How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x^2+4)/(6x-5x^2)#?

1 Answer
Apr 27, 2017

Answer:

#"vertical asymptotes at " x=0" and " x=6/5#

#"horizontal asymptote at " y=-1/5#

Explanation:

#"let " f(x)=(x^2+4)/(6x-5x^2)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes

#"solve " 6x-5x^2=0rArrx(6-5x)=0#

#rArrx=0" and " x=6/5" are the asymptotes"#

Horizontal asymptotes occur as.

#lim_(xto+-oo),f(x)toc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(x^2/x^2+4/x^2)/((6x)/x^2-(5x^2)/x^2)=(1+4/x^2)/(6/x-5)#

as #xto+-oo,f(x)to(1+0)/(0-5)#

#rArry=-1/5" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 2 ) Hence there are no oblique asymptotes.
graph{(x^2+4)/(6x-5x^2) [-10, 10, -5, 5]}