Determine function $\displaystyle y=\frac{2x^3+6x-1}{4-x^2}$ graphic's asymptotes.
The verical asymptotes occuer at the zero's of the denominator
$\displaystyle 4-x^2=0 \iff (2-x)(2+x)=0$
so we get $\displaystyle x= \pm 2$
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
$\displaystyle \frac{2x^3+6x-1}{4-x^2}=\frac{2x^3-8x+14x-1}{4-x^2}=\frac{-2x(4-x^2)+14x-1}{4-x^2}=$
$\displaystyle \frac{-2x(4-x^2)}{4-x^2}+\frac{14x-1}{4-x^2}=-2x+\frac{14x-1}{4-x^2}$
So the slant asymptotes is $\displaystyle y=-2x$
Hello,
You has just to make a division :
$\displaystyle 2x^3+6x-1=-2x(4-x^2)+14x-1$
Hence $\displaystyle \color{blue}\boxed{y+2x=\frac{14x-1}{4-x^2}\underset{x\to \pm \infty}{\longrightarrow} 0}$
So $\displaystyle \color{red} x\mapsto -2x$ is the asymptote.
For vertical Asymptote, Denominator = 0
$\displaystyle 4 - x^2 =0$
$\displaystyle \Rightarrow x=-2,2$
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 $\displaystyle y = - 2x$ is answer, so the linear-oblique asymptote (LOA) is $\displaystyle y=-2x$
See graph attached. The green lines show asymptote.