# How do you find vertical, horizontal and oblique asymptotes for (3x^2+x-4) / (2x^2-5x)?

##### 1 Answer
Apr 4, 2016

vertical asymptotes x = 0 , $x = \frac{5}{2}$
horizontal asymptote $y = \frac{3}{2}$

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve :  2x^2 - 5x = 0 → x(2x-5) = 0

$\Rightarrow x = 0 , x = \frac{5}{2} \text{ are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide all terms on numerator/denominator by ${x}^{2}$

$\frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{x}{x} ^ 2 - \frac{4}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 - \frac{5 x}{x} ^ 2} = \frac{3 + \frac{1}{x} - \frac{4}{x} ^ 2}{2 - \frac{5}{x}}$

As $x \to \pm \infty , \frac{1}{x} , \frac{4}{x} ^ 2 \text{ and } \frac{5}{x} \to 0$

$\Rightarrow y = \frac{3}{2} \text{ is the asymptote }$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(3x^2+x-4)/(2x^2-5x) [-10, 10, -5, 5]}