find the vertical asymptote(s) of the graph of the given function.\nf(x) = \\frac{x^{2}+2x - 8}{x^{2}-4x…

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+2x - 8=(x + 4)(x - 2)$. The denominator $x^{2}-4x - 12=(x - 6)(x+2)$. So $f(x)=\frac{(x + 4)(x - 2)}{(x - 6)(x + 2)}$.

Step2: Find values that make denominator zero

Vertical asymptotes occur at values of $x$ that make the denominator of a rational - function equal to zero while the numerator is non - zero. Set the denominator equal to zero: $(x - 6)(x + 2)=0$. Using the zero - product property, $x-6 = 0$ gives $x = 6$ and $x+2 = 0$ gives $x=-2$. The numerator is non - zero at $x = 6$ and $x=-2$.

Answer:

C. $x=-2,x = 6$