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

Mar 12, 2016

Vertical asymptote at $x = 4$ and slanting asymptote at $y = 2 x$

#### Explanation:

In $y = \frac{2 {x}^{3} - 5 x + 3}{{x}^{2} - 5 x + 4}$, the factors of denominator are $\left(x - 4\right) \left(x - 1\right)$ and numerator is also divisible by $x - 1$, as such

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

Hence, as denominator will be zero for $x = 4$, we will have vertical asymptote at $x = 4$.

Further as degree of numerator in $\frac{2 {x}^{2} + 2 x - 3}{x - 4}$ is greater than that of denominator and ratio of highest degrees by one in the two is $2 {x}^{2} / x = 2 x$, we will have a slanting asymptote at $y = 2 x$

graph{(2x^3-5x+3)/(x^2-5x+4) [-15, 15, -5, 5]}