How do you graph y=(3x^3+1)/(4x^2-32)y=3x3+14x232 using asymptotes, intercepts, end behavior?

2 Answers
Nov 23, 2016

The vertical asymptotes are x=sqrt8x=8 and x=-sqrt8x=8
The slant asymptote is y=3/4xy=34x
No horozontal asymptote.

Explanation:

Let's factorise the denominator (4x^2-32)(4x232)

=4(x^2-8)=4(x+sqrt8)(x-sqrt8)=4(x28)=4(x+8)(x8)

The domain of yy 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+1color(white)(aaaa)4x^2-32

color(white)(aaaa)3x^3-24xcolor(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

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.