How do you find the vertical, horizontal and slant asymptotes of: #g(x)=(3x^2+2x-1)/(x^2-4)#?

1 Answer
Jul 21, 2016

Answer:

vertical asymptotes x = ± 2
horizontal asymptote y = 3

Explanation:

The denominator of g(x) cannot be zero as this is 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^2-4=0rArr(x-2)(x+2)=0rArrx=±2#

#rArrx=-2,x=2" are the asymptotes"#

Horizontal asymptotes occur as

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

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

#((3x^2)/x^2+(2x)/x^2-1/x^2)/((x^2)/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)#

as #xto+-oo,g(x)to(3+0-0)/(1-0)#

#rArry=3" 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{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}