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did you mean
f(x) = (x^3 + 8)/(x^2 - 6)
or is it the way you typed it?
What is with the +-6 ?
Given the function f(x)= x^3+8/x^2+-6, determine the eqaution of the asymptotes and state the end behaviours of the graph near the asymptotes.
3 answers
it is f(x)=x^3+8/x^2+x-6
assuming you really do mean
(x^3 + 8)/(x^2 + x - 6)
= (x+2)(x^2 - 2x + 4) / (x+3)(x-2)
It crosses the x-axis at x = -2
there are vertical asymptotes where the denominator is zero:
x = -3 or x=2
as x gets large, y -> x^3/x^2 = x
(x^3 + 8)/(x^2 + x - 6)
= (x+2)(x^2 - 2x + 4) / (x+3)(x-2)
It crosses the x-axis at x = -2
there are vertical asymptotes where the denominator is zero:
x = -3 or x=2
as x gets large, y -> x^3/x^2 = x
3 answers