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 Expressions - 3.6 Exercises - Page 309: 39

Answer

Vertical asymptote: $x=0$ Horizontal asymptote: $y=3$

Work Step by Step

$r(x)=\dfrac{6x^{3}-2}{2x^{3}+5x^{2}+6x}$ Vertical asymptote A rational function has vertical asymptotes where the function is undefined, that is, where the denominator is zero. Set the denominator equal to 0 and solve for x to find the vertical asymptotes of this function: $2x^{3}+5x^{2}+6x=0$ $x(2x^{2}+5x+6)=0$ $x=0$ $2x^{2}+5x+6=0$ Use the quadratic formula, which is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ to solve this second equation. Here, $a=2$, $b=5$ and $c=6$: $x=\dfrac{-5\pm\sqrt{5^{2}-4(2)(6)}}{2(2)}=\dfrac{-5\pm\sqrt{25-48}}{4}$ The operation inside the square root will yield a negative number, so this equation has complex roots. The only vertical asymptote this function has is $x=0$ Horizontal asymptote The degrees of the numerator and the denominator are the same, divide the leading coefficient of the numerator by the leading coefficient of the denominator to obtain the horizontal asymptote: $y=\dfrac{6}{2}=3$
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