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

We can long divide the coefficients like this:

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#

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To find the slant asymptote of a rational function, we need to perform polynomial long division to divide the numerator by the denominator. The result will be a quotient and a remainder. The quotient will represent the polynomial part of the slant asymptote.

Given the rational function {-9x^3 - 9x^2 + 28x + 9} / {3x^2 + 6x + 2}, we perform polynomial long division to find the quotient:

```
-3x + 1
____________________________
```

3x^2 + 6x + 2 | -9x^3 - 9x^2 + 28x + 9 -(-9x^3 - 18x^2 - 6x) ____________________ 9x^2 + 34x + 9 - (9x^2 + 18x + 6) __________________ 16x + 3

The quotient is -3x + 1. Therefore, the slant asymptote of the rational function is y = -3x + 1.

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