Give the equations of any vertical asymptotes.
-$$h ( x ) = \frac { ( x - 4 ) ( x + 1 ) } { x ^ { 2 } - 9 }$$
A) $x = - 4 , x = 1$
B) $x = 4 , x = - 1$
C) none
D) $x = 3 , x = - 3$

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). For $h(x) = \frac{(x - 4)(x + 1)}{x^2 - 9}$, the denominator is $x^2 - 9 = (x - 3)(x + 3)$, which is zero at $x = 3$ and $x = -3$, hence the vertical asymptotes are at $x = 3$ and $x = -3$.

Give the Equations of Any Vertical Asymptotes h(x)=(x−4)(x+1)x2−9h ( x ) = \frac { ( x - 4 ) ( x + 1 ) } { x ^ { 2 } - 9 }h(x)=x2−9(x−4)(x+1)​

Give the Equations of Any Vertical Asymptotes $h ( x ) = \frac { ( x - 4 ) ( x + 1 ) } { x ^ { 2 } - 9 }$