Answer
There is a hole at point $\left(-2,\frac{4}{3}\right)$
The $x$-intercept: $(2,0)$
The $y$-intercept: $(0,2)$
The vertical asymptote: $x=1$
The horizontal asymptote: $y=1$
See graph
Work Step by Step
Simplify the function:
$$f(x)=\frac{(x+2)(x-2)}{(x+2)(x-1)}$$ $$f(x)=\frac{x-2}{x-1}$$
Equate $x+2$ to $0$:
$$x+2=0$$ $$x=-2$$ $$f(-2)=\frac{-2-2}{-2-1}=\frac{4}{3}$$
Thus, there is a hole at point $\left(-2,\frac{4}{3}\right)$.
Equate the numerator to $0$ to find the $x$-intercept:
$$x-2=0$$ $$x=2$$
Thus, the $x$-intercept is at $(2,0)$.
Set $x=0$ to find the $y$-intercept:
$$f(0)=\frac{0-2}{0-1}=2$$
Thus, the $y$-intercept is also at $(0,2)$.
The function is undefined when the denominator is $0$:
$$x-1=0$$ $$x=1$$
Thus, the vertical asymptote is $x=1$.
Since the degree of the numerator is equal to the degree of the denominator, divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote is:
$$y=\frac{1}{1}$$ $$y=1$$
Thus, the horizontal asymptote is $y=1$.
The sketch of the graph of the function is as shown.
