Find \(g'(x)\).

Show that \(k = 6\).

Find the equation of the tangent to the graph of \(y = g(x)\) at \(x = 2\). Give your answer in the form \(y = mx + c\).

Use your answer to part (a) and the value of \(k\), to find the \(x\)-coordinates of the stationary points of the graph of \(y = g(x)\).

Find \(g’( - 1)\).

Hence justify that \(g\) is decreasing at \(x =  - 1\).

Find the \(y\)-coordinate of the local minimum.

\(3{x^2} + 2kx - 15\)     (A1)(A1)(A1)

Note:     Award (A1) for \(3{x^2}\), (A1) for \(2kx\) and (A1) for \( - 15\). Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

\(21 = 3{(2)^2} + 2k(2) - 15\)     (M1)(M1)

Note:     Award (M1) for equating their derivative to 21. Award (M1) for substituting 2 into their derivative. The second (M1) should only be awarded if correct working leads to the final answer of \(k = 6\).

Substituting in the known value, \(k = 6\), invalidates the process; award (M0)(M0).

\(k = 6\)     (AG)

[2 marks]

\(g(2) = {(2)^3} + (6){(2)^2} - 15(2) + 5{\text{ }}( = 7)\)     (M1)

Note:     Award (M1) for substituting 2 into \(g\).

\(7 = 21(2) + c\)     (M1)

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient intercept form.

OR

\(y - 7 = 21(x - 2)\)     (M1)

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient point form.

\(y = 21x - 35\)     (A1)     (G2)

[3 marks]

\(3{x^2} + 12x - 15 = 0\) (or equivalent)     (M1)

Note:     Award (M1) for equating their part (a) (with \(k = 6\) substituted) to zero.

\(x =  - 5,{\text{ }}x = 1\)     (A1)(ft)(A1)(ft)

Note:     Follow through from part (a).

[3 marks]

\(3{( - 1)^2} + 12( - 1) - 15\)     (M1)

Note:     Award (M1) for substituting \( - 1\) into their derivative, with \(k = 6\) substituted. Follow through from part (a).

\( =  - 24\)     (A1)(ft)     (G2)

[2 marks]

\(g’( - 1) < 0\) (therefore \(g\) is decreasing when \(x =  - 1\))     (R1)

[1 marks]

\(g(1) = {(1)^3} + (6){(1)^2} - 15(1) + 5\)     (M1)

Note:     Award (M1) for correctly substituting 6 and their 1 into \(g\).

\( =  - 3\)     (A1)(ft)     (G2)

Note:     Award, at most, (M1)(A0) or (G1) if answer is given as a coordinate pair. Follow through from part (c).

[2 marks]