Topic 3 of 8

Rational functions

Step 13 · Algebra 2 topics

Find the vertical and horizontal asymptotes, holes and zeros of rational functions, and solve rational equations, checking for answers that do not work.

Asymptotes

A rational function is a fraction of two polynomials. Where the bottom is 0 and the top is not, the graph shoots up or down along a vertical asymptote.

y = 1 ÷ x: a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. The curve gets ever closer to the axes without touching them.

A horizontal asymptote shows where the graph levels off as x gets very large. Compare the degrees of the top and the bottom.

Degrees	Horizontal asymptote	Example
top lower than bottom	y = 0	$\frac{3x}{x^{2} + 1}$
top equal to bottom	y = ratio of the leading coefficients	$\frac{4x + 1}{2x - 3}$ : y = 2
top higher than bottom	none	$\frac{x^{2}}{x + 1}$

Example: Finding asymptotes

Find the asymptotes of $f(x) = \frac{2x - 1}{x + 3}$ .

The bottom is 0 at x = −3: a vertical asymptote.
The degrees are equal, so y = 2 ÷ 1 = 2.

Answerx = −3 and y = 2

Holes and zeros

Zeros are where the top is 0 and the bottom is not. If a factor cancels from the top and the bottom, there is a hole at that x instead of an asymptote.

Example: A hole

Describe $f(x) = \frac{x^{2} - 4}{x - 2}$ .

Factor the top: (x + 2)(x − 2) over (x − 2).
It equals x + 2 everywhere except x = 2, where it is undefined.

AnswerThe line y = x + 2 with a hole at (2, 4)

Example: Zeros, holes and asymptotes

Find the zeros, holes and vertical asymptotes of $g(x) = \frac{x(x - 5)}{(x + 1)(x - 5)}$ .

x − 5 cancels, leaving a hole at x = 5.
The zero is x = 0, and the vertical asymptote is x = −1.

AnswerZero at 0, asymptote x = −1, hole at x = 5

Solving rational equations

To solve a rational equation, multiply every term by the common denominator to clear the fractions. Then check each answer: any value that makes a denominator 0 must be thrown out.

Example: Clearing fractions

Solve $\frac{3}{x} + \frac{1}{2} = \frac{5}{x}$ .

Multiply every term by 2x: 6 + x = 10.
x = 4, which does not make any denominator 0.

Answerx = 4

Example: An answer that fails

Solve $\frac{x}{x - 3} = \frac{3}{x - 3} + 2$ .

Multiply by x − 3: x = 3 + 2(x − 3).
x = 2x − 3, so x = 3.
But x = 3 makes the denominators 0.

AnswerNo solution: x = 3 is extraneous

Example: A work problem

One pipe fills a tank in 6 hours and another in 3 hours. How long do they take together?

Each hour they fill $\frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ of the tank.
So the whole tank takes 2 hours.

Answer2 hours

Try it yourself

Work each one on paper first, then open the answer to check.

Find the vertical asymptote of $y = \frac{5}{x - 7}$ .
Show answer
x = 7
Find the horizontal asymptote of $y = \frac{6x^{2} + 1}{3x^{2} - x}$ .
Show answer
y = 2
Find the horizontal asymptote of $y = \frac{x + 1}{x^{2} + 4}$ .
Show answer
y = 0
Where is the hole in $y = \frac{x^{2} - 9}{x + 3}$ ?
Show answer
At x = −3, the point (−3, −6)
Solve $\frac{4}{x} = \frac{2}{x - 1}$ .
Show answer
x = 2
Solve $\frac{1}{x} + \frac{1}{2x} = \frac{1}{4}$ .
Show answer
x = 6
One worker paints a room in 4 hours and another in 12 hours. How long do they take together?
Show answer
3 hours