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

Question

Lily Anderson · Accepted Answer

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

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]}

Caleb Alexander · Answer

undefined To find the vertical asymptotes, factor the denominator and identify any values of $ x $ that make the denominator equal to zero. These values represent vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

For oblique asymptotes, perform polynomial long division. If the division results in a non-zero remainder, there is an oblique asymptote given by the quotient. If the remainder is zero, there is no oblique asymptote.