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

1 Answer
Sep 15, 2016

Answer:

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]}