How do you find the asymptotes for y = (3x^2+x-4) / (2x^2-5x) ?

Nov 25, 2016

The vertical asymptotes are $x = 0$ and x=5/2
No hole, No slant asymptote

The horizontal asymptote is $y = \frac{3}{2}$

Explanation:

Let's factorise the denominator

$2 {x}^{2} - 5 x = x \left(2 x - 5\right)$

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{0 , \frac{5}{2}\right\}$

As you cannot divide by $0$, $x \ne 0$ and $x \ne \frac{5}{2}$

Therefore, the vertical asymptotes are $x = 0$ and x=5/2

As the degree of the numerator is $=$ than the degree of the denominator, we have no slant asymptote.

For the limits, we take the terms of highest degree

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} \frac{3 {x}^{2}}{2 {x}^{2}} = \frac{3}{2}$

The horizontal asymptote is $y = \frac{3}{2}$

graph{(y-((3x^2+x-4)/(2x^2-5x)))(y-3/2)=0 [-14.24, 14.24, -7.12, 7.12]}