# How do you find the asymptotes for  f(x)=(x^2+x-2)/(x^2-x+2)?

Feb 14, 2016

Re-express as $f \left(x\right) = 1 + \frac{2 x - 4}{{x}^{2} - x + 2}$ to find horizontal asymptote $y = 1$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + x - 2}{{x}^{2} - x + 2}$

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

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

Now ${x}^{2} - x + 2$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 1$ and $c = 2$, so has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - \left(4 \cdot 1 \cdot 2\right) = 1 - 8 = - 7$

So this quadratic has no Real zeros.

The numerator $\left(2 x - 4\right)$ is of lower degree than the denominator, so $\frac{2 x - 4}{{x}^{2} - x + 2} \to 0$ as $x \to \pm \infty$

Therefore $f \left(x\right) \to 1 + 0 = 1$ as $x \to \pm \infty$

So the only asymptote is a horizontal one: $y = 1$

graph{(x^2+x-2)/(x^2-x+2) [-10, 10, -5, 5]}