# How do you find oblique asymptote of (x^2 + x + 3) /(x-1)?

Jun 20, 2015

Partially divide the numerator by the denominator to get:

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

Hence the oblique asymptote is $y = x + 2$

#### Explanation:

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

with exclusion $x \ne 1$

$\frac{5}{x - 1} \to 0$ as $x \to \pm \infty$

So the oblique asymptote is $y = x + 2$

graph{(y - (x^2+x+3) / (x-1))(y - x - 2) = 0 [-40, 40, -20, 20]}