How do you find the vertical, horizontal or slant asymptotes for #(x^2+4)/(6x-5x^2)#?

1 Answer
Ally A.
Nov 30, 2017

Horizontal asymptote of #-1/5#
No slant asymptotes
Vertical asymptotes are #x=1.2# and #x=0#

enter image source here

Explanation:

HA of #-1/5# because the degrees are the same in the denominator and numerator, so you divide the coefficients.

No slant asymptotes because the numerator degree must be one higher than the denominator, and it is not.

Vertical asymptotes are #x=1# and #x=5# because #6x-5x^2# factored is #(x-1.2)# and #(x)#

Related questions
See all questions in Asymptotes
Impact of this question
1587 views around the world
You can reuse this answer
Creative Commons License