SOLUTION: How do I find out what the Vertical Asymptote is for this function: 2x^2-6/x^3-8?

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Question 429773: How do I find out what the Vertical Asymptote is for this function:
2x^2-6/x^3-8?

Found 2 solutions by stanbon, richard1234:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find out what the Vertical Asymptote is for this function:
2x^2-6/x^3-8?
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Factor:
[2(x^2-6)]/[(x-2)(x^2+2x+4)]
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You have a vertical asymptote when the denominator
of this problem is zero.
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Solve x^2+2x+4 = 0
x = [-4+-sqrt(4-4*4)]/2
Solution is imaginary since the discriminant is negative.
---
Vertical asymptote at x = 2 only.
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graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C%282x%5E2-6%29%2F%28x%5E3-8%29%29
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Cheers,
Stan H.

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
A vertical asymptote occurs when the denominator is zero and the numerator is nonzero. Here, we want x%5E3+-+8+=+0 --> x%5E3+=+8. The only real solution is x=2. The numerator is equal to 2%282%5E2%29+-+6+=+2+%3C%3E+0 so we can say that x=2 produces a vertical asymptote.