Wondering if anyone else finds this strangely difficult. Find the zeros of the graph of f(x)= x^3 - 3x^2 - 9x. I've been able to narrow everything down except the 3 x-intercepts. Just wondering why it is giving me such a bad time...

Question

anonymous · Answer

Factor out an x to start x(x^2 - 3x - 9). From here one solution right off the bat is x = 0. Now use quadratic formula on the inside factor of x^2 - 3x - 9 to get the remainign twosolutions.

anonymous · Answer

And I get $$\frac{ 3\pm \sqrt{9+36} }{ 2 }$$ Which should give me intercepts at $$\frac{ 3 \pm 3\sqrt{5} }{ 2 }$$

anonymous · Answer

Right. So those two intercepts + 0 would be the 3yourelooking for : )

anonymous · Answer

Ok, so I got the ones at 0, and 6ish. But I can't seem to pin down the one at neg 2. I only know it's there because my maximum is at (-1, 5).

anonymous · Answer

Well how exact do you have to be with it? The answers you gave via quadratic formula were just fine for the x-intercepts.

anonymous · Answer

|dw:1387173896836:dw| For the most part not very right now. I'm just trying to make sure there aren't any surprises for finals.