How do you find the asymptotes for (x^2 + 9) / (9 x - 5 x^2)x2+99x−5x2?
1 Answer
vertical asymptotes at x = 0 ,
x =9/5x=95 horizontal asymptote at
y = -1/5y=−15
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 = 09x−5x2=0 'take out common factor' ie x (9 - 5x ) = 0
rArr x = 0 , x= 9/5 color(black)(" are vertical asymptotes")⇒x=0,x=95 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 isy = -1/5 Here is the graph of the function as an illustration.
graph{(x^2+9)/(9x-5x^2) [-10, 10, -5, 5]}