Find all points of discontinuity, and determine whether there is a hole or a vertical asymptote at each point.
-$f ( x ) = \frac { x - 8 } { x ^ { 2 } - 16 }$
A)-4: vertical asymptote
B)8: vertical asymptote
C)4: vertical asymptote; -4: vertical asymptote
D)4: vertical asymptote

Question

Quizplus · Accepted Answer

The function $f(x) = \frac{x - 8}{x^2 - 16}$ can be simplified to $f(x) = \frac{x - 8}{(x - 4)(x + 4)}$. The denominator equals zero when $x = 4$ and $x = -4$, which are the points of discontinuity. Since the numerator does not cancel out the factors $x - 4$ or $x + 4$, these points are vertical asymptotes, not holes.

Find All Points of Discontinuity, and Determine Whether There Is f(x)=x−8x2−16f ( x ) = \frac { x - 8 } { x ^ { 2 } - 16 }f(x)=x2−16x−8​

Find All Points of Discontinuity, and Determine Whether There Is $f ( x ) = \frac { x - 8 } { x ^ { 2 } - 16 }$