Question

identify the vertical asymptote(s) of each function. check all of the boxes that apply.

left function:
(f(x) = \frac{x - 8}{x^2 - 3x + 2})
(x = -8)
(x = -2)
(x = -1)
(x = 1)
(x = 2)
(x = 8)

right function:
(f(x) = \frac{3x}{x^2 - 16})
(x = -16)
(x = -4)
(x = 0)
(x = 4)
(x = 16)

Explanation:

⚡ Using what you learned: Identifying Restrictions and Asymptotes

Step 1: Analyze the first function

The first function is:

$$ f(x) = \frac{x - 8}{x^2 - 3x + 2} $$

Factor the denominator:

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

Since there are no common factors between the numerator and the denominator, the vertical asymptotes occur where the denominator is zero:

$$ (x - 1)(x - 2) = 0 \implies x = 1 \text{ and } x = 2 $$

Step 2: Analyze the second function

The second function is:

$$ f(x) = \frac{3x}{x^2 - 16} $$

Factor the denominator:

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

Since there are no common factors between the numerator and the denominator, the vertical asymptotes occur where the denominator is zero:

$$ (x - 4)(x + 4) = 0 \implies x = -4 \text{ and } x = 4 $$

Answer:

• For the first function \( f(x) = \frac{x - 8}{x^2 - 3x + 2} \):
• \( x = 1 \)
• \( x = 2 \)

• For the second function \( f(x) = \frac{3x}{x^2 - 16} \):
• \( x = -4 \)
• \( x = 4 \)