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

1 Answer
Aug 21, 2016

Answer:

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

Explanation:

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: #2x^2+3x-2=0rArr(2x-1)(x+2)=0#

#rArrx=-2" and " x=1/2" 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^2-1/x^2)/((2x^2)/x^2+(3x)/x^2-2/x^2)=(1-1/x^2)/(2+3/x-2/x^2)#

as #xto+-oo,f(x)to(1-0)/(2+0-0)#

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