# How do you identify the oblique asymptote of #f(x) = (x^2+6x-9)/(x-2)#?

Below

graph{(x^2+6x-9)/(x-2) [-10, 10, -5, 5]}

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To identify the oblique asymptote of the function ( f(x) = \frac{x^2 + 6x - 9}{x - 2} ), you can perform polynomial long division to divide the numerator ( x^2 + 6x - 9 ) by the denominator ( x - 2 ). After performing the division, the quotient will represent the linear function part of the oblique asymptote, and any remainder will represent the non-linear part.

The result of the polynomial long division for this function yields ( x + 8 ) as the quotient and a remainder of ( 7 ). Therefore, the oblique asymptote of the function ( f(x) ) is the linear function ( y = x + 8 ).

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To identify the oblique asymptote of the function ( f(x) = \frac{x^2 + 6x - 9}{x - 2} ), we perform polynomial long division to divide ( x^2 + 6x - 9 ) by ( x - 2 ):

```
x + 8
___________________
x - 2 | x^2 + 6x - 9
- (x^2 - 2x)
_____________
8x - 9
- (8x - 16)
__________
7
```

Therefore, ( f(x) = x + 8 + \frac{7}{x - 2} ) after performing the division. As ( x ) approaches infinity or negative infinity, ( \frac{7}{x - 2} ) approaches 0. So, the oblique asymptote of ( f(x) ) is ( y = x + 8 ).

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