Question

find the vertical asymptotes and holes (removable discontinuities) of the rational function.

f(x)=\frac{x^{2}-12 x + 32}{x^{2}-4 x - 32}

the holes are at ( x = )

the vertical asymptotes are ( x = )

if there is more than one of each, enter your answers separated by a comma. if there is not a hole or asymptote, enter dne.

question help:

Explanation:

Step1: Factor numerator and denominator

Factor \(x^{2}-12x + 32=(x - 4)(x - 8)\)
Factor \(x^{2}-4x-32=(x + 4)(x - 8)\)
So \(f(x)=\frac{(x - 4)(x - 8)}{(x + 4)(x - 8)}\)

Step2: Simplify the function

Cancel out the common factor \((x - 8)\) (for \(x
eq8\)). The simplified function is \(f(x)=\frac{x - 4}{x + 4}\), \(x
eq8\)

Step3: Find holes and vertical asymptotes

• Holes: A hole occurs when a factor cancels out. The hole is at \(x = 8\)
• Vertical asymptotes: Set the denominator of the simplified function \(x+4 = 0\), so \(x=-4\)

Answer:

The holes are at \(x = 8\). The vertical asymptotes are \(x=-4\)