posted by Pegs .
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 non-vertical 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 non-vertical 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.