Given that the graph of $y = {x^3} - 6{x^2} + kx - 4$ has exactly one point at which the gradient is zero, find the value of k .

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{x^2} - 12x + k$     M1A1

For use of discriminant ${b^2} - 4ac = 0$ or completing the square $3{(x - 2)^2} + k - 12$     (M1)

 $144 - 12k = 0$     (A1)

 Note: Accept trial and error, sketches of parabolas with vertex (2,0) or use of second derivative.

$k = 12$     A1

[5 marks] 

Generally candidates answer this question well using a diversity of methods. Surprisingly, a small number of candidates were successful in answering this question using the discriminant of the quadratic and in many cases reverted to trial and error to obtain the correct answer.