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

Feb 1, 2016

Horizontal asymptote:
$y = \frac{3}{2}$

Vertical asymptote:
$x = 0$
$x = \frac{5}{2}$

Explanation:

Here is a graph of the expression in question.
graph{(3x^2 + x - 4)/(2x^2 - 5x) [-10, 10, -5, 5]}
You can try to simplify the expression first.

$\frac{3 {x}^{2} + x - 4}{2 {x}^{2} - 5 x} = \frac{3}{2} + \frac{17 x - 8}{4 {x}^{2} - 10 x}$

When $x$ is very large (or small), observe that the second term tends to zero. Therefore,

${\lim}_{x \to - \infty} \frac{3 {x}^{2} + x - 4}{2 {x}^{2} - 5 x} = \frac{3}{2}$ and

${\lim}_{x \to \infty} \frac{3 {x}^{2} + x - 4}{2 {x}^{2} - 5 x} = \frac{3}{2}$.

Hence, there is a horizontal asymptote with equation $y = \frac{3}{2}$.

Also note that the expression is not defined for $x = \frac{5}{2}$ and $x = 0$, as they result in division by zero.

Hence, there are 2 vertical asymptotes, with equation $x = \frac{5}{2}$ and $x = 0$.