How do you find the Vertical, Horizontal, and Oblique Asymptote given #g(x)=(3x^2+2x-1 )/( x^2-4)#?

1 Answer
Jul 10, 2016

Answer:

vertical asymptotes x = ± 2
horizontal asymptote y = 3

Explanation:

The denominator of g(x) cannot be zero. This would give division by zero which is undefined. Setting the denominator equal to zero and solving for x 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 the highest exponent 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"#

Oblique 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 oblique asymptotes.
graph{(3x^2+2x-1)/(x^2-4) [-20, 20, -10, 10]}