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

Question

Elijah Allen · Accepted Answer

Below

#f(x)=(x^2+6x-9)/(x-2)# #f(x)=((x-2)(x+8)+7)/(x-2)# #f(x)=(x+8)+7/(x-2)# Therefore, the oblique asymptote is #y=x+8# and the vertical asymptote is #x=2# graph{(x^2+6x-9)/(x-2) [-10, 10, -5, 5]}

Caleb Alexander · Answer

undefined 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 $.

Lincoln Anderson · Answer

undefined 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 $.