How do you find the asymptotes for f(x) =(x^2 - 4)/(9 - x^2)?

May 6, 2016

Vertical asymptotes at $x = 3$ and $x = - 3$ and horizontle asymptote at $y = - 1$
To find all the asymptotes for function $y = \frac{{x}^{2} - 4}{9 - {x}^{2}}$ let us first start with vertical asymptotes, which are given by putting denominator equal to zero i.e. here $9 - {x}^{2} = 0$ i.e. $\left(3 - x\right) \left(3 + x\right) = 0$. Hence vertical asymptotes are $x = 3$ and $x = - 3$.
Further as in $y = \frac{{x}^{2} - 4}{9 - {x}^{2}}$, highest degree in numerator is for ${x}^{2}$ and for denominator $- {x}^{2}$, hence we have a horizontal asymptote at $y = {x}^{2} / \left(- {x}^{2}\right) = - 1$