# How do you find the vertical, horizontal and slant asymptotes of: h(x)= (x^2-4)/(x)?

Aug 15, 2016

vertical asymptote at x = 0
slant asymptote y = x

#### Explanation:

The denominator of h(x) cannot be zero as this would make h(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $x = 0 \Rightarrow x = 0 \text{ is the asymptote}$

$\textcolor{b l u e}{\text{Horizontal asymptotes}}$ occur when the degree of the numerator ≤ degree of the denominator. This is not the case here (numerator degree 2 , denominator degree 1 ) Hence there are no horizontal asymptotes.

$\textcolor{b l u e}{\text{slant asymptotes}}$ occur when the degree of the numerator > degree of denominator. Hence there is a slant asymptote.

divide numerator by denominator.

$\Rightarrow h \left(x\right) = \frac{{x}^{2}}{x} - \frac{4}{x} = x - \frac{4}{x}$

as $x \to \pm \infty , h \left(x\right) \to x - 0$

$\Rightarrow y = x \text{ is the asymptote}$
graph{(x^2-4)/x [-10, 10, -5, 5]}