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

1 Answer
Feb 4, 2016

Answer:

vertical asymptotes at x = 0 , # x =9/5#

horizontal asymptote at # y = -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: # 9x - 5x^2 = 0#

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

# rArr x = 0 , x= 9/5 color(black)(" 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 : # (x^2+9)/(-5x^2+9x)#

then # y = 1/(-5) = -1/5 #
horizontal asymptote is # y = -1/5 #

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