# Math Help - Asymptote

1. ## Asymptote

Determine function $y=\frac{2x^3+6x-1}{4-x^2}$ graphic's asymptotes.

2. Originally Posted by Bernice
Determine function $y=\frac{2x^3+6x-1}{4-x^2}$ graphic's asymptotes.

The verical asymptotes occuer at the zero's of the denominator

$4-x^2=0 \iff (2-x)(2+x)=0$

so we get $x= \pm 2$

It does not have any horizontal asymptotes because the degree of the numerator is larger than the denominator.

It does have a slant asymptote( the degree of the numerator is one larger than the denominator) This can be calculated by dividing the denominator into the numerator. (long division)

Or by this trick

$\frac{2x^3+6x-1}{4-x^2}=\frac{2x^3-8x+14x-1}{4-x^2}=\frac{-2x(4-x^2)+14x-1}{4-x^2}=$

$\frac{-2x(4-x^2)}{4-x^2}+\frac{14x-1}{4-x^2}=-2x+\frac{14x-1}{4-x^2}$

So the slant asymptotes is $y=-2x$

3. Hello,

You has just to make a division :

$2x^3+6x-1=-2x(4-x^2)+14x-1$

Hence $\color{blue}\boxed{y+2x=\frac{14x-1}{4-x^2}\underset{x\to \pm \infty}{\longrightarrow} 0}$

So $\color{red} x\mapsto -2x$ is the asymptote.

Originally Posted by Bernice
Determine function $y=\frac{2x^3+6x-1}{4-x^2}$ graphic's asymptotes.
For vertical Asymptote, Denominator = 0

$4 - x^2 =0$

$\Rightarrow x=-2,2$

Vertical asymptote (VA) are x = -2, and x = 2

The degree of Numerator is one more than Denominator, so, it will also have linear-oblique asymptote.

Divide Numerator with Denominator, using long division.

long divide polynomials

Since $y = - 2x$ is answer, so the linear-oblique asymptote (LOA) is $y=-2x$

See graph attached. The green lines show asymptote.

5. Thank you all for explanations. They are very useful!

6. TheEmptySet > Is it possible to calculate the integral of your signature without the residue theorem ?