for the function below, find all x - values for which the function is discontinuous and state what kind of…

Explanation:

Step1: Factor the numerator and denominator

Factor $2x^{2}-7x - 15=(2x + 3)(x-5)$ and $x^{2}-x - 20=(x - 5)(x+4)$. So $f(x)=\frac{(2x + 3)(x - 5)}{(x - 5)(x + 4)}$.

Step2: Find the values that make the denominator zero

Set $x^{2}-x - 20 = 0$, i.e., $(x - 5)(x + 4)=0$. Solving gives $x = 5$ and $x=-4$.

Step3: Simplify the function and analyze the discontinuities

Cancel out the common factor $(x - 5)$ (for $x\neq5$), $f(x)=\frac{2x + 3}{x + 4},x\neq5$. For $x = 5$, $\lim_{x\rightarrow5}f(x)=\frac{2\times5+3}{5 + 4}=\frac{13}{9}$, but the original function is undefined at $x = 5$, so there is a hole at $x = 5$. For $x=-4$, $\lim_{x\rightarrow - 4^{-}}f(x)=-\infty$ and $\lim_{x\rightarrow - 4^{+}}f(x)=\infty$, so there is a vertical - asymptote at $x=-4$.

Answer:

The function is discontinuous at $x = 5$ (hole) and $x=-4$ (vertical asymptote).