slope = 3 x^2 -12 x =3 x (x-4)
second derivative (curvature)
= 6 x - 12
so the slope is 0 at x = 0 and at x = 4
at x = 0
is that a min or a max or in inflection point?
second derivative = -12
That is a maximum, starts down from there
at x = 4 is that a min or a max or an inflection point?
second derivative = +12
so that is a minimum, starts up fro there.
how do I graph this using the first and second derivative and three sign charts.
i have to label the inflection point, extreme values and the intercepts also.
f(x)= x^3-6x^2
2 answers
Oh, intercepts
y = x^3-6x^2
0 = x^2 (x-6)
bounced off x axis at x = 0
crossed x axis at x = 6
y = x^3-6x^2
0 = x^2 (x-6)
bounced off x axis at x = 0
crossed x axis at x = 6