How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x^3+4)/(2x^2+x-1)#?

1 Answer
Feb 20, 2017

Answer:

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

Explanation:

Let start by factorising the denominator

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

Let #f(x)=(x^3+4)/(2x^2+x-1)=(x^3+4)/((2x-1)(x+1))#

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

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

As the degree of the numerator is #># than the degree of the denominator, we have an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##x^3##color(white)(aaaaaaaaaaaa)##+4##color(white)(aaaa)##|##2x^2+x-1#

#color(white)(aaaa)##x^3+x^2/2-x/2##color(white)(aaaa)####color(white)(aaaaa)##|##1/2x-1/4#

#color(white)(aaaaa)##0-x^2/2+x/2##color(white)(aaaa)##+4#

#color(white)(aaaaaaa)##-x^2/2-x/4##color(white)(a)##+1/4#

#color(white)(aaaaaaa)##-0+3/4x##color(white)(aaaaa)##+15/4#

Therefore,

#f(x)=x/2-1/4+(3/4x+15/4)/(2x^2+x-1)#

#lim_(x->-oo)f(x)-(x/2-1/4)=lim_(x->-oo)(3/4x)/(2x^2)=0^-#

#lim_(x->+oo)f(x)-(x/2-1/4)=lim_(x->+oo)(3/4x)/(2x^2)=0^+#

The oblique asymptote is #y=x/2-1/4#

graph{(y-(x^3+4)/(2x^2+x-1))(y-x/2+1/4)=0 [-18.02, 18.03, -9, 9.02]}