SOLUTION: what is the oblique asymptote of f(x)=(7x^3-5x^2-9x+7)/(-6x^2-5x-9)

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Question 709092: what is the oblique asymptote of f(x)=(7x^3-5x^2-9x+7)/(-6x^2-5x-9)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%287x%5E3-5x%5E2-9x%2B7%29%2F%28-6x%5E2-5x-9%29
Find find the oblique asymptote of a rational function you divide the numerator by the denominator so you can rewrite the function in the form:
f(x) = quotient + remainder/(f(x)'s denominator)

We will not be able to use synthetic division since our function's denominator is quadratic. So we must use long division. [Note: Since -6x^2 does not divide evenly into 7x^3 (and probably other terms too) this is going to get messy with fractions.]
              (-7/6)x + (-65/36)                  
-6x^2 -5x -9 / 7x^3       -5x^2            -9x + 7
             - 7x^3 + (35/6)x^2    +   (42/6)x
                        (-65/6)x^2 + (-294/6)x + 7
                      - (-65/6)x^2 + (325/36)x + 65/4
                                   (-2089/36)x + (-37/4)

So f(x) in the desired form is:

As x get to be very large (positive or negative), the fraction at the end becomes closer and closer to zero. This makes the quotient part the oblique asymptote for large values of x. So the oblique asymptote for f(x) is the line:
y+=+%28-7%2F6%29x+%2B+%28-65%2F36%29