Indicate the x- and y-intercepts. (

Equation: f(x) = 4x^3 + 6x^2 − 24x + 1

So I set it=0 but I don't know how to solve this :/ please help!

4x^3 + 6x^2 − 24x + 1=0

Question

Indicate the x- and y-intercepts. (

Equation: f(x) = 4x^3 + 6x^2 − 24x + 1

So I set it=0 but I don't know how to solve this :/ please help!

4x^3 + 6x^2 − 24x + 1=0

anonymous · Answer

Are you in calculus?

anonymous · Answer

yes

anonymous · Answer

Ok.  Getting the first derivative, y' = 12x^2 + 12x - 24 = 12(x + 2)(x - 1), we have extrema at x= 1 or -2.  With me so far?

anonymous · Answer

yes, but I thought all you had to do for finding the x-intercept was to set the equation=0?

anonymous · Answer

Yes, that's correct, and then you use synthetic division to look for rational zeros and also use Descartes' Rule of signs.  But there are no rational zeros, just irrational but real zeros.  I was going into the derivative so that you could roughly graph it so that you see the curve, but we don't have to.

anonymous · Answer

oh I see, extrema would the y value just be 0? so -2,0? 1,0?

anonymous · Answer

|dw:1351547029691:dw|  Well, the extrema are (-2, 9)  and (1, -13).  You have a y-intercept at (0, 1).  You have an irrational zero at slightly less than x = 1/4.