Question

The directions say to use a grapking utility to approximate (to two-decimal-place accuracy) any relative minimum or maximum values of the function:

y=x^3 - 6x^2 + 15

Can someone please help me because my graphing calculator is not working!!!

Answer 1

y' = 3 x^2 - 12 x
y' = 0 at max or min
3x (x-4) = 0
x = 0 or x = 4 for slope horizontal
y" = 6 x - 12
at x = 0, y"=-12 so maximum
at x = 4, y" = +12 so minimum

Answer 2

then at x = 0, y = 15
and at x = 4, y = -17

Answer 3

So if you do not know calculus sketch the graph starting at lower left, rising to 15 at x = 0, dropping to -17 at x = 4, then rising up to upper right.
sketch some points right around x = 0 and right around x = 4