Chapter 2 – Analyzing Polynomial and Rational Functions AnswerKey2.1 Methods for Solving Quadratic Functions

Answers1. a) standard form

b) vertex formc) factored form.

2. x = {7, -2}3. x = 24. x = {-1/2, -3/2}

5. , 6. x = {4, -4}

7. , 8. An equation with a power of 4 can often be FOILed and you can then solve eachbinomial

and get your answers for the equation. So it is possible to use quadratic methods on anequation with a power of 4.

9. x = {2, 4}10 x = {4, 8}11. x = {2, -4}

12.

13.

14.

15. 16. The discriminant is positive17. The discriminant is zero18. The discriminant is negative.

2.2 Graphs of Quadratic Functions

Answers1. The graph of a quadratic is a parabola2. A parabola opens up if is positive

2. A parabola opens up if is positive3. If the coefficient of is positive in , the parabola opens right4. The vertex is the extreme point of a parabola5. The line of symmetry divides a parabola in two symmetrical parts6. The graph is a parabola with the vertex at (0, 3)7. The parabola opens down, with the vertex at (2, 0)8. The parabola opens up, is narrower than the reference, and has a vertex at (-2, -8)9. a) The parabola opens down b) The vertex is at (-1, 2) c) It is not stretched10. is the narrowest11.

12.

13.

14.

15.

16. Maximum height is 11.25yds17. At max height, the ball is 15yds down the field18.

19. By approximating the vertex (15, 11)

2.3 Graphs of Polynomials Using Transformations

Answers1. a)

b) 4th degree polynomialc) 7

2. Shift up 33. Shift up 44. Shift up 65. Vertical compression (horizontal stretch) by factor of 106. Function should resemble

Graph of function should resemble: , ,

Graph of function should resemble:

Step 1: , Step 2: , Step 3: , Step 4: 9. Reflected over -axis10. Reflected over -axis11. Vertical and horizontal compression12. Horizontal and vertical translation

2.4 Graphs of Polynomials Using Zeroes

Answers1. -intercept2. the original polynomial3. The zeroes (-intercepts) are 0, 2, and 34. There are 4 intercepts5. The leading term is6. -intercepts: (-1, 0) and (3, 0) : -intercept: (0, 6)

1.

2.

7. -intercepts: (-3, 0), (-2, 0), and (2, 0) : -intercept (0, -12)

8. -intercepts 3 in number: (0, 0), : -intercept (0, 0)

9. -intercepts (2 in number): apx (-2.1, 0) and apx (0.36, 0) : -intercept (0, 3)

10. and -intercepts both at (0, 0)

11. -intercepts (2 of them):

12. -intercepts (3 of them): (-1, 0), (1, 0) and (0, 0) : -intercept (0, 0)

13. a) b) 6th degree c) 218714. 15.

2.5 Horizontal and Vertical AsymptotesAnswers1. A line or curve to which a function's graph draws closer without touching

2. Vertical asymptote

3. Discontinuities: −3, 2

Vertical Asym.: = −3, = 2

Holes: None

Horz. Asym.: = 0

-intercepts: None

-intercepts: None

4. Discontinuities: 4

Vertical Asym.: = 4

Holes: None

Horz. Asym.: = 1

-intercepts: 2

5. Discontinuities: 2, −3

Vertical Asym.: = 2, = −3

Holes: None

Horz. Asym.: None

-intercepts: 0, −2, 3

6. Discontinuities: −1, −3Vertical Asym.: = −1Holes: = −3Horz. Asym.: = −1/4-intercepts: 2

7. Discontinuities: 0, 3

Vertical Asym.: = 0, = 3Holes: NoneHorz. Asym.: = 0

8. Discontinuities: −4

8. Discontinuities: −4Vertical Asym.: = −4Holes: NoneHorz. Asym.:

9. Discontinuities: −4Vertical Asym.: = −4Holes: NoneHorz. Asym.:

10. Discontinuities: −1, 3

Vertical Asym.: = −1Holes:x = 3Horz. Asym.: None

11. Discontinuities: −1, 3Vertical Asym.:= −1, = 3Holes: NoneHorz. Asym.:= 3

12. Discontinuities: 3, −2Vertical Asym.: = 3, = −2Holes: NoneHorz. Asym.: None

13. Discontinuities: 2Vertical Asym.: = 2Holes: NoneHorz. Asym.: None

14. Discontinuities: −3

Vertical Asym.:Holes: NoneHorz. Asym.:

15. Discontinuities: −1, −2Vertical Asym.:Holes:Horz. Asym.:

16. Discontinuities: 2

Vertical Asym.:Holes: NoneHorz. Asym.:

17. 0+

18. 0-

19. +infinity20. -infinity2.6 Oblique Asymptotes

Answers1. The numerator is a higher degree than the denominator.2. 3. 4. 5. 6.

6. 7. 8. 9. 10.

11.

12.

13.

14.

The and -intercepts are both at (0, 0)There is a vertical asymptote atThe slant asymptote is

15.

and intercept at (0, 0)Vertical asymptote atSlant asymptote at

2.7 Graphs of Rational Functions

Answers1. A point of discontinuity 2. Polynomial graphs are continuous as a rule, rational graphs the opposite 3. Holes and/or asymptotes 4. Simplifying the fraction hides the discontinuities

5. Correct order:

a. Factor numerator and denominator completely and put in lowest terms. Identifyany holes.

b. Find all intercepts.c. Find all asymptotes.c. Determine the behavior around the vertical asymptotes using a table of signs.e. Find the places where the function crosses the horizontal asymptote/oblique

asymptote.f. Sketch a smooth graph based on the above information.

