Give the equations of any vertical asymptotes.
-$h ( x ) = \frac { ( x - 6 ) ( x + 3 ) } { x ^ { 2 } - 1 }$
A) $x = 6 , x = - 3$
B) $x = - 6 , x = 3$
C) $x = 1 , x = - 1$
D) none

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 $h(x) = \frac{(x - 6)(x + 3)}{x^2 - 1}$, the denominator is $x^2 - 1$, which factors to $(x + 1)(x - 1)$. Setting each factor equal to zero gives $x = -1$ and $x = 1$, hence the vertical asymptotes are at $x = -1$ and $x = 1$.

Give the Equations of Any Vertical Asymptotes h(x)=(x−6)(x+3)x2−1h ( x ) = \frac { ( x - 6 ) ( x + 3 ) } { x ^ { 2 } - 1 }h(x)=x2−1(x−6)(x+3)​

Give the Equations of Any Vertical Asymptotes $h ( x ) = \frac { ( x - 6 ) ( x + 3 ) } { x ^ { 2 } - 1 }$