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

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 Click here to see ALL problems on Functions 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(57328)   (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)] ---- 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. ========================================== =========== Cheers, Stan H. Answer by richard1234(5390)   (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.