Answer
The vertical asymptote: $x=1$.
The horizontal asymptote: $y=1$.
Work Step by Step
Simplify the equation:
$$f(x)=\frac{(x-2)(x+2)}{(x-2)(x-1)}=\frac{x+2}{x-1},x\not=2$$
Finding the vertical asymptotes, equate the denominator to $0$:
$$x-1=0$$ $$x=1$$
Thus, the vertical asymptote is $x=1$.
Since the degree of the numerator and the degree of the denominator are equal, taking the quotient of the leading coefficient of the numerator and the leading coefficient of the denominator, the horizontal asymptote is:
$$y=\frac{1}{1}$$ $$y=1$$
