# How do you graph f(x)=(2x^3-2x^2)/(x^3-9x) using holes, vertical and horizontal asymptotes, x and y intercepts?

Dec 30, 2017

See explanation

#### Explanation:

f(x)=(2x^3−2x^2)/(x^3−9x)=(2x^2(x-1))/(x(x^2-9))=(2x(x-1))/((x-3)(x+3)

Domain: $\mathbb{R} - \left\{0 , \pm 3\right\}$

Vertical asymptote: $x = 3$ and $x = - 3$

Horizontal asymptote: $y = 2$

y=Lim_(xrarr+-oo)f(x)=Lim_(xrarr+-oo)(2x^3−2x^2)/(x^3−9x)=2/1=2

Intercepts:
f(x) doesn't intercept y axis because a point $x = 0$ in not in the domain(We can't divide by 0)

$0 = 2 x \left(x - 1\right)$

$x = 0$ makes no sense (it will just seem like it goes through [0,0])
$x = 1 \quad \quad \quad {X}_{2} \left[1 , 0\right]$

I personally wouldn't dare to end it here. Let's find how f(x) behaves in undefined points.

$L i {m}_{x \rightarrow - {3}^{-}} \frac{2 {x}^{2} - 2 x}{{x}^{2} - 9} = \left(\text{plus")/("plus}\right) = + \infty$

$L i {m}_{x \rightarrow - {3}^{+}} \frac{2 {x}^{2} - 2 x}{{x}^{2} - 9} = \left(\text{plus")/("minus}\right) = - \infty$

$L i {m}_{x \rightarrow {3}^{-}} \frac{2 {x}^{2} - 2 x}{{x}^{2} - 9} = \left(\text{plus")/("minus}\right) = - \infty$

$L i {m}_{x \rightarrow {3}^{+}} \frac{2 {x}^{2} - 2 x}{{x}^{2} - 9} = \left(\text{plus")/("plus}\right) = + \infty$

Notice how function is getting closer in $+ \infty$ to value 2. It is getting closer from below where as in $- \infty$ is getting closer from above. Therefore it is better to always find that too.

We said that horizontal asymptote is $y = 2$. and $L i {m}_{x \rightarrow {3}^{+}} f \left(x\right) = + \infty$ that means: If the fuction approaches y=2 from below it must intercept a line: y=2 in the first place. If we can evaluate a point where function intercepts y=2 that means it is getting closer from below:

$2 = f \left(x\right) = \frac{2 {x}^{2} - 2 x}{{x}^{2} - 9}$

$\cancel{2 {x}^{2}} - 18 = \cancel{2 {x}^{2}} - 2 x$

$x = 9$

Note that $x = 9 > x = 3$ (an asymptote) That means we have only one point crossing y=2 which is on right (in $+ \infty$).

(Note: there are some functions which are getting closer from below in $+ \infty$ and also in $- \infty$. If that was the case we would have quadratic funtion)

don't forget undefined point [0,0]