Expert: Sherry Wallin - 1/10/2009

Question
i am stuck on finding the vertical asymptote of 3/(27x^3-8). I understand that it factors to (3x-2)(9x^2+6x+4), Therefore one vertical asymptote is 2/3. However, for the (9x^2+6x+4), would there be a vertical asymptote. It does not factor, but i could use quadratic but i do not believe that would be correct. so would there be a vertical asymptote for (9x^2+6x+4)

Answer
Daniel~
    We refer to the product of the sum or difference of two cubes as having the quadratic part 'prime' because it does not factor over the real numbers. You would find out if you used the quadratic formula that your 'roots' or 'solutions' are [-1+-i*sqrt3]/3 which are complex and not real. When looking for asymptotes you are looking for those 'real' values that make the denominator 0 because on the x-axis which is the 'real' axis this is where the function will go to either positive or negative infinity and never cross that value of a 'real' x. I hope this gives you a better understanding of what an asymptote is.

Math Prof