Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.6 - Rational Functions - 3.6 Exercises - Page 308: 23

Answer

$y$-intercept: $\dfrac{1}{3}$ $x$-intercepts: $2$ and $-1$

Work Step by Step

$t(x)=\dfrac{x^{2}-x-2}{x-6}$ Substitute $t(x)$ by $y$: $y=\dfrac{x^{2}-x-2}{x-6}$ To find the $y$-intercept, set $x$ equal to $0$ and solve for $y$: $y=\dfrac{(0)^{2}-0-2}{0-6}=\dfrac{-2}{-6}=\dfrac{1}{3}$ To find the $x$-intercept, set $y$ equal to $0$ and solve for $x$: $0=\dfrac{x^{2}-x-2}{x-6}$ $(0)(x-6)=x^{2}-x-2$ $x^{2}-x-2=0$ Solve by factoring: $(x-2)(x+1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x+1=0$ $x=-1$ $y$-intercept: $\dfrac{1}{3}$ $x$-intercepts: $2$ and $-1$
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