$\text { Find the vertical asymptotes (if any) of the function } f ( x ) = \frac { x ^ { 2 } - 4 } { x ^ { 2 } + 3 x + 2 } \text {. }$
A) $x = 2$
B) $x = - 1$
C) $x = 1$
D) $x = - 2$.
E) $x = - 2$

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). Factoring the denominator $x^2 + 3x + 2 = (x + 1)(x + 2)$, we see that $x = -1$ and $x = -2$ make the denominator zero. However, the numerator $x^2 - 4 = (x - 2)(x + 2)$ is also zero when $x = -2$, which means it's not a vertical asymptote due to the factor canceling out. Thus, the only vertical asymptote is at $x = -1$.

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$\text { Find the vertical asymptotes (if any) of the function } f ( x ) = \frac { x ^ { 2 } - 4 } { x ^ { 2 } + 3 x + 2 } \text {. }$