How do you find the slant asymptote of ${-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}$ ?

Question

1398 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-29T17:31:52

$y = -3x+3$

We can long divide the coefficients like this:

enter image source here

to find that:

$(-9x^3-9x^2+28x+9)/(3x^2+6x+2) = -3x+3+(16x+3)/(3x^2+6x+2)$

So the slant asymptote is:

$y = -3x+3$

How do you find the slant asymptote of {-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}{-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}?