SOLUTION: How do I find out what the Vertical Asymptote is for this function:
2x^2-6/x^3-8?
Algebra.Com
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) (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?
------------
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.
---
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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Cheers,
Stan H.
Answer by richard1234(7193) (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 --> . The only real solution is x=2. The numerator is equal to so we can say that x=2 produces a vertical asymptote.
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