Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `x /( 4x^2+7x-2)`?

Answer 1

vertical asymptotes at x = -2 , x `=1/4`
horizontal asymptote at y = 0

Explanation:

The denominator of the function cannot be zero as this would make the function 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: `4x^2+7x-2=0rArr(4x-1)(x+2)=0`

`rArrx=-2" and " x=1/4" 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/x^2)/((4x^2)/x^2+(7x)/x^2-2/x^2)=(1/x)/(4+7/x-2/x^2)`

as `xto+-oo,f(x)to0/(4+0-0)`

`rArry=0" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 ,denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{x/(4x^2+7x-2) [-10, 10, -5, 5]}