# How do you find the domain, VA, HA, Zeros and intercepts of y=(x^2-16)/(x^2-4)?

Jul 9, 2017

$\text{see explanation}$

#### Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } {x}^{2} - 4 = 0 \to \left(x - 2\right) \left(x + 2\right) = 0$

$\Rightarrow x = \pm 2 \text{ are the asymptotes}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 2$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , y \to c \text{ (a constant )}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$y = \frac{{x}^{2} / {x}^{2} - \frac{16}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{1 - \frac{16}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

as $x \to \pm \infty , y \to \frac{1 - 0}{1 - 0}$

$\Rightarrow y = 1 \text{ is the asymptote}$

$\textcolor{b l u e}{\text{Intercepts}}$

$x = 0 \to y = \frac{- 16}{- 4} = 4 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to {x}^{2} - 16 = 0 \to \left(x - 4\right) \left(x + 4\right) = 0$

$\Rightarrow x = \pm 4 \leftarrow \textcolor{red}{\text{ x-intercepts}}$

$\text{which, of course are also the zeros}$
graph{(x^2-16)/(x^2-4) [-10, 10, -5, 5]}