Chalk−1

Math · College algebra · Worked example

Graph a rational function with a hole

Find the domain, hole, asymptotes and intercepts of f(x) = (x² − 1)/(x² − x − 2).

f⁡(x)=x2−1x2−x−2

Factor and find the domain

The numerator is (x − 1)(x + 1) and the denominator (x − 2)(x + 1), which is zero at x = 2 and x = −1. The domain excludes both.

x2−1x2−x−2(x−1)⁢(x+1)(x−2)⁢(x+1)

Cancel to find the hole

x + 1 cancels, so there is a hole at x = −1. Its height is the simplified expression’s value there.

−1−1−1−2=23

Find the vertical asymptote

x − 2 stays in the denominator, so x = 2 is a vertical asymptote. Just left of 2 the values fall toward −∞; just right of 2 they rise toward +∞.

Find the horizontal asymptote

Numerator and denominator both have degree 2 and leading coefficient 1, so the horizontal asymptote is y = 1/1 = 1.

Find the intercepts

The simplified numerator x − 1 is zero at x = 1, the x-intercept. The y-intercept is f(0).

0−10−2=12

Result

Domain x ≠ −1, 2. Hole at (−1, 2/3), vertical asymptote x = 2, horizontal asymptote y = 1, intercepts (1, 0) and (0, 1/2).

Your turn

Find the vertical and horizontal asymptotes of g(x) = (3x + 1)/(x − 4).

Show the answer and explanation

x = 4 and y = 3.

The denominator is zero at x = 4 and the numerator is not (3·4 + 1 = 13), so x = 4 is a vertical asymptote. Both degrees are 1, so the horizontal asymptote is the ratio of leading coefficients, 3/1 = 3.

g⁡(x)=3⁢x+1x−4g⁡(0)=−14

Keep exploring

In the limit explorer, move the approach from −1 to 2. The left and right limits become −∞ and +∞: that is what a vertical asymptote looks like.

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See the hole and asymptotes in Graph Explore the hole in the limit explorer Check the simplification in Math Open worked example on a board Asymptotes and holes in Math Reference

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