How do you graph #y=(3x^3+1)/(4x^2-32)# using asymptotes, intercepts, end behavior?

2 Answers
Nov 23, 2016

Answer:

The vertical asymptotes are #x=sqrt8# and #x=-sqrt8#
The slant asymptote is #y=3/4x#
No horozontal asymptote.

Explanation:

Let's factorise the denominator #(4x^2-32)#

#=4(x^2-8)=4(x+sqrt8)(x-sqrt8)#

The domain of #y# is #D_y=RR- {sqrt8,-sqrt8} #

As we cannot divide by #0#,

So #x!=sqrt8# and #x!=-sqrt8#

The vertical asymptotes are #x=sqrt8# and #x=-sqrt8#

As the degree of the numerator is #># the degree of the denominator, we expect a slant asymptote.

Let's do a long division

#color(white)(aaaa)##3x^3+1##color(white)(aaaa)##∣##4x^2-32#

#color(white)(aaaa)##3x^3-24x##color(white)(aa)##∣##(3x)/4#

#color(white)(aaaaa)##0-24x+1#

So, #(3x^3+1)/(4x^2-32)=(3x)/4-(24x+1)/(4x^2+32)#

The slant asymptote is #y=3/4x#

To calculate the limits, we use the terms of highest degree.

#lim_(x->+oo)y=lim_(x->+oo)(3x^3)/(4x^2)=lim_(x->+oo)(3x)/4=+oo#

#lim_(x->-oo)y=lim_(x->-oo)(3x^3)/(4x^2)=lim_(x->-oo)(3x)/4=-oo#

There are no horizontal asymptote

#lim_(x->-sqrt8^(-))=-oo#

#lim_(x->-sqrt8^(+))=+oo#

#lim_(x->sqrt8^(-))=-oo#

#lim_(x->sqrt8^(+))=+oo#

When #x=0#, #=>#, #y=-1/32#

When #y=0#, #0>#, #x=(-1/3)^(1/3)#

graph{(y-(3x^3+1)/(4x^2-32))(y-x3/4)=0 [-28.86, 28.9, -14.43, 14.43]}

Nov 23, 2016

Answer:

Asymptotes: slant #y=3/4x# and vertical #x=+-sqrt8#
y-intercept #= -1/32# and x-intercept =#-1/3^(1/3)=-6934#, nearly

Explanation:

Resolving into partial fractions,

#y = 3/4x+(6x+1/4)/(x^2-8)#

Rearranging.

#y-3/4x = (6x+1/4)/((x-sqrt8)(x+sqrt8)#

The form reveals that the asymptotes are given by

the slant #y = 3/4x# and the vertical ones# x=+-sqrt8#.

Easily from the given equation, the intercepts can be obtained, as

given in the answer.

Also, as #x to +-oo, y to +-oo, observing that the second fraction

tends to 0.