f(x)=(-2x² + 8)/(x - 2)\nanswer attempt 1 out of 2\nhorizontal asymptote: y = \nno horizontal…

Explanation:

Step1: Simplify the function

First, factor the numerator: $-2x^{2}+8=-2(x^{2} - 4)=-2(x + 2)(x - 2)$. Then $f(x)=\frac{-2(x + 2)(x - 2)}{x - 2}=-2(x + 2)$ for $x\neq2$.

Step2: Find horizontal asymptote

Since the simplified function $y=-2x - 4$ is a linear - function, there is no horizontal asymptote.

Step3: Find vertical asymptote

The original function $f(x)=\frac{-2x^{2}+8}{x - 2}$ is undefined when $x - 2=0$, so $x = 2$ is a removable discontinuity (a hole), not a vertical asymptote. So, no vertical asymptote.

Step4: Find x - intercept

Set $y = 0$ in the simplified function $y=-2(x + 2)$. Then $0=-2(x + 2)$, which gives $x=-2$. So the x - intercept is $(-2,0)$.

Step5: Find y - intercept

Set $x = 0$ in the original function $f(0)=\frac{-2\times0^{2}+8}{0 - 2}=\frac{8}{-2}=-4$. So the y - intercept is $(0,-4)$.

Step6: Find the hole

The function is undefined at $x = 2$. Substitute $x = 2$ into the simplified function $y=-2(x + 2)$. When $x = 2$, $y=-2(2 + 2)=-8$. So the hole is at $(2,-8)$.

Answer:

Horizontal Asymptote: No horizontal asymptote Vertical Asymptote: No vertical asymptote x - Intercept: $(-2,0)$ y - Intercept: $(0,-4)$ Hole: $(2,-8)$