Precalculus Honors · Study guide

1.3 · Rational and radical functions: restrictions and asymptotes

Preserve exclusions, distinguish holes from asymptotes, and check proposed solutions after algebraic manipulation.

Precalculus Honors · Reading: Demana 10th ed. §§2.6–2.8; Stewart 8th ed. §§1.5–1.8, 3.6–3.7

Learning objectives

Classify excluded inputs of a rational function.
Find rational end behavior.
Check radical and rational equation solutions.

Factor before classifying a discontinuity

A zero denominator gives an excluded input in the original formula. If its factor cancels fully, nearby behavior may have a finite limit, producing a hole. A remaining denominator factor can produce a vertical asymptote.

For $\displaystyle (x^2-1)/(x-1)$ , the graph agrees with $\displaystyle x+1$ away from $\displaystyle 1$ , but the original function still excludes $\displaystyle 1$ . Canceling changes the convenient formula, not the original domain.

Use degree to predict rational end behavior

For a proper rational function whose numerator degree is lower, the horizontal asymptote is $\displaystyle y=0$ . Equal degrees give the ratio of leading coefficients. Higher numerator degree requires division to describe the polynomial end behavior.

Asymptotes describe limiting behavior, not barriers that curves can never cross. A graph should also show intercepts, holes, and any restricted intervals.

Squaring and clearing denominators need checks

Squaring a radical equation can introduce extraneous solutions. First isolate the radical, identify any sign restrictions, square, and substitute every candidate into the original equation.

Clearing denominators also requires the original excluded inputs to remain excluded. For inequalities, use a sign chart with both zeros and denominator exclusions; multiplying by an unknown-sign expression can reverse an inequality.

Worked examples

A hole and an asymptote

Analyze $\displaystyle f(x)=\frac{x^2-1}{(x-1)(x-2)}$ .

The original domain excludes $\displaystyle 1$ and $\displaystyle 2$ .
For allowed inputs, the formula simplifies to $\displaystyle (x+1)/(x-2)$ .
There is a hole at $\displaystyle (1,-2)$ and a vertical asymptote at $\displaystyle x=2$ ; the horizontal asymptote is $\displaystyle y=1$ .

Check a squared equation

Solve $\displaystyle \sqrt{x+2}=x$ .

The right side must be nonnegative, so $\displaystyle x\geq0$ .
Squaring gives $\displaystyle x^2-x-2=0$ , with candidates $\displaystyle 2$ and $\displaystyle -1$ .
Only $\displaystyle x=2$ satisfies the original equation.

Common mistakes

Restoring an excluded input after canceling a factor.
Calling a canceled factor a vertical asymptote.
Accepting all roots of a squared equation.

Try it yourself

No calculator required.

1. Classify the excluded inputs and asymptotes of $\displaystyle f(x)=\frac{x^2-1}{(x-1)(x-2)}$ .

The curve (x plus 1) over (x minus 2) has a hole at (1, minus 2) and a vertical asymptote at x = 2. — Preserve both exclusions when canceling the common factor.

Show worked solution

The exclusions are $\displaystyle 1$ and $\displaystyle 2$ . Canceling $\displaystyle x-1$ gives a hole at $\displaystyle (1,-2)$ ; the remaining denominator gives $\displaystyle x=2$ as a vertical asymptote. Equal degrees give horizontal asymptote $\displaystyle y=1$ .

The curve (x plus 1) over (x minus 2) has a hole at (1, minus 2) and a vertical asymptote at x = 2. The canceled input gives a hole at (1, minus 2); asymptotes are x = 2 and y = 1. — The canceled input gives a hole at (1, minus 2); asymptotes are x = 2 and y = 1.

2. Solve $\displaystyle \sqrt{x+2}=x$ and check every candidate.

Show worked solution

Squaring gives $\displaystyle (x-2)(x+1)=0$ . The candidate $\displaystyle -1$ fails the nonnegative right-side requirement; $\displaystyle 2$ works. The solution is $\displaystyle x=2$ .

3. Solve $\displaystyle \frac{x-1}{x+2}>0$ .

Show worked solution

The critical inputs are $\displaystyle -2$ and $\displaystyle 1$ . A sign chart is positive on $\displaystyle (-\infty,-2)$ and $\displaystyle (1,\infty)$ . The denominator exclusion and strict inequality remove both endpoints.

Download lesson worksheet Download solutions

FRQ practice & solutions

Need help with this topic?

Work through questions with a tutor and get help with the steps you find difficult.

Get help with this topic