Question

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

Answer 1

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:

`(v^2+4)/(v^2-3v-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^2-3v-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 `color(red)(-)18`, we need to multiply a positive number by a negative number.

To find the factors, let's ignore the signs for now:
`color(white)(xx)+ 3`
`color(white)(+)xx 18`
........................
`1 xx 18`
`2 xx 9`
`color(red)(3 xx 6)`

`3+ -6=-3` and `-6 xx 3=18`

Now we have our factors:

`(v-6)(v+3)`

`(v^2+4)/((v-6)(v+3))`

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 and -3`

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