Ch. 9.3 Rational Functions and Their Graphs. For each rational function, find any points of...

Ch. 9.3 Rational Functions and Their Graphs

For each rational function, find any points of discontinuity.

ALGEBRA 2 LESSON 9-3ALGEBRA 2 LESSON 9-3

Rational Functions and Their GraphsRational Functions and Their Graphs

The function is undefined at values of x for which x2 – x – 12 = 0.

x2 – x – 12 = 0 Set the denominator equal to zero.

(x – 4)(x + 3) = 0 Solve by factoring or using the Quadratic Formula.

x – 4 = 0 or x + 3 = 0 Zero-Product Property

x = 4 or x = –3 Solve for x.

There are points of discontinuity at x = 4 and x = –3.

9-3

a. y = 3 x2 – x –12

b. y =

(continued)



2x 3x2 + 4

The function is undefined at values of 3x2 + 4 = 0.

3x2 + 4 = 0 Set the denominator equal to zero.

x2 = – Solve for x.43

x = = ± –4 3

± 2i 3

9-3

Since is not a real number, there is no real value for x for

which the function y = is undefined. There is no point

of discontinuity.

± 2i 3

2x 3x2 + 4

Try ThisTry This

For each rational function, find any points of discontinuity

16

12

x

y

x = 4 and x = -4

Try ThisTry This


3

12

2

x

xy

No point of discontinuity

Try ThisTry This


82

12

xx

xy

x = -4 and x = 2

Vertical AsymptotesVertical Asymptotes

The rational function has a point of discontinuity for each real zero of Q(x).

If P(x) and Q(x) have no common real zeros, then the graph of f(x) has a vertical asymptote at each real zero of Q(x).

If P(x) and Q(x) have a common real zero a, then there is a hole in the graph or a vertical asymptote at x = a

P x

f xQ x

Describe the vertical asymptotes and holes for the graph of

each rational function.



Since –1 and –5 are the zeros of the denominator and neither is a zero of the numerator, x = –1 and x = –5 are vertical asymptotes.

–3 is a zero of both the numerator and the denominator. The graph of this function is the same as the graph y = x, except it has a hole at x = –3.

a. y = x – 7 (x + 1)(x + 5)

b. y = (x + 3)x x + 3

c. y = (x – 6)(x + 9) (x + 9)(x + 9)(x – 6)

6 is a zero of both the numerator and the denominator.

9-3

The graph of the function is the same as the graph y =

which has a vertical asymptote at x = –9, except it has a hole at x = 6.

1 (x + 9)

,

Try ThisTry This

Describe the vertical asymptotes and holes for the graph of each rational function.

31

2

xx

xy

Vertical asymptotes: x = 1 and x = -3

Try ThisTry This


32

2

xx

xy

Hole : x = 2; V.A.: x = -3

Try ThisTry This


1

12

x

xy

Hole: x = -1; no V.A.

Horizontal AsymptotesHorizontal Asymptotes

1. If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y = 0 (x-axis)

Ex.

H.A.: y = 0

2. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the fraction formed by the coefficients of the terms with the highest degree.

Ex.

H.A.:

32

12

xx

xy Degree of 1

Degree of 2

1554

132

2

xx

xy Degree of 2

Degree of 2

4

3y

Horizontal AsymptotesHorizontal Asymptotes

3. If the degree of the numerator is greater than the degree of the denominator, then the graph has no horizontal asymptote.

Ex.

No horizontal asymptote

1

12

x

xy

Find the horizontal asymptote of y =



–4x + 3 2x + 1

The horizontal asymptote is y = –2.

9-3

.

The degree of the numerator and denominator are equal

2

4.;.

yah

Try this Try this

Find the horizontal asymptote

86

12

3

xx

xy

No H.A.

Try this Try this


86

1233

2

xx

xxy

H.A.: y = 0

Try this Try this


338

1242

2

xx

xxy

2

1: yHA

Sketch the graph y = .



x + 1 (x – 3)(x + 2)

The degree of the denominator is greater than the degree of the numerator, so the x-axis is the horizontal asymptote.

When x > 3, y is positive. So as x increases, the graph approaches the y-axis from above.

When x < –2, y is negative. So as x decreases, the graph approaches the y-axis from below.

Since –1 is the zero of the numerator, the x-intercept is at –1.

Since 3 and –2 are the zeros of the denominator, the vertical asymptotes are at x = 3 and x = –2.

Calculate the values of y for values of x near the asymptotes. Plot those points and sketch the graph.

9-3

Try ThisTry This

Sketch the graph of 51

3

xx

xy

H.A.:

V.A.:

x-int.:

y-int.:

, Exercises: #2 – 18 e, 26 – 30 all

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