Graphing Rational Functions

2

x f(x)

2 0.5

1 1

0.5 2

0.1 10

0.01 100

0.001 1000

x f(x)

-2 -0.5

-1 -1

-0.5 -2

-0.1 -10

-0.01 -100

-0.001 -1000

As x → 0–, f(x) → -∞.As x → 0+, f(x) → +∞.

A rational function is a function of the form f(x) = ,

where P(x) and Q(x) are polynomials and Q(x) = 0.)(

)(

xQ

xP

f(x) =x

1

Example: f (x) = is defined for all real numbers except x = 0.x

1

3

x

x = a

as x → a –

f(x) → + ∞

x

x = a

as x → a –

f(x) → – ∞

x

x = a

as x → a +

f(x) → + ∞

x

x = a

as x → a +

f(x) → – ∞

The line x = a is a vertical asymptote of the graph of y = f(x), if and only if f(x) → + ∞ or f(x) → – ∞ as x → a + or as x → a –.

4

Example: Show that the line x = 2 is a vertical asymptote of the

graph of f(x) = .2)2(

4

x

x f(x)

1.5 16

1.9 400

1.99 40000

2 -

2.01 40000

2.1 400

2.5 16

Observe that:x→2–, f (x) → – ∞

x→2+, f (x) → + ∞

This shows that x = 2 is a vertical asymptote.

y

x100

0.5

f (x) = 2)2(

4

x

x = 2

5

Set the denominator equal to zero and solve. Solve the quadratic equation x2 + 4x – 5. (x – 1)(x + 5) = 0

Therefore, x = 1 and x = -5 are the values of x for which f may have a vertical asymptote.

As x →1– , f(x) → – ∞.

As x →1+, f(x) → + ∞.

As x → -5–, f(x) → + ∞.

As x →-5+, f(x) → – ∞.

x = -5 is a vertical asymptote.x = 1 is a vertical asymptote.

A rational function may have a vertical asymptote at

x = a for any value of a such that Q(a) = 0.)(

)(

xQ

xP

Example:Find the vertical asymptotes of the graph of f(x) = .

)54(

12 xx

6

1. Find the roots of the denominator. 0 = x2 – 4 = (x + 2)(x – 2) Possible vertical asymptotes are x = -2 and x = +2.

2. Calculate the values approaching -2 and +2 from both sides. x → -2, f(x) → -0.25; so x = -2 is not a vertical asymptote.

x → +2–, f(x) → – ∞ and x →+2+, f(x) → + ∞. So, x = 2 is a vertical asymptote.

f is undefined at -2

A hole in the graph of f at (-2, -0.25) shows a removable singularity.

x = 2

Example: Find the vertical asymptotes of the graph of f(x) = .

)4(

)2(2

x

x

x

y

(-2, -0.25)

7

y

y = b

as x → + ∞ f(x) → b –

y

y = b

as x → – ∞ f(x) → b –

y

y = b

as x → + ∞ f(x) → b +

y

y = b

as x → – ∞ f(x) → b +

The line y = b is a horizontal asymptote of the graph of y = f(x) if and only if f(x) → b + or f(x) → b – as x → + ∞ or as x → – ∞.

8

x f(x)

10 0.1

100 0.01

1000 0.001

0 –-10 -0.1

-100 -0.01

-1000 -0.001

As x becomes unbounded positively, f(x) approaches zero from above; therefore, the line y = 0 is a horizontal asymptote of the graph of f. As f(x) → – ∞, x → 0 –.

Example: Show that the line y = 0 is a horizontal asymptote of the graph of the function f(x) = .

x

1

x

y

f(x) =x

1

y = 0

9

y

x

Similarly, as x → – ∞, f(x) →1–.

Therefore, the graph of f has y = 1 as a horizontal asymptote.

Example: Determine the horizontal asymptotes of the graph of

f(x) = .)1( 2

2

x

x

Divide x2 + 1 into x2. f(x) = 1 – )1(

12 x

As x → +∞, → 0– ; so, f(x) = 1 – →1 –.)1(

12 x )1(

12 x

y = 1

10

Finding Asymptotes for Rational Functions

• If c is a real number which is a root of both P(x) and Q(x), then there is a removable singularity at c.

• If c is a root of Q(x) but not a root of P(x), then x = c is a vertical asymptote.

• If m > n, then there are no horizontal asymptotes.

• If m < n, then y = 0 is a horizontal asymptote.

• If m = n, then y = am is a horizontal asymptote.bn

Given a rational function: f (x) = P(x) am xm + lower degree terms

Q(x) bn xn + lower degree terms=

11

Factor the numerator and denominator.

The only root of the numerator is x = -1. The roots of the denominator are x = -1 and x = 2 .

Since -1 is a common root of both, there is a hole in the graph at -1 .

Since 2 is a root of the denominator but not the numerator, x = 2 will be a vertical asymptote.

Since the polynomials have the same degree, y = 3 will be a horizontal asymptote.

Example: Find all horizontal and vertical asymptotes of f (x) = .

2

3632

2

xx

xx

y = 3

x = 2

x

y

12

A slant asymptote is an asymptote which is not vertical or horizontal.

The slant asymptote is y = 2x – 5.

As x → + ∞, → 0+. 3

14

x

Example: Find the slant asymptote for f(x) = .3

12 2

x

xx

x

yx = -3

y = 2x - 5

Divide:3

12 2

x

xx3

1452

xx

Therefore as x → ∞, f(x) is more like the line y = 2x – 5.

3

14

xAs x → – ∞, → 0–.