# How do you identify all asymptotes or holes for g(x)=(2x^2+3x-1)/(x+1)?

Dec 17, 2016

There is no hole. We have vertical asymptote at $x = - 1$ and oblique asymptote at $y = 2 x + 1$

#### Explanation:

Let us examine $g \left(x\right) = \frac{2 {x}^{2} + 3 x - 1}{x + 1}$.

As for the $x = - 1$, although denominator $x + 1 = 0$ but $2 {x}^{2} + 3 x - 1 \ne 0$, there is no common factor between numerator and denominator and hence we do not have any hole.

It is also apparent that as $x \to - 1$, $\frac{2 {x}^{2} + 3 x - 1}{x + 1} \to \infty$

and as we have $f \left(x\right) = \frac{{x}^{2} + 11 x + 18}{2 x + 1}$, $f \left(x\right) \to \infty$ or $f \left(x\right) \to - \infty$, as we approach $x = - 1$ from negative or positive side.

Hence, we have a vertical asymptote at $x = - \frac{1}{2}$.

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

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

Hence, as $x \to \infty$. $f \left(x\right) \to 2 x + 1$

and we have an oblique asymptote at $y = 2 x + 1$
graph{(2x^2+3x-1)/(x+1) [-10, 10, -5, 5]}