Show that \(8a + 4b + c = 9\) .

The graph of f has a local minimum at \((1{\text{, }}4)\) .

Find two other equations in a , b and c , giving your answers in a similar form to part (a).

Find the value of a , of b and of c .

attempt to substitute coordinates in f     (M1)

e.g. \(f(2) = 9\)

correct substitution     A1

e.g. \(a \times {2^3} + b \times {2^2} + c = 9\)

\(8a + 4b + c = 9\)    AG     N0

[2 marks]

recognizing that \((1{\text{, }}4)\) is on the graph of f     (M1)

e.g. \(f(1) = 4\)

correct equation     A1

e.g. \(a + b + c = 4\)

recognizing that \(f' = 0\) at minimum (seen anywhere)     (M1)

e.g. \(f'(1) = 0\)

\(f'(x) = 3a{x^2} + 2bx\) (seen anywhere)     A1A1

correct substitution into derivative     (A1)

e.g. \(3a \times {1^2} + 2b \times 1 = 0\)

correct simplified equation     A1

e.g. \(3a + 2b = 0\)

[7 marks]

valid method for solving system of equations     (M1)

e.g. inverse of a matrix, substitution

\(a = 2\) , \(b = - 3\) , \(c = 5\)     A1A1A1     N4

[4 marks]

Part (a) was generally well done, with a few candidates failing to show a detailed substitution. Some substituted 2 in place of x, but didn't make it clear that they had substituted in y as well.

A great majority could find the two equations in part (b). However there were a significant number of candidates who failed to identify that the gradient of the tangent is zero at a minimum point, thus getting the incorrect equation \(3a + 2b = 4\) .

A considerable number of candidates only had 2 equations, so that they either had a hard time trying to come up with a third equation (incorrectly combining some of the information given in the question) to solve part (c) or they completely failed to solve it.

Despite obtaining three correct equations many used long elimination methods that caused algebraic errors. Pages of calculations leading nowhere were seen.

Those who used matrix methods were almost completely successful.