Find the asymptotes, if any.
f(x)=(x^2-1)/(x+4).

I understand the vertical asymptote is x=-4, but I know the horizontal has to do with the degree of the numerator and denominator, but I'm not sure of how it works D: Find the asymptotes, if any.
f(x)=(x^2-1)/(x+4).

I understand the vertical asymptote is x=-4, but I know the horizontal has to do with the degree of the numerator and denominator, but I'm not sure of how it works D: @Mathematics

Question

Find the asymptotes, if any.
f(x)=(x^2-1)/(x+4).

I understand the vertical asymptote is x=-4, but I know the horizontal has to do with the degree of the numerator and denominator, but I'm not sure of how it works D: Find the asymptotes, if any.
f(x)=(x^2-1)/(x+4).

I understand the vertical asymptote is x=-4, but I know the horizontal has to do with the degree of the numerator and denominator, but I'm not sure of how it works D: @Mathematics

asnaseer · Answer

vertical asymptote is when denominator is zero, so you correctly identified x=-4 as the vertical asymptote.
since numerator is of a greater degree than the denominator (x^2 vs x), then it must have an oblique asymptote as well. you get this by long division - see next drawing.

asnaseer · Answer

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