How do you find the slant asymptotes for #(x^2-6x+7)/(x+5)#?

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How do you find the slant asymptotes for #(x^2-6x+7)/(x+5)#?

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Answer 1 · 2015-02-14T17:23:33

Your function has a first asymptote (vertical) at #x=-5#. This values of #x# makes the denominator equal to #0#.
The other asymptote can be found as follows:
enter image source here

And graphically:
enter image source here

hope it helps

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How do you find the slant asymptotes for #(x^2-6x+7)/(x+5)#?

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