# How do you find the asymptotes for (x^2 + 9) / (9 x - 5 x^2)?

Feb 4, 2016

vertical asymptotes at x = 0 , $x = \frac{9}{5}$

horizontal asymptote at $y = - \frac{1}{5}$

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero.
To find the equation of an asymptote let the denominator equal zero.

solve: $9 x - 5 {x}^{2} = 0$

'take out common factor' ie x (9 - 5x ) = 0

$\Rightarrow x = 0 , x = \frac{9}{5} \textcolor{b l a c k}{\text{ are vertical asymptotes}}$

Horizontal asymptotes occur as  lim_(x→±∞) f(x) → 0

If the degree of the numerator and denominator are equal then the equation can be found by taking the ratio of leading coefficients.

for this function they are equal , both of degree 2

rewriting as : $\frac{{x}^{2} + 9}{- 5 {x}^{2} + 9 x}$

then $y = \frac{1}{- 5} = - \frac{1}{5}$
horizontal asymptote is $y = - \frac{1}{5}$

Here is the graph of the function as an illustration.
graph{(x^2+9)/(9x-5x^2) [-10, 10, -5, 5]}