Functions — Question 7

Question 7 - Solution

We are given $$f(x)=\frac{x^2+2x-3}{x^2-4}.$$ Factor the numerator and denominator: $$f(x)=\frac{(x+3)(x-1)}{(x-2)(x+2)}.$$

(a) Vertical asymptotes occur where the denominator is zero and no factor cancels. Since $$x^2-4=(x-2)(x+2),$$ the denominator is zero at $$x=2 \quad \text{and} \quad x=-2.$$ There are no common factors with the numerator, so both are vertical asymptotes: $$\boxed{x=-2,\ x=2}.$$

Since the numerator and denominator have the same degree, the horizontal asymptote is the ratio of leading coefficients: $$y=\frac{1}{1}=1.$$ Thus, $$\boxed{y=1}.$$

(b) Holes occur when a factor cancels from the numerator and denominator. Since $$\frac{(x+3)(x-1)}{(x-2)(x+2)}$$ has no common factors, there are $$\boxed{\text{no holes}}.$$

(c) The domain excludes the values that make the denominator zero: $$x\neq -2,\quad x\neq 2.$$ Therefore, $$\boxed{(-\infty,-2)\cup(-2,2)\cup(2,\infty)}.$$