Question

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

Answer 1

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

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero, To find the equation/s set the denominator equal to zero.

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

`rArrx=-1/2,x=2" are the asymptotes"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by `x^2`

`(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 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]}