” ( )x t ( )0.414 0.828x = , or 0.828 units to the right of the origin Section 7–4 Multiple Choice

1. ( ) ( ) ( )f x f af a

x a−′ =−

( ) ( ) ( )222

f x ff

x−′ =−

Using trial and error: Answer (A):

( )( )2 2 24 2 42 2

x xx xx x

+ −− = = + =− −

(No)

Answer (B): ( ) ( )4 8 4 24 8 42 2 2

x xxx x x

− − − − −− += = = −− − −

(No)

Answer (C): ( )2 24 4 4 42 2

x x x xx x

− − − − +=− −

(Yes)

(C) is the correct response 3.

( )( ) ( )( )( )

( )

( )

2

2

2

1 1 1 11

1 112

1

x xy

xx xyx

yx

+ − − −′ =+

− − − +′ =+

−′ =+

22

yx

′= −

= −

Tangent Line: ( )3 2 2

3 2 42 7 0

y xy xx y

+ = − ++ = − −+ + =

(D) is the correct response 5. ( )f x is increasing, then decreasing, then increasing, so ( )f x′ must be positive, then negative, then positive again over the same intervals. The only graph that fits the criteria is graph (A). (A) is the correct response 7. The solution set matches areas on the sign pattern that are either negative or zero,

so the symbol used in the solution inequality must be ≤ . Since there is no sign switch around 2x = , the factor ( )2x − must appear an even number of times. Finally, since there is a negative sign to the right of 2x = , there must be a leading negative sign on the function. (C) is the correct reponse 9. A local maximum is characterized by a change in the sign pattern of ( )f x′ from + to – around an x value where the derivative exists. This occurs at 1x = − . (B) is the correct response Section 7 – 5: General Rational Curve Sketching Section 7–5 Free Response 1.

3.

−8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

7

x

y

5. 2

44 5xy

x x=

+ −

( )( )

45 1xy

x x=

+ −

( )( ) ( )( )( )

( )

( )

2

22

2 2

22

2

22

4 5 4 4 2 4

4 5

4 16 20 8 16

4 5

4 20

4 5

x x x xy

x x

x x x xyx x

xyx x

+ − − +′ =

+ −

+ − − −′ =+ −

− −′ =+ −

(i) 24 20 0x− − = No solutions (ii) ( )22 4 5 0x x+ − = 5, 1x = − (but these are not in the domain) (iii) EoASD: none No Extreme Points End Behavior: n m< , so there is an HA at 0y =

Domain: x ≠ −5, 1 Range: All Reals Zeros:

0, 0( )

y-int: 0, 0( )

VA: x = −5, x =1 POE: None

HA: y = 0 Extreme Points: None

7. 2

2

4 36

x xyx x− +=− −

( )( )( )( )

3 13 212

x xy

x xxyx

− −=

− +−≈+

( )( ) ( )( )( )

( )

( )

2

2

2

2 1 1 12

2 12

32

x xy

xx xyx

yx

+ − −′ =+

+ − +′ =+

′ =+

(i) 3 0= No solutions

(ii) ( )22 0x + =

2x = − but this is not in the domain (iii) EoASD: none No Extreme Points End Behavior: n m= , so there is an HA at 1y =

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10 11 12

−4

−3

−2

−1

1

2

3

4

5

6

x

y

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−5

−4

−3

−2

−1

1

2

3

4

5

x

y

There is a POE at 3x = ; find the y–coordinate by plugging 3x = into the simplified version of the original function:

13 2

yx

==

, so the POE is 13, 2

⎛ ⎞⎜ ⎟⎝ ⎠

.

Domain: x ≠ −2, 3

Range: y ≠1, 2

5

Zero: 1, 0( )

y-int:

0, − 12

⎛⎝⎜

⎞⎠⎟

VA: x = −2

POE:

3, 25

⎛⎝⎜

⎞⎠⎟

HA: y = 1 Extreme Points: None

9. 3 2

2

3 6 86 5

x x xyx x+ − −=

+ +

Use calculator and/or synthetic to factor the numerator:

( )( )( )( )( )4 1 25 1

x x xy

x x+ + −

=+ +

( )( )

2

4 25

2 85

x xy

xx xyx

+ −≈

++ −≈+

( )( ) ( )( )( )

( )

( )

2

2

2 2

2

2

2

5 2 2 2 8 1

5

2 12 10 2 85

10 185

x x x xy

x

x x x xyx

x xyx

+ + − + −′ =

+

+ + − − +′ =+

+ +′ =+

(i) 2 10 18 0x x+ + = 2.354, 7.646x =− − (ii) ( )25 0x + = 5x = − , but this is not in the domain (iii) EoASD: none Extreme Points: ( ) ( ) 2.354, 2.708 , 7.646, 13.292− − − −

End Behavior: 1n m= + , so there is a slant asymptote. Use long division.

