How do you find the asymptotes for #R (x) = (7x) /( 4x^2 -1)#?

1 Answer
Dec 9, 2016

Answer:

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

Explanation:

we use #a^2-b^2=(a+b)(a-b)#

Let's factorise the denominator

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

The domain of #R(x)# is #D_R(x)=RR-{-1/2,1/2}#

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

The vertical asymptotes are #x=-1/2# and #x=1/2#

The degree of the numerator is #<# than the degree of the denominator, there is no slant asymptotes.

To calculate the limits as #x->+-oo#, we take the terms of highets degree in the denominator

#lim_(x->-oo)R(x)=lim_(x->-oo)(7x)/(4x^2)=lim_(x->-oo)7/(4x)=0^(-)#

#lim_(x->+oo)R(x)=lim_(x->+oo)(7x)/(4x^2)=lim_(x->+oo)7/(4x)=0^(+)#

The horizontal asymptote is #y=0#

graph{7x/(4x^2-1) [-6.24, 6.244, -3.12, 3.12]}