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

Oct 29, 2017

By considering asymptotes, yields;
graph{y = (x^2-12x)/(x^2-2x-3) [-19.82, 19.83, -9.9, 9.92]}

#### Explanation:

First we can factorise to give $f \left(x\right)$ = $\frac{x \left(x - 12\right)}{\left(x - 3\right) \left(x + 1\right)}$

Now we can consider verticle asymptotes, hence at $x = 3 \mathmr{and} x = - 1$, Where the function is undefined, when the denominator $= 0$

Now we can cosnider Horizontal asymptotes:

${\lim}_{x \to + / - \infty} f \left(x\right)$ = ${\lim}_{x \to + / - \infty} \frac{1 - \frac{12}{x}}{1 - \frac{2}{x} - \frac{3}{{x}^{2}}}$

Hence ${\lim}_{x \to + / - \infty} f \left(x\right)$ = 1
as $\frac{a}{\infty} \to 0$

Hence Horizontal asymptote ; $y = 1$

We now can consider roots; $f \left(x\right) = 0$

Hence $f \left(x\right) = 0 \to x \left(x - 12\right) = 0$

Hence has roots, $x = 0 \mathmr{and} x = 12$

Now can cosnider the nature of function as they approach the verticle asymptotes from possitive and negative direction:

We can compute the follwoing limits, via letting $x$ be close to the limit value $x$, e.g. lim_(x->3^-) f(x) ~ f(2.9999)  to give us a general idea to weather it tends to pos or neg $\infty$

${\lim}_{x \to {3}^{-}} f \left(x\right) = \infty$
${\lim}_{x \to {3}^{+}} f \left(x\right) = - \infty$
${\lim}_{x \to - {1}^{-}} f \left(x\right) = \infty$
${\lim}_{x \to - {1}^{+}} f \left(x\right) = - \infty$

We this is efficiant information to be able to sketch this function,

So we note that as $x \to {3}^{+} : y \to - \infty$, one of the roots is $x = 12$, and how as $x \to \infty : y \to 1$ yielding, graph{(x^2-12x)/(x^2-2x-3) [2.76, 23.975, -9.42, 1.19]}

We can repeat this for the other side of the asymptote, noting $f \left(0\right) = 0$

Hence;

graph{y = (x^2-12x)/(x^2-2x-3) [-19.82, 19.83, -9.9, 9.92]}