Question

identify the vertical asymptote(s) of each function. check all of the boxes that apply. \\( f(x) = \frac{x - 8}{x^2 - 3x + 2} \\) \\( \square \\ x = -8 \\) \\( \square \\ x = -2 \\) \\( \square \\ x = -1 \\) \\( \square \\ x = 1 \\) \\( \square \\ x = 2 \\) \\( \square \\ x = 8 \\) done \\( f(x) = \frac{3x}{x^2 - 16} \\) \\( \square \\ x = -16 \\) \\( \square \\ x = -4 \\) \\( \square \\ x = 0 \\) \\( \square \\ x = 4 \\) \\( \square \\ x = 16 \\) done

Explanation:

Step1: Factor first function's denominator

$x^2 - 3x + 2 = (x-1)(x-2)$

Step2: Find zeros of denominator

Set $(x-1)(x-2)=0$, so $x=1$ and $x=2$

Step3: Factor second function's denominator

$x^2 - 16 = (x-4)(x+4)$

Step4: Find zeros of denominator

Set $(x-4)(x+4)=0$, so $x=4$ and $x=-4$

Answer:

For $f(x)=\frac{x-8}{x^2-3x+2}$:
x = 1, x = 2
For $f(x)=\frac{3x}{x^2-16}$:
x = -4, x = 4