Question: Show that, when k is constant, the curve

$\displaystyle y = 3x^4- 8x^3-6x^2+24x+k $

has a stationary point when x =1 and find the values of x at the other two stationary points on the curve. Find the values of k for which the curve touches the x axis.

My attempt:

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=12x^{3}-24x^{2}-12x+24 $

when $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=12x^{3}-24x^{2}-12x+24 = 0 $

$\displaystyle 12x^{3}-24x^{2}-12x+24 = 0 $

$\displaystyle 12(x^{3}-2x^{2}-x+2) = 0 $

try x = 1

$\displaystyle 12(1^{3}-2(1)^{2}-1(1)+2) = 0 $ Thus since f'(1) =0 thus at x =1 there is a stationary point.

The other values of x are x =-1 and x =2

my problem is how to find the values of k

please can you help me

thank you