( )

( )

2 3 2

3 2

2

2

36 5 3 6 8

6 5

3 11 8

3 18 15

7 7

xx x x x x

x x x

x x

x x

x

−+ + + − −

− + +

− − −

− − − −

+

Slant Asymptote: 3y x= +

There is a POE at 1x = − ; find the y–coordinate by plugging 1x = − into the simplified version of the original function:

91 4

yx

= −= −

, so the POE is

91, 4

⎛ ⎞− −⎜ ⎟⎝ ⎠.

Domain: x ≠ −5, −1

−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−5

−4

−3

−2

−1

1

2

3

4

5

x

y

Range:

y∈ −∞, −13.292( ⎤⎦ ∪ −2.708, ∞⎡⎣ )Zeros:

2, 0( ), −4, 0( )

y-int:

0, − 85

⎛⎝⎜

⎞⎠⎟

VA: x = −5

POE: −1, − 9

4⎛⎝⎜

⎞⎠⎟

SA: y = x − 3 Extreme Points:

−2.354, − 2.708( ) ,

−7.645, −13.292( )

11. 2 2 4 0x xy y x− + − =

( )

( )

2

2

2

2 42 4

4242

xy y x xy x x x

x xyxx x

yx

− = −− = −

−=−−

=−

( )( ) ( )( )( )

( )

( )

2

2

2 2

2

2

2

2 2 4 4 1

2

2 8 8 42

4 82

x x x xy

x

x x x xyx

x xyx

− − − −′ =

−

− + − +′ =−

− +′ =−

(i) 2 4 8 0x x− + = No solutions (ii) ( )22 0x − = 2x = , but this is not in the domain (iii) EoASD: none No Extreme Points End Behavior: 1n m= + , so there is a slant asymptote. Use long division.

( )

( )

2

2

22 4 0

2

2 0 2 4

4

xx x x

x x

xx

−− − +

− −

− +− − +

−

Slant Asymptote at 2y x= − Domain: x ≠ 2 Range: All Reals Zeros:

0, 0( ), 4, 0( )

y-int: 0, 0( )

VA: x = 2 POE: None SA: y = x − 2 Extreme Points: None

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−24

−20

−16

−12

−8

−4

4

x

y

−11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10 11 12

−16

−12

−8

−4

4

8

12

x

y

13. 3 2

2

6 5 3 22 3 2x x xyx x+ − −=

− −

Use synthetic and/or the graphing calculator to factor the numerator:

( )( )( )( )( )

( )( )

2

1 2 1 3 22 1 2

1 3 22

3 22

x x xy

x x

x xy

xx xyx

+ + −=

+ −

+ −≈

−+ −≈−

( )( ) ( )( )( )

( )

( )( )

( )

2

2

2 2

2

2

2

2

2 6 1 3 2 1

2

6 11 2 3 22

3 122

3 42

x x x xy

x

x x x xyx

x xyx

x xy

x

− + − + −′ =

−

− − − − +′ =−

−′ =−

−′ =−

(i) ( )3 4 0x x − =

0, 4x =

(ii) ( )22 0x − =

2x = , but this is not in the domain (iii) EoASD: none

Extreme Points: ( ) ( )0, 1 , 4, 25

End Behavior: 1n m= + , so there is a slant asymptote. Use long division.

( )

( )

2 3 2

3 2

2

2

3 72 3 2 6 5 3 2

6 9 6

14 3 2

14 21 14

24 12

xx x x x x

x x x

x x

x x

x

+− − + − −

− − −

+ −

− − −

+

Slant Asymptote: 3 7y x= +

Domain: x ≠ − 1

2 , 2

Range: y∈ −∞, 1( ⎤⎦ ∪ 25, ∞⎡⎣ )

Zeros:

23, 0⎛

⎝⎜⎞⎠⎟, −1, 0( )

y-int: 0, 1( )

VA: x = 2

POE: − 1

2 , 710

⎛⎝⎜

⎞⎠⎟

SA: y = 3x + 7 Extreme Points:

0, 1( ), 4, 25( )

15. One possible equation is

y = −xx 3x −1( )

17. One possible equation is

y =2 x +1( ) x +5( )

x −1( )

−16 −12 −8 −4 4 8 12 16 20

−40

−35

−30

−25

−20

−15

−10

−5

5

10

15

20

25

30

35

x

y

Section 7–5 Multiple Choice 1. 2 10x xy+ =

( )( ) ( )( )

2

2

2

2

2 2

2

2

2

1010

2 10 1

2 10

10

72 2

xy xxyxx x xdy

dx xdy x xdx xdy xdx xdydxx

= −−=

− − −=

− − +=

− −=

= −=

(A) is the correct response

3. 1xyx

=+

( ) ( )( ) ( )( )

( )( )

( ) ( )( ) ( )

( )( )

( )

2

2

2

3

3

1 1 11

11

1

2 1 121

0 2

x xf x

x

f xx

f x x

f x x

f xx

f

−

−

+ −′ =+

′ =+

′ = +

′′ = − + •−′′ =+

′′ = −

(A) is the correct response

5. 2

2

3lim4 1x

x xx x→∞

− −+ −

is found by analysis of

the highest powers in the numerator and

denominator. n m= , so the limit represents

an HA at 14

y = − .

(D) is the correct response