6. =

7. -intercepts of are (-4, 0) and (2, 0) – there are no -intercepts, since the -axis is thevertical asymptote. There is a slant asymptote at . End behavior: as and as .

8. The domain is all real numbers except for 0. The graph looks like:

9. The function simplifies as: There is a discontinuity at x = 3, an asymptote at and another atThe function looks similar to #8, except the curves stay entirely in QII and QIV

10. The function simplifies as: There is a discontinuity atThe graph looks like the line

11. Discontinuity at , the graph is quadratic 12. Discontinuity at , the graph is linear 13. Discontinuities at and , the graph is linear

14.

15 The function simplifies to: Vertical asymptotes are atSlant asymptote atGraph looks like:

Graph looks like:

2.8 Analysis of Rational Functions

Answers

1. : x ≠ 2

2. y = : x ≠ 4

3. f(x) = : x ≠ 4

4. f(x) = : x ≠ -4/5

5. y = : x ≠ -7 6. Asymptotes: and (not required to show)

Holes: none – intercept(s): apx (-4.7, 0) – intercept: (0, -7)

Sketch:

7. Asymptotes: and

Holes: – intercept(s): None – intercept: None

Sketch:

8. Asymptotes:

Holes: None – intercept(s): apx (.8, 0) – intercept: (0, 4)

Sketch:

9. Asymptotes: and and (not required to show)Holes: None– intercept(s) - apx: ={-.68, .12, 2.36}– intercept: (0, -.25)

Sketch:

10. Asymptotes: and (not required to show)

10. Asymptotes: and (not required to show)Holes: none– intercept(s) apx: {-7, 2.23} – intercept: (0, -7)

Sketch:

11. Asymptotes:Holes:– intercept(s): (-.14, 0)– intercept: (0, 7/4)

Sketch:

12. Asymptotes: and

Holes: None– intercept(s) apx: {-.87, .-47, 1.05}– intercept: None

Sketch:

13. Asymptotes: and (not required)

13. Asymptotes: and (not required)Holes:– intercept(s): (1.725, 0)– intercept: None

Sketch:

14. Asymptotes:

Holes: None– intercept(s):– intercept:

Sketch:

15. Asymptotes:

Holes: None– intercept(s): (1, 0)– intercept: (0, ½)

Sketch:

2.9 Quadratic Inequalities

Answers1.

3 4

2. -4 3

3.

-2 6 4.

-4 7 5.

-8 3

6. -7 -2Critical points

7. -5 10Critical points

8. -8 6

-8 6Critical Points

9. -2 0 10Critical points

10. -7 -3

0 Critical points

11. -6 3Critical points

12. -1 8Critical points

13. -9 10Critical points

14. -1/3 8Critical points

15. -3 2Critical points

2.10 Polynomial and Rational Inequalities

Answers

1. -3 1

1.

1/3 2

-5/2 1/3

1/5 2

0 1/2

6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

16.

a) 60 Ohms

b) The total resistance will always be less than 20 because is the horizontal asymptote of

18. Width must be greater than 4 meters, .

2.11 Synthetic Division of Polynomials

1.

2.

3.

4.

1.

2.11 Synthetic Division of Polynomials

Answers

1.

2.

3.

4.

5.

Factored polynomial:

6.

7.

8.

9.

10.

2.12 Real Zeros of Polynomials

Answers

a) 21b) 31c) 10d) 66e) TO BE REVISED (Could not find any roots for the polynomial f(x)

k = -10a)

11.

12.

13.

14.

15.

16.

17.

1.2.3.4.5.

6.7.

b)8.

a)

2.13 Intermediate Value Theorem

Answers

9.

10.11.12.13.14.b)15.

16.

Answers1. a) L.C. = 3 Degree = 5 b) 1 Real Zero c) 4 ImaginaryZeroes2. a) L.C. = -1 Degree = 3 b) 1 Real Zero c) 2 ImaginaryZeroes3. a) L.C. = 1/2 Degree = 4 b) 2 Real Zeroes c) 2 Imaginary Zeroes4. a) L.C. = -3/2 Degree = 6 b) 0 Real Zeroes c) 6 Imaginary Zeroes5. a) L.C. = 1 Degree = 5 b) 3 Real Zeroes c) 2 Imaginary Zeroes6. a) L.C. = 1 Degree = 4 b) 3 Real Zeroes c) 1 Imaginary Zero7. There is a zero in [-2, -1]8. There is a zero in [-1, 0] and [2, 3]9. There is a zero in [-2, -1], [-1, 0], [0, 1], and [3, 4]10. There is a zero in [-2, -1] and [1, 2]11. There is a zero in [-2, -1], [-1, 0], [0, 1] and [1, 2]

12. has a vertical asymptote at , so it is not continuous in theinterval . Therefore we cannot use the Bounds on Zeros theorem to claim thereis a zero in that interval.

13. The zero is in [0.3125, 0.34825]14. The zero is in [1.5000, 1.5625]15. Zero is in the interval: [0.8125, 0.875]

2.14 Fundamental Theorem of Algebra

Answers1. 2. 3. 4. 5. 6. 7. 8. Yes; 9. Yes; 10. No11.

12.

13.

14.

15.

CK-12 Math Analysis Concepts 1