Find the vertical asymptotes of the rational function.
-$$f ( x ) = \frac { - 2 x ( x + 2 ) } { 4 x ^ { 2 } - 5 x - 9 }$$
A) $x = \frac { 4 } { 9 } , x = - 1$
B) $x = - \frac { 9 } { 4 } , x = 1$
C) $x = \frac { 9 } { 4 } , x = - 1$
D) $x = - \frac { 4 } { 9 } , x = 1$

Question

Quizplus · Accepted Answer

Vertical asymptotes occur where the denominator of a rational function is zero (and the numerator is not zero at those points). The denominator is $4x^2 - 5x - 9$. Factoring or using the quadratic formula, we find the roots of the denominator to be $x = \frac{5 \pm \sqrt{(-5)^2 - 4(4)(-9)}}{2(4)} = \frac{5 \pm \sqrt{25 + 144}}{8} = \frac{5 \pm \sqrt{169}}{8} = \frac{5 \pm 13}{8}$. This gives $x = \frac{18}{8} = \frac{9}{4}$ and $x = \frac{-8}{8} = -1$, which are the points where the function has vertical asymptotes.

Find the Vertical Asymptotes of the Rational Function f(x)=−2x(x+2)4x2−5x−9f ( x ) = \frac { - 2 x ( x + 2 ) } { 4 x ^ { 2 } - 5 x - 9 }f(x)=4x2−5x−9−2x(x+2)​

Find the Vertical Asymptotes of the Rational Function $f ( x ) = \frac { - 2 x ( x + 2 ) } { 4 x ^ { 2 } - 5 x - 9 }$