# How do you identify all asymptotes or holes for f(x)=(x^2+11x+18)/(2x+1)?

Dec 16, 2016

There is no hole. We have vertical asymptote at $x = - \frac{1}{2}$ and oblique asymptote at $2 x - 4 y + 21 = 0$

#### Explanation:

Let us examine $f \left(x\right) = \frac{{x}^{2} + 11 x + 18}{2 x + 1}$.

As ${x}^{2} + 11 x + 18 = {x}^{2} + 9 x + 2 x + 18$

= $x \left(x + 9\right) + 2 \left(x + 9\right) = \left(x + 2\right) \left(x + 9\right)$

As a common factor cannot be crossed out from both the numerator and the denominators, we do not have a hole in $f \left(x\right)$.

It is also apparent that as $x \to - \frac{1}{2}$, $\left(2 x + 1\right) \to 0$

and as we have $f \left(x\right) = \frac{{x}^{2} + 11 x + 18}{2 x + 1}$, $f \left(x\right) \to \infty$

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

Further, $f \left(x\right) = \frac{{x}^{2} + 11 x + 18}{2 x + 1}$

= $\frac{\frac{x}{2} \left(2 x + 1\right) + \frac{21}{4} \left(2 x + 1\right) + \frac{51}{4}}{2 x + 1}$

= x/2+21/4+51/(4(2x+1)

Hence, as $x \to \infty$. $f \left(x\right) \to \frac{x}{2} + \frac{21}{4}$

and we have an oblique asymptote at $y = \frac{x}{2} + \frac{21}{4}$ or $2 x - 4 y + 21 = 0$
graph{(y-(x^2+11x+18)/(2x+1))(x+1/2)(2x-4y+21)=0 [-21.08, 18.92, -5.32, 14.68]}