# How do you simplify and find the excluded values of (v^2+4) / (v^2-3v-18)?

May 26, 2017

Asymptote at $v = - 3$ and at $6$.

#### Explanation:

Excluded values mean asymptotes and holes. So let's look for them:

First, let's expand all our components:

$\frac{{v}^{2} + 4}{{v}^{2} - 3 v - 18}$

The numerator almost looks like a differnce of squares, but it's adding instead of subtracting. That means that the expanded version of it will have imaginary numbers ($i$) in it. Let's not expand it if we don't have to.

The denominator is easier. We just need to factor.
${v}^{2} - 3 v - 18$

We are looking for two numbers that add to $- 3$ and multiply to $- 18$. We know that one of the numbers will be negative, because to get $\textcolor{red}{-} 18$, we need to multiply a positive number by a negative number.

To find the factors, let's ignore the signs for now:
$\textcolor{w h i t e}{\times} + 3$
$\textcolor{w h i t e}{+} \times 18$
........................
$1 \times 18$
$2 \times 9$
$\textcolor{red}{3 \times 6}$

$3 + - 6 = - 3$ and $- 6 \times 3 = 18$

Now we have our factors:

$\left(v - 6\right) \left(v + 3\right)$

$\frac{{v}^{2} + 4}{\left(v - 6\right) \left(v + 3\right)}$

Asymptotes are values of $v$ that eqaul a division by $0$. To solve for them, we set each component in the denominator equal to $0$ and solve for $v$:

Case 1

$v - 6 = 0$

$v = 6$

Case 2

$v + 3 = 0$

$v = - 3$

So, when $v = 6$ or $- 3$, the denominator becomes $0$, creating an asymptote.
there are no holes (no identical factors in the numerator and denominator), so the only excluded values are $v = 6 \mathmr{and} - 3$

Just to check our work, let's graph the equation and see
graph{y=(x^2+4)/((x-6)(x+3))}