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

1 Answer
Sep 20, 2016

Answer:

vertical asymptotes at x = - 1 , x = 6
horizontal asymptote at y = 2

Explanation:

The denominator of g(x) cannot be zero as this would make g(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: #x^2-5x-6=0rArr(x-6)(x+1)=0#

#rArrx=-1" and " x=6" are the asymptotes"#

Horizontal asymptotes occur as

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

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

#g(x)=((2x^2)/(x^2))/(x^2/x^2-(5x)/x^2-6/x^2)=2/(1-5/x-6/x^2)#

as #xto+-oo,g(x)to2/(1-0-0)#

#rArry=2" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no slant asymptotes.
graph{(2x^2)/(x^2-5x-6) [-20, 20, -10, 10]}