$y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes?

Question

$y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes?

My attempt is as follows:-

Horizontal asymptotes are the horizontal lines which signify the values of $y$ which graph cannot ever attain.

There are two ways to find horizontal asymptotes for the rational function:

$1)$ If degree of numerator is equal to degree of denominator, then $y=\dfrac{\text {leading coefficient of numerator }}{\text{leading coefficient of denominator}}$

So here we have degree of denominator and numerator as equal, so $y=-\dfrac{1}{2}$ will be the horizontal asymptote.

It means graph cannot ever touch $y=-\dfrac{1}{2}$

$2)$ Another way is by finding range, elements which are not in the range will correspond to horizontal asymptotes.

Let's find out the range,

$$-4x^2y+3xy+2y=2x^2+3x-4$$ $$x^2(2+4y)+3x(1-y)-4-2y=0$$

$$D\ge0$$ $$9(1+y^2-2y)+4(4+2y)(2+4y)\ge0$$ $$9+9y^2-18y+4(8+20y+8y^2)\ge0$$ $$41y^2+62y+41\ge0$$

This is always greater than equal to zero because $D=62^2-4\cdot1681<0$

So this indicates that the range is $\left(-\infty,\infty\right)$. So it means there should be no horizontal asymptotes, but by the first way we got $y=-\dfrac{1}{2}$ as horizontal aysmptote.

What am I missing here?

Answer 1

I believe the question is a bit ill-stated, you must be wanting to find a horizontal asymptote on a specific branch of the above polynomial. Plotting the above function in Desmos, we obtain this picture.

This clearly shows that if you took into picture the different branches, the range of the above would in fact be $(- \infty, \infty)$. However, if you focus on a single branch, the asymptote would infact be $\frac{-1}{2}$ as obtained.

Answer 2

Another way of finding the horizontal asymptotes is by multiplying by $\frac{\frac1{x^2}}{\frac1{x^2}}$

$$ y = \dfrac{2x^2+3x-4}{-4x^2+3x+2} \sim \dfrac{2x^2+3x-4}{-4x^2+3x+2}\frac{\frac1{x^2}}{\frac1{x^2}} = \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}} $$

and find the limit as $x\to\infty$

$$ \lim_{x\to\infty} y \sim \lim_{x\to\infty} \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}} = \dfrac{2 + 0 + 0}{-4 + 0 + 0} = -\frac24 = \color{green}{\boxed{-\frac12}} $$