How do you find the vertical, horizontal or slant asymptotes for #(x )/(4x^2+7x-2) #?

1 Answer
Nov 29, 2016

Answer:

The vertical asymptotes are #x=-2# and #x=1/4#
No slant asymptote
The horizontal asymptote is #y=0#

Explanation:

Let's factorise the denominator

#4x^2+7x-2=(4x-1)(x+2)#

Let #f(x)=x/(4x^2+7x-2)=x/((4x-1)(x+2))#

The domain of #f(x)#is #D_f(x)=RR-{-2,1/4} #

As we cannot divide by #0#, #x!=-2# and #x!=1/4#

So, the vertical asymptotes are #x=-2# and #x=1/4#

As the degree of the numerator is #<# than the degree of the denominator, there is no slant asymptote.

To calculate the limits as x tends to #oo#, we take the terms of highest degree in the numerator and the denominator.

#lim_(x->+oo)f(x)=lim_(x->+oo)x/(4x^2)=lim_(x->+oo)1/(4x)=0^(+)#

#lim_(x->-oo)f(x)=lim_(x->-oo)x/(4x^2)=lim_(x->-oo)1/(4x)=0^(-)#

So, the horizontal asymptote is, #y=0#
graph{x/(4x^2+7x-2) [-6.244, 6.243, -3.12, 3.123]}