# How do you graph f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2) using holes, vertical and horizontal asymptotes, x and y intercepts?

##### 1 Answer
Oct 8, 2017

See below for more details about holes, intercepts, asymptotes, and stationary points too!

#### Explanation:

We have $f \left(x\right) = g \frac{x}{h \left(x\right)} = \frac{2 {x}^{3} - {x}^{2} - 2 x + 1}{{x}^{2} + 3 x + 2}$.

Function $h \left(x\right)$ quite easily factorises down to $\left(x + 1\right) \left(x + 2\right)$.

Divide $g \left(x\right)$ by $x + 1$ to factorise the cubic, and it can be found through polynomial division that $g \left(x\right) = \left(x + 1\right) \left(2 {x}^{2} - 3 x + 1\right)$, the latter quadratic of which factorises down to $\left(2 x - 1\right) \left(x - 1\right)$.

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

By further polynomial division, that is $\frac{2 {x}^{2} - 3 x + 1}{x + 2}$, it can be found that $f \left(x\right) = 2 x - 7 + \frac{15}{x + 2}$.

Clearly $x$ cannot equal $- 2$ as it leaves the fraction undefined, so there is an asymptote at $x = - 2$.

Secondly $f \left(x\right)$ cannot equal $2 x - 7$, as this would suggest that $\frac{15}{x + 2} = 0 \Rightarrow 15 = 0$, so there is an oblique asymptote at $y = 2 x - 7$.

To solve for $x$-intercept(s) return to the factorised function, from which it can be seen easily that for $f \left(x\right)$ to equal zero, $\left(2 x - 1\right) \left(x - 1\right) = 0$, therefore $x = \frac{1}{2}$ or $1$ for the $x$-intercepts.

Let $x = 0$ to find $y$-intercepts. Thus $f \left(x\right) = \frac{\left(- 1\right) \left(- 1\right)}{2}$, therefore $y = \frac{1}{2}$.

Lastly, the 'holes'. Two $x + 1$ expressions were cancelled early on, hence there is a hole at $x = - 1$. So, a discontinuity is found at $\left(- 1 , f \left(- 1\right)\right) \equiv \left(- 1 , 6\right)$.

To find stationary points, calculate $f ' \left(x\right)$ and let it equal zero:
$f ' \left(x\right) = x - \frac{15}{{\left(x + 2\right)}^{2}} = 0$.
$\therefore x {\left(x + 2\right)}^{2} = 15$
$\therefore {x}^{3} + 4 {x}^{2} + 4 x - 15 = 0$
So you end up with a stationary point at approximately $\left(1.34 , f \left(1.34\right)\right)$.

Hope this helps!