# How do you graph f(x)=(2x^2+10x+12)/(x^2+3x+2) using holes, vertical and horizontal asymptotes, x and y intercepts?

Sep 10, 2017

Vertical asymptote at $x = - 1$, horizontal asymptote at $y = 2$, hole at $x = - 2$, $y$-intercept at $\left(0 , 6\right)$ and $x$-intercept at $x = - 3$.

#### Explanation:

When $x = 0$ $f \left(x\right) = \frac{2 {x}^{2} + 10 x + 12}{{x}^{2} + 3 x + 2} = 6$, so we have a $y$-intercept at $\left(0 , 6\right)$.

As $f \left(x\right) = \frac{2 {x}^{2} + 10 x + 12}{{x}^{2} + 3 x + 2}$

= $\frac{2 \left({x}^{2} + 5 x + 6\right)}{{x}^{2} + 3 x + 2}$

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

= $2 \frac{x + 3}{x + 1}$

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

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

And when $y = 0$, $2 + \frac{4}{x + 1} = 0$ or $\frac{4}{x + 1} = - 2$ leading to $x = - 3$ and we have a $x$-intercept at $\left(- 3 , 0\right)$.

Observe that although $x + 2$ cancels out $\frac{2 {x}^{2} + 10 x + 12}{{x}^{2} + 3 x + 2}$ is not defined for $x + 2 = 0$ i.e. $x = - 2$. Hence we have a hole at $x = - 2$.

Also for $x \to - 1$, from left $f \left(x\right) \to - \infty$ and from right $f \left(x\right) \to \infty$, so we have a vertical asymptote at $x = - 1$

Further, as $x \to \infty$ or $x \to - \infty$, $f \left(x\right) \to 2$ and hence we have a horizontal asymptote at $y = 2$.

The graph using these points and trends appears as follows:

graph{(2x^2+10x+12)/(x^2+3x+2) [-9.79, 10.21, -4.24, 5.76]}