How do you simplify and find the excluded values of v2+4v23v18?

1 Answer
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:

v2+4v23v18

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.
v23v18

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 18, we need to multiply a positive number by a negative number.

To find the factors, let's ignore the signs for now:
×+3
+×18
........................
1×18
2×9
3×6

3+6=3 and 6×3=18

Now we have our factors:

(v6)(v+3)

v2+4(v6)(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

v6=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=6and3

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