How do you find the asymptotes for #(x^2 + 9) / (9 x - 5 x^2)#?
1 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]}