Notes 3-7: Rational Functions

Warm Up Find the zeros of each function.

1. f(x) = x2 + 2x – 15

2. f(x) = x2 – 49

3.

4.

x ≠ ± 1

x ≠ 6

Simplify. Identify any x-values for which the expression is undefined.

x2 – 8x + 12 x2 – 12x + 36

x – 2 x – 6

x + 4

x – 1

–5, 3

±7

x2 + 5x + 4 x2 – 1

A rational function is a function thatcan be written as a ratio of two polynomials. The parent rational

function is 𝑓 𝑥 =1

𝑥 .

Like logarithmic and exponential

functions, rational functions may

have asymptotes. The function

𝑓 𝑥 =1

𝑥 has a vertical asymptote

at

x = 0 and a horizontal asymptote

at y = 0.

I. Rational Functions

The rational function 𝑓 𝑥 =1

𝑥 can be transformed by

using methods similar to those used to transform other

types of functions.

II. Transformations of Rational Functions

a. g(x) =

Because h = –4, translate f 4 units left.

1 x + 4

b. g(x) =

Because k = 1, translate f 1 unit up.

1 x

+ 1

Ex 1: Using the graph of 𝒇 𝒙 =𝟏

𝒙 as a guide, describe

the transformation and graph each function. Identify the location of the vertical and horizontal asymptotes.

Vertical asymptote: x = -4 Horizontal asymptote: y = 0

Vertical asymptote: x = 0 Horizontal asymptote: y = 1

A. Vertical Asymptotes and Holes: To find vertical asymptotes, find the zeros of the denominator. If there is the same zero in the numerator, the discontinuity is a hole. If there is NOT the same zero in the numerator, the discontinuity is a vertical asymptotes. Compare:

𝑓 𝑥 =𝑥+1

𝑥−2

There is a vertical asymptote at x = 2

g 𝑥 =(𝑥−2)(𝑥+1)

𝑥−2 There

is a hole at x = 2

III. Identifying Discontinuities for Other Types of Rational Functions:

For functions like f(x) = 𝟑𝒙−𝟏

𝒙−𝟐, we can use the following strategies to

find discontinuities:

B. Horizontal Asymptotes If the highest degree is in the denominator, the horizontal asymptote is at y = 0. If the degrees in the numerator and denominator are the same, the horizontal asymptote is the ratio of leading coefficients. If the highest degree is in the numerator, there is no horizontal asymptote.

𝑓 𝑥 =𝑥 − 1

𝑥2 + 1: horizontal asympotote: y = o

𝑓 𝑥 =3𝑥 − 1

2𝑥 + 1: horizontal asympotote: y =

3

2

𝑓 𝑥 =𝑥2 − 1

𝑥 + 1: no horizontal asymptote

Ex 1: Determine the discontinuities for the graph of

f(x) = . (x2 + 7x + 6) x + 3

(x + 6)(x + 1) x + 3

f(x) =

Step 1 Vertical asymptotes/holes.

No Holes; Vertical asymptote: x = –3

The denominator is 0 when x = –3. (x + 3) is not in the

numerator, so it is a vertical

asymptote and not a hole.

Step 2 Horizontal asymptotes.

None: The exponent in the numerator is the largest, so there is no horizontal asymptote.

Ex 2: Determine the discontinuities for the graph of

f(x)=

Step 1 Vertical asymptotes/holes.

No vertical asymptote. A hole is at x = 2.

The denominator is 0

when x = 2. Since (x -

2) is also in the

numerator, it is a hole,

not a vertical asymptote.

Step 2 Horizontal asymptotes.

None: The exponent in the numerator is the largest.

x2 + x – 6 x – 2

(x – 2)(x + 3) x – 2

f(x) =

Remember

This is the same as the graph of y = x + 3, except for the hole at x = 2. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 2}.

Hole at x = 2

Ex 3: Determine the discontinuities for the graph of

.

Step 1 Vertical asymptotes/holes.

No Holes; Vertical asymptotes: x = -1 and x = 0.

The denominator is 0 when x = -1

or 0. Since neither of those factors

are also in the numerator, they are

vertical asymptotes and not a

holes.

Step 2 Horizontal asymptotes.

y = 0. The exponent in the denominator is the largest.

x – 2 x2

+ x f(x) = x – 2

x(x + 1) f(x) =

Ex 4: Determine the discontinuities for the graph of

f(x) = .

Step 1 Vertical asymptotes/holes.

No Holes; Vertical asymptote: x = 1

The denominator is 0 when x = 1. (x - 1) is not in the numerator,

so it is a vertical asymptote and

not a hole.

Step 2 Horizontal asymptotes.

None: The exponent in the numerator is the largest, so there is no horizontal asymptote.

x2 + 2x – 15 x – 1 (x – 3)(x + 5)

x – 1 f(x) =

C. Slant Asymptotes.

The graph of a rational function has a slant asymptote if the degree of the numerator is exactly one more than the degree of the denominator. Long division is used to find slant asymptotes.

The only time you have an oblique asymptote is when there is no horizontal asymptote. You cannot have both.

When doing long division, we do not care about the remainder.

Example 1 • Determine if the following function has a slant

asymptote. If it does, find the equation for it.

2 2( )

1

xf x

x

n > d by exactly one, so no horizontal asymptote, but there is an oblique (slant) asymptote.

The slant asymptote is the line y = x + 1

1 1 0 -2

1 1

1 1 1

Example 2 • Determine if the following function has a slant

asymptote. If it does, find the equation for it.

n > d by exactly one, so no horizontal asymptote, but there is an oblique (slant) asymptote.

The slant asymptote is the line y = x

1 1 -1 -2

1 0

1 0 -2

2 2

1

x xf x

x