Answer:
[tex]\boxed{y=2x-\dfrac{1}{2}}.[/tex]
Step-by-step explanation:
We may start by noticing that [tex]h[/tex] doesen't have horizontal asymptotes:
[tex]\lim\limits_{x\to\pm\infty} h(x) = \lim\limits_{x\to\pm\infty} \dfrac{4x^2+5x-2}{2x+3} \overset{\frac{\infty}{\infty}}{=} \lim\limits_{x\to\pm\infty} \dfrac{8x+5}{2} = \pm\infty,[/tex]
where we have used L'Hôpital's rule. Let's now check if it has any oblique asymptotes of the form [tex]y = mx+b[/tex], with:
[tex]m = \lim\limits_{x\to\pm\infty}\dfrac{h(x)}{x} \quad\textrm{and}\quad b = \lim\limits_{x\to\pm\infty} (h(x)-mx),[/tex]
provided that the limits exist. Let's compute them using L'Hôpital's rule:
[tex]m = \lim\limits_{x\to\pm\infty} \dfrac{h(x)}{x} = \lim\limits_{x\to\pm\infty} \dfrac{4x^2+5x-2}{x(2x+3)}= \lim\limits_{x\to\pm\infty} \dfrac{4x^2+5x-2}{2x^2+3x}\overset{\frac{\infty}{\infty}}{=}\\\\= \lim\limits_{x\to\pm\infty} \dfrac{8x+5}{4x+3} \overset{\frac{\infty}{\infty}}{=} \lim\limits_{x\to\pm\infty} \dfrac{8}{4} = 2[/tex]
[tex]b=\lim\limits_{x\to\pm\infty} (h(x) - 2x) = \lim\limits_{x\to\pm\infty} \left(\dfrac{4x^2+5x-2}{2x+3} - 2x\right) =\\\\=\lim\limits_{x\to\pm\infty} \dfrac{4x^2+5x-2-4x^2-6x}{2x+3} =\lim\limits_{x\to\pm\infty} \dfrac{-x-2}{2x+3} \overset{\frac{\infty}{\infty}}{=} \lim\limits_{x\to\pm\infty} \dfrac{-1}{2}= -\dfrac{1}{2}.[/tex]
So we conclude that [tex]h[/tex] has an oblique asymptote, for [tex]y\to\pm\infty[/tex], given by the equation:
[tex]\boxed{y=2x-\dfrac{1}{2}}.[/tex]
