How do you find the asymptotes for (x^2 + x + 3 )/( x-1)?

Jun 21, 2016

Vertical asymptote $x = 1$

Oblique asymptote $y = x + 2$

Explanation:

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

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

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

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

As $x \to \pm \infty$ the term $\frac{5}{x - 1} \to 0$

So there is an oblique (slant) asymptote $y = x + 2$

When $x = 1$ the denominator of $f \left(x\right)$ becomes $0$ while the numerator is non-zero. So $f \left(x\right)$ is undefined at $x = 1$ and has a vertical asymptote there.

graph{(y - (x^2+x+3)/(x-1))(y - (x+2)) = 0 [-39, 41, -17.04, 22.96]}