How do you find the asymptotes for #y = (3x^2+x-4) / (2x^2-5x) #?

1 Answer
Nov 25, 2016

Answer:

The vertical asymptotes are #x=0# and x=5/2#
No hole, No slant asymptote

The horizontal asymptote is #y=3/2#

Explanation:

Let's factorise the denominator

#2x^2-5x=x(2x-5)#

The domain of #y# is #D_y=RR-{0,5/2} #

As you cannot divide by #0#, #x!=0# and #x!=5/2#

Therefore, the vertical asymptotes are #x=0# and x=5/2#

As the degree of the numerator is #=# than the degree of the denominator, we have no slant asymptote.

For the limits, we take the terms of highest degree

#lim_(x->+-oo)y=lim_(x->+-oo)(3x^2)/(2x^2)=3/2#

The horizontal asymptote is #y=3/2#

graph{(y-((3x^2+x-4)/(2x^2-5x)))(y-3/2)=0 [-14.24, 14.24, -7.12, 7.12]}