use the dropdown menu to indicate whether each statement is true or false about the graph of the function…

Explanation:

Step1: Simplify the function

First, factor the denominator $x^{2}-36=(x + 6)(x - 6)$. Then $f(x)=\frac{x - 6}{(x + 6)(x - 6)}=\frac{1}{x + 6},x\neq6$.

Step2: Analyze vertical asymptote

Vertical asymptotes occur where the denominator of a rational - function is zero after simplification. For $y = \frac{1}{x + 6}$, the denominator is zero when $x=-6$. So the statement "The graph has a vertical asymptote at $x = 6$" is false.

Step3: Analyze continuity

The original function $f(x)=\frac{x - 6}{x^{2}-36}$ is not defined at $x = 6$ and $x=-6$. After simplification, $y=\frac{1}{x + 6}$ is not defined at $x=-6$. There is a removable discontinuity at $x = 6$ and an infinite discontinuity at $x=-6$. So the graph is not continuous, and the statement "The graph is continuous" is false.

Step4: Analyze point of discontinuity

Since $f(x)=\frac{x - 6}{x^{2}-36}$ is not defined at $x = 6$ (but $\lim_{x\rightarrow6}\frac{x - 6}{x^{2}-36}=\lim_{x\rightarrow6}\frac{1}{x + 6}=\frac{1}{12}$), there is a point of discontinuity at $x = 6$. So the statement "The graph has a point of discontinuity at $x = 6$" is true.

Step5: Analyze y - intercept

To find the y - intercept, set $x = 0$. Then $f(0)=\frac{0 - 6}{0^{2}-36}=\frac{-6}{-36}=\frac{1}{6}$. So the statement "The graph has a y - intercept of $(0,-\frac{1}{6})$" is false.

Answer:

a. FALSE b. FALSE c. TRUE d. FALSE