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

Question

Anna Allen · Accepted Answer

#y=x+8#

The asymptotes are a tendency that occur at particular point. In this case these points will be #+-oo#

As #x->oo# the constants become insignificant so we virtually end up with:#" "(x^2+6x)/x# Which is the equivalent of #y=x+6#

However. If this is investigated further it soon becomes evident that this is only a very rough estimate and is slightly out.

If you actually carry out a polynomial division you end up with:

#y=x+8 + 7/(x-2)#

As x increases in magnitude then #7/(x-2)# becomes smaller and smaller until you virtually have 0

So you end up with the asymptote being #y=x+8#

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They do not ask for the vertical asymptote. So just for the sake of interest. You are not allowed to divide by 0. Mathematically this is given the name of being 'undefined'

So the denominator becomes 0 at #x=2# Thus the vertical asymptote is #x=2#

So 2 is the 'excluded value'

Lily Anderson · Answer

undefined To find the slant asymptote of the function $ f(x) = \frac{x^2 + 6x - 9}{x - 2} $, perform polynomial long division.

Divide $ x^2 + 6x - 9 $ by $ x - 2 $ to find the quotient and remainder. The quotient represents the equation of the slant asymptote.

Performing polynomial long division, you'll find that $ f(x) $ simplifies to $ x + 8 $ with a remainder of $ 7 $.

Therefore, the slant asymptote of $ f(x) $ is $ y = x + 8 $.