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November 24, 2014

November 24, 2014

Posted by **logan** on Monday, December 5, 2011 at 10:40pm.

- grade 12 math -
**Reiny**, Monday, December 5, 2011 at 11:13pmcheck you typing.

did you mean

f(x) = (x^3 + 8)/(x^2 - 6)

or is it the way you typed it?

What is with the +-6 ?

- grade 12 math -
**logan**, Monday, December 5, 2011 at 11:38pmit is f(x)=x^3+8/x^2+x-6

- grade 12 math -
**Anonymous**, Tuesday, December 6, 2011 at 10:34amassuming 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

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