# How do you find the vertical, horizontal or slant asymptotes for y=(3x^2+x-4) / (2x^2-5x) ?

Nov 29, 2016

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

#### Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : $2 {x}^{2} - 5 x = 0 \Rightarrow x \left(2 x - 5\right) = 0$

$\Rightarrow x = 0 \text{ and " x=5/2" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , y \to c \text{ ( a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

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

as $x \to \pm \infty , y \to \frac{3 + 0 - 0}{2 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no slant asymptotes.
graph{(3x^2+x-4)/(2x^2-5x) [-10, 10, -5, 5]}