Question

For what values of x does the curve y= -x^3+6x^2 have a slope of -12 ?

Answer 1

\[-3x^2+12x=-12\] solve for x

Answer 2

y'=-3x^2+12x
-3x^2+12x=-12
-3x^2+12x+12=0
x^2-4x-4=0
\[x=\frac{4 \pm \sqrt{16-4(1)(-4)}}{2}=\frac{4 \pm \sqrt{32}}{2}=\frac{4 \pm 4\sqrt{2}}{2}=2 \pm 2\sqrt{2}\]

Answer 3

thank you so much :)