The function can be written in the form \(f(x) = a{(x - h)^2} + k\) .

(i)     Write down the value of h and of k .

(ii)    Show that \(a = 3\) .

Find \(f(x)\)  , giving your answer in the form \(A{x^2} + Bx + C\) .

Calculate the area enclosed by the graph of f , the x-axis, and the lines \(x = 2\) and \(x = 4\) .

(i) \(h = 2\) , \(k = 1\)     A1A1     N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f     M1

e.g. \(13 = a{(0 - 2)^2} + 1\)

working towards solution     A1

e.g. \(13 = 4a + 1\)

\(a = 3\)     AG     N0

[4 marks]

attempting to expand their binomial     (M1)

e.g. \(f(x) = 3({x^2} - 2 \times 2x + 4) + 1\) , \({(x - 2)^2} = {x^2} - 4x + 4\)

correct working     (A1)

e.g. \(f(x) = 3{x^2} - 12x + 12 + 1\)

\(f(x) = 3{x^2} - 12x + 13\) (accept \(A = 3\) , \(B = - 12\) , \(C = 13\) )     A1     N2

[3 marks]

METHOD 1

integral expression     (A1)

e.g. \(\int_2^4 {(3{x^2}}  - 12x + 13)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{x^3} - 6{x^2} + 13x]_2^4\)     A1A1A1

Note: Award A1 for \({x^3}\) , A1 for \( - 6{x^2}\) , A1 for \(13x\) .

 

correct substitution of correct limits into their expression     A1A1

e.g. \(({4^3} - 6 \times {4^2} + 13 \times 4) - ({2^3} - 6 \times {2^2} + 13 \times 2)\) , \(64 - 96 + 52 - (8 - 24 + 26)\)

Note: Award A1 for substituting 4, A1 for substituting 2.

 

correct working     (A1)

e.g. \(64 - 96 + 52 - 8 + 24 - 26,20 - 10\)

\({\rm{Area}} = 10\)     A1     N3

[8 marks]

METHOD 2

integral expression     (A1)

e.g. \(\int_2^4 {(3{{(x - 2)}^2}}  + 1)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{(x - 2)^3} + x]_2^4\)     A2A1

Note: Award A2 for \({(x - 2)^3}\) , A1 for \(x\) .

 

correct substitution of correct limits into their expression     A1A1

e.g. \({(4 - 2)^3} + 4 - [{(2 - 2)^3} + 2]\) , \({2^3} + 4 - ({0^3} + 2)\) , \({2^3} + 4 - 2\)

Note: Award A1 for substituting 4, A1 for substituting 2.

 

correct working     (A1)

e.g. \(8 + 4 - 2\)

\({\rm{Area}} = 10\)    A1     N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4     (R1)

e.g. sketch, \(\int_2^4 {f = \int_0^2 f } \)

integral expression     (A1)

e.g. \(\int_0^2 {(3{x^2}}  - 12x + 13)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{x^3} - 6{x^2} + 13x]_0^2\)     A1A1A1

Note: Award A1 for \({x^3}\) , A1 for \( - 6{x^2}\) , A1 for \(13x\) .

 

correct substitution of correct limits into their expression     A1(A1)

e.g. \(({2^3} - 6 \times {2^2} + 13 \times 2) - ({0^3} - 6 \times {0^2} + 13 \times 0)\) , \(8 - 24 + 26\)

Note: Award A1 for substituting 2, (A1) for substituting 0.

\({\rm{Area}} = 10\)     A1     N3

[8 marks]

In part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able to successfully show that \(a = 3\) . Unfortunately, a few candidates did not understand the "show that" command, and simply verified that \(a = 3\) would work, rather than showing how to find \(a = 3\) .

In part (b), most candidates were able to find \(f(x)\) in the required form. For a few candidates, algebraic errors kept them from finding the correct function, even though they started with correct values for a, h and k.

In part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic and arithmetic errors prevented many from finding the correct area.