calculus
posted by Pegs on .
f(x) = (9x^3  1x^2 + 8x + 4)/(2x^2  4x  2)
What is the smallest value of x at which f(x) intersects its nonvertical asymptote?
I took the slant asymptote which is 9/2x+19/2 and set it equal to the function and tried to solve it as a quadratic equation. But any of my answers I enter into my online homework are wrong apparently. Help?

Did you really mean to write 1x^2 as the second term of the numerator? or did you mean to write 19 x^2?
For very large x, the f(x) ratio becomes 9x^3/2x^2 = (9/2)x
That means the nonvertical asymptote is fasym(x) = (9/2) x
At an intersection with the asymptote, you can subsitute (9/2)x for f(x).
If you make that substition and multiply both sides by (2x^2  4x  2), the cubic terms drop out and you are left wth a quadratic to solve. Is that what you did? 
No, I wrote everything correctly.
Yes I did that and then I did multiply both sides by the denominator. The answer I got from the quadratic equation was not correct, however, according to my online homework.