Precalculus Rational Functions and Asymptotes Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Rational Functions and Asymptotes – Maths PDF

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This rational functions answers for Precalculus covers rational functions and asymptotes (Common Core A-APR.D.7, F-IF.C.7d, A-CED.A.1), with a free PDF to print.

Use this key to check each rational function. Factoring, cancelling and the extraneous-solution checks are shown, with final answers in bold.

Precalculus · Common Core State Standards: A-APR.D.7, F-IF.C.7d, A-CED.A.1

Rational functions answers: exercises 1 to 10

Exercise 1 – Domain

$x^{2}-5x+6=(x-2)(x-3)$, so $x\neq 2$ and $x\neq 3$.

$(-\infty,2)\cup(2,3)\cup(3,\infty)$

Exercise 2 – Hole and asymptote

$f(x)=\frac{(x-3)(x+3)}{(x-3)(x+2)}$. The factor $x-3$ cancels: hole at $x=3$ with $y=\frac{3+3}{3+2}=\frac65$. The factor $x+2$ remains: asymptote $x=-2$.

Hole $\left(3,\frac65\right)$; vertical asymptote $x=-2$

Exercise 3 – Equal degrees

Equal degrees: divide leading coefficients, $\frac62$.

$y=3$

Exercise 4 – Lower numerator degree

The numerator has degree 1 and the denominator has degree 2.

$y=0$

Exercise 5 – No horizontal asymptote

The degree difference is $4-2=2$, more than 1.

Neither

Exercise 6 – Intercepts

$2x-6=0$ gives $x=3$. $f(0)=\frac{-6}{4}=-\frac32$.

$x$-intercept $(3,0)$; $y$-intercept $\left(0,-\frac32\right)$

Exercise 7 – Read a graph

Asymptotes x = 3 and y = 2.

The denominator is 0 at $x=3$. The degrees are equal with leading coefficients 2 and 1. $f(0)=0$.

$x=3$; $y=2$; intercept $(0,0)$

Exercise 8 – Slant asymptote

The dashed slanted line is y = x + 2.

Divide with synthetic division, $c=3$: $1,-1,-2\to 1,2,4$. The quotient is $x+2$.

$y=x+2$

Exercise 9 – Find the hole

Hole at (2, 4).

Cancel $x-2$: $f(x)=x+2$. At $x=2$, $y=4$.

$(2,4)$

Exercise 10 – Solve a proportion

Cross multiply: $3(x+2)=2(x-1)$, $3x+6=2x-2$, so $x=-8$. Check: $\frac{3}{-9}=-\frac13$ and $\frac{2}{-6}=-\frac13$.

$x=-8$

Exercises 11 and up

Exercise 11 – Solve with two fractions

Multiply by $x(x+1)$: $2(x+1)+3x=2x(x+1)$, so $5x+2=2x^{2}+2x$ and $2x^{2}-3x-2=0$. Factor: $(2x+1)(x-2)=0$. Both $x=-\frac12$ and $x=2$ keep the denominators nonzero.

$x=2$ or $x=-\frac12$

Exercise 12 – Extraneous solution

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

No solution (3 is extraneous)

Exercise 13 – Rational inequality

Solution: left of -2 and from 1 rightward.

Critical values: $x=1$ (numerator 0, included) and $x=-2$ (denominator 0, excluded). Test $x=-3$: $\frac{-4}{-1}>0$. $x=0$: $\frac{-1}{2}< 0$. $x=2$: $\frac{1}{4}>0$.

$(-\infty,-2)\cup[1,\infty)$

Exercise 14 – Write an equation

The denominator is $x-2$. Equal degrees with ratio 3: numerator $3(x+1)$.

$f(x)=\frac{3(x+1)}{x-2}$

Exercise 15 – Behavior near an asymptote

$f(2.01)=\frac{1}{0.01}=100$. $f(1.99)=\frac{1}{-0.01}=-100$.

$100$ and $-100$; $f\to\infty$ as $x\to2^{+}$, $f\to-\infty$ as $x\to2^{-}$

Exercise 16 – Hole and asymptote together

Cancel $x-1$: $\frac{x+3}{x-5}$. The hole is at $x=1$, $y=\frac{4}{-4}=-1$. The asymptote is $x=5$.

Hole $(1,-1)$; asymptote $x=5$

Exercise 17 – Average cost of shirts

The average cost falls toward 12 dollars.

$A(50)=\frac{1100}{50}=22$. The asymptote $y=12$: as $n$ grows, the fixed cost is spread out and the average approaches $\$12$. Solve $500+12n\le 15n$: $n\ge\frac{500}{3}\approx 166.7$.

$\$22$; $y=12$; $n=167$ shirts

Exercise 18 – Salt mixture

Half salt at x = 6.

Equal degrees, ratio $\frac11$: $y=1$ (the fraction can never reach 100%). Solve $2+x=0.5(10+x)$: $2+x=5+0.5x$, $0.5x=3$, $x=6$.

$y=1$; $x=6$ gallons

Exercise 19 – Painting a fence

Multiply by $2x(x-3)$: $2(x-3)+2x=x(x-3)$, so $4x-6=x^{2}-3x$ and $x^{2}-7x+6=0$, giving $x=6$ or $x=1$. If $x=1$, then $x-3< 0$, which is impossible. So $x=6$. Check: $\frac16+\frac13=\frac12$.

Mei $6$ hours; Jamal $3$ hours

Exercise 20 – Complete analysis

Hole at (-2, 0.8), asymptotes x = 3 and y = 1.

$f(x)=\frac{(x-2)(x+2)}{(x-3)(x+2)}$. Cancel $x+2$: hole at $x=-2$, $y=\frac{-4}{-5}=\frac45$. Vertical asymptote $x=3$. Equal degrees: $y=1$. $x$-intercept: $2$. $y$-intercept: $f(0)=\frac{-4}{-6}=\frac23$.

Hole $\left(-2,\frac45\right)$; $x=3$; $y=1$; $(2,0)$; $\left(0,\frac23\right)$

Key facts behind this rational functions answers

Key facts

Domain: all real numbers except zeros of the denominator.
Vertical asymptote: a denominator zero that does not cancel. Hole: a cancelled factor.
Degree of numerator < denominator: y = 0. Equal degrees: y = ratio of leading coefficients.
Degree of numerator = degree of denominator + 1: slant asymptote from division.
Always check rational equation answers for extraneous solutions.

How to use this rational functions answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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