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dcgirl16
Jun11-07, 05:01 PM
I'm having trouble finding x intercepts i have the question x^3-9x^2+15x+30 and by setting y=0 i got -30=x(x^2-9x+15) I used the quadratic formula to find that x^2-9x+15 give me no real roots, so the only intercept would be x=0. But this would also mean that the yintercept is 0 and i found that it is 30. What am i doing wrong is it something to do with moving the 30 to the left side?

malawi_glenn
Jun11-07, 05:12 PM
-30=x(x^2-9x+15)

You can not solve the right hand side to be = 0 if you have left hand side = -30 ...

dcgirl16
Jun11-07, 05:20 PM
so how do i do it. i was moving that over to solve for x

malawi_glenn
Jun11-07, 05:23 PM
test solution, then rewrite x^3-9x^2+15x+30 as (x+n)(x-b)(x-4) or something in that style.

If c solves the p(x) = then p(x) can be divided with (x-c)..
and vice versa, (x+r) ; P(-r) = 0