Question

determine each feature of the graph of the given function. f(x) = (-2(x - 2))/((x - 2)^3) answer attempt 1 out of 4 # of horizontal asymptotes: none (dropdown with none, one, two, three) # of holes: none (dropdown) # of vertical asymptotes: none (dropdown) # of x-intercepts: none (dropdown) # of y-intercepts: none (dropdown)

Explanation:

Step1: Simplify the function

First, cancel the common factor $(x-2)$ (for $x
eq 2$):

$$f(x)=\frac{-2(x-2)}{(x-2)^3}=\frac{-2}{(x-2)^2}, \quad x eq 2$$

Step2: Find horizontal asymptotes

Compare degrees of numerator/denominator:
Numerator degree = 0, denominator degree = 2. Since $0<2$, the horizontal asymptote is $y=0$. So there is one horizontal asymptote.

Step3: Find holes

Holes occur when a factor cancels, and the canceled factor gives the x-value of the hole. Here, $(x-2)$ cancels, so $x=2$ is a hole. So there is one hole.

Step4: Find vertical asymptotes

Set simplified denominator to 0:
$$(x-2)^2=0 \implies x=2$$
But $x=2$ is a hole, not a vertical asymptote. So there are none vertical asymptotes.

Step5: Find x-intercepts

Set $f(x)=0$:
$$\frac{-2}{(x-2)^2}=0$$
No solution (numerator is non-zero). So there are none x-intercepts.

Step6: Find y-intercepts

Set $x=0$:
$$f(0)=\frac{-2}{(0-2)^2}=\frac{-2}{4}=-\frac{1}{2}$$
So there is one y-intercept.

Answer:

of Horizontal Asymptotes: One

of Holes: One

of Vertical Asymptotes: None

of x-intercepts: None

of y-intercepts: One