Question

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

Answer 1

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]}

Answer 2

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.