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

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

As

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:

As x increases in magnitude then

So you end up with the asymptote being

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

So 2 is the 'excluded value'

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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 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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