Curve Sketching: Rational FunctionsMATH 151 Calculus for Management

J. Robert Buchanan

Department of Mathematics

Fall 2018

Objectives

Recall: a rational function is a function of the form

f (x) =anxn + an−1xn−1 + · · ·+ a1x + a0

bmxm + bm−1xm−1 + · · ·+ b1x + b0.

The numerator and denominator are polynomials.

After this lesson we will be able to:I find vertical, horizontal, and oblique (slant) asymptotes,

andI sketch the graphs of rational functions.

Vertical Asymptotes

DefinitionThe function f (x) has a vertical asymptote at x = a if

limx→a−

f (x) = ±∞ or limx→a+

f (x) = ±∞

Remark: vertical asymptotes are lines of the form x = a andoccur where the denominator of a rational function is 0 and thenumerator is not zero.

Example

Find all the vertical asymptotes of the following function.

f (x) =1− 2xx2 − 1

The denominator is zero at x = ±1. Checking the numeratorwith x = ±1, we see the numerator takes on the values −1 or3, both if which are non-zero. Thus f has vertical asymptotes atx = ±1.

Example

Find all the vertical asymptotes of the following function.

f (x) =1− 2xx2 − 1

The denominator is zero at x = ±1. Checking the numeratorwith x = ±1, we see the numerator takes on the values −1 or3, both if which are non-zero. Thus f has vertical asymptotes atx = ±1.

Graph

f (x) =1− 2xx2 − 1

-4 -2 2 4x

-4

-2

2

4

y

Horizontal Asymptotes

DefinitionThe function f (x) has a horizontal asymptote at y = L if

limx→−∞

f (x) = L or limx→∞

f (x) = L.

Horizontal Asymptotes of Rational Functions

Suppose that f (x) =p(x)q(x)

is a rational function.

1. If the degree of the numerator is less than the degree ofthe denominator then y = 0 is a horizontal asymptote ofthe graph of f (both to the left and the right).

2. If the degree of the numerator equals the degree of thedenominator, then y =

ab

is a horizontal asymptote of thegraph of f (both to the left and the right), where a and b arethe leading coefficients of p(x) and q(x) respectively.

3. If the degree of the numerator is greater than the degree ofthe denominator, then the graph of f has no horizontalasymptote.

Example

Find the horizontal asymptotes of the function

f (x) =4x2 − x + 53x2 − 2x + 1

Since the degrees of the numerator and the denominator arethe same (both are of degree 2) then

limx→−∞

f (x) =43

and limx→∞

f (x) =43

which implies there is a horizontal asymptote at y = 4/3.

Example

Find the horizontal asymptotes of the function

f (x) =4x2 − x + 53x2 − 2x + 1

Since the degrees of the numerator and the denominator arethe same (both are of degree 2) then

limx→−∞

f (x) =43

and limx→∞

f (x) =43

which implies there is a horizontal asymptote at y = 4/3.

Graph

f (x) =4x2 − x + 53x2 − 2x + 1

-4 -2 0 2 4x

2

4

6

8

10

y

Oblique (Slant) Asymptotes

If the denominator of a rational function has degree 1 or largerand if the degree of the numerator of the rational function isexactly 1 more than the degree of the denominator, the graph ofthe rational function has an oblique asymptote (or slantasymptote).

Example

f (x) =x3 + 4x2 − 4x − 13

x2 − 4= x + 4︸ ︷︷ ︸

slant asymptote

+3

x2 − 4

Oblique (Slant) Asymptotes

If the denominator of a rational function has degree 1 or largerand if the degree of the numerator of the rational function isexactly 1 more than the degree of the denominator, the graph ofthe rational function has an oblique asymptote (or slantasymptote).

Example

f (x) =x3 + 4x2 − 4x − 13

x2 − 4= x + 4︸ ︷︷ ︸

slant asymptote

+3

x2 − 4

Illustration

f (x) =x3 + 4x2 − 4x − 13

x2 − 4= x + 4 +

3x2 − 4

-10 -5 5 10x

-10

-5

5

10

y

Finding Slant Asymptotes

We may find slant asymptotes by using polynomial longdivision.

x + 4x2 − 4

)x3 + 4x2 − 4x − 13− x3 + 4x

4x2 − 13− 4x2 + 16

3

f (x) =x3 + 4x2 − 4x − 13

x2 − 4= x + 4 +

3x2 − 4

Example (1 of 4)

Sketch the graph of the following function.

f (x) =3x2 + 12

x

f ′(x) =3(x2 − 4)

x2

f ′′(x) =24x3

I Vertical asymptote at x = 0.I Slant asymptote: y = 3x .

Example (1 of 4)

Sketch the graph of the following function.

f (x) =3x2 + 12

x

f ′(x) =3(x2 − 4)

x2

f ′′(x) =24x3

I Vertical asymptote at x = 0.I Slant asymptote: y = 3x .

Example (1 of 4)

Sketch the graph of the following function.

f (x) =3x2 + 12

x

f ′(x) =3(x2 − 4)

x2

f ′′(x) =24x3

I Vertical asymptote at x = 0.I Slant asymptote: y = 3x .

Example (2 of 4)

1. Critical numbers: x = ±2.2. f is increasing on the set (−∞,−2) ∪ (2,∞).3. f is decreasing on the set (−2,2).4. Local maximum at x = −2 and local minimum at x = 2.5. No points of inflection.6. f is concave down on the set (−∞,0) and concave up on

the set (0,∞).

Example (3 of 4)

Sketch the asymptotes and the points representing the localextrema.

-10 -5 5 10x

-20

-10

10

20

y

Example (4 of 4)

Fill in the rest of the graph using the increasing/decreasing andconcavity information found.

-10 -5 5 10x

-20

-10

10

20

y