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Date May 2014 Marks available 4 Reference code 14M.1.sl.TZ2.8
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 8 Adapted from N/A

Question

Let \(f(x) = 3{x^2} - 6x + p\). The equation \(f(x) = 0\) has two equal roots.

Write down the value of the discriminant.

[2]
a(i).

Hence, show that \(p = 3\).

[1]
a(ii).

The graph of \(f\)has its vertex on the \(x\)-axis.

Find the coordinates of the vertex of the graph of \(f\).

[4]
b.

The graph of \(f\) has its vertex on the \(x\)-axis.

Write down the solution of \(f(x) = 0\).

[1]
c.

The graph of \(f\) has its vertex on the \(x\)-axis.

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

[1]
d(i).

The graph of \(f\) has its vertex on the \(x\)-axis.

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

[1]
d(ii).

The graph of \(f\) has its vertex on the \(x\)-axis.

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

[1]
d(iii).

The graph of \(f\) has its vertex on the \(x\)-axis.

The graph of a function \(g\) is obtained from the graph of \(f\) by a reflection of \(f\) in the \(x\)-axis, followed by a translation by the vector \(\left( \begin{array}{c}0\\6\end{array} \right)\). Find \(g\), giving your answer in the form \(g(x) = A{x^2} + Bx + C\).

[4]
e.

Markscheme

correct value \(0\), or \(36 - 12p\)     A2     N2

[2 marks]

a(i).

correct equation which clearly leads to \(p = 3\)     A1

eg     \(36 - 12p = 0,{\text{ }}36 = 12p\)

\(p = 3\)     AG     N0

[1 mark]

a(ii).

METHOD 1

valid approach     (M1)

eg     \(x =  - \frac{b}{{2a}}\)

correct working     A1

eg     \( - \frac{{( - 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

METHOD 2

valid approach     (M1)

eg     \(f(x) = 0\), factorisation, completing the square

correct working     A1

eg     \({x^2} - 2x + 1 = 0,{\text{ }}(3x - 3)(x - 1),{\text{ }}f(x) = 3{(x - 1)^2}\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

METHOD 3

valid approach using derivative     (M1)

eg     \(f'(x) = 0,{\text{ }}6x - 6\)

correct equation     A1

eg     \(6x - 6 = 0\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

[4 marks]

b.

\(x = 1\)     A1     N1

[1 mark]

c.

\(a = 3\)     A1     N1

[1 mark]

d(i).

\(h = 1\)     A1     N1

[1 mark]

d(ii).

\(k = 0\)     A1     N1

[1 mark]

d(iii).

attempt to apply vertical reflection     (M1)

eg     \( - f(x),{\text{ }} - 3{(x - 1)^2}\), sketch

attempt to apply vertical shift 6 units up     (M1)

eg     \( - f(x) + 6\), vertex \((1, 6)\)

transformations performed correctly (in correct order)     (A1)

eg     \( - 3{(x - 1)^2} + 6,{\text{ }} - 3{x^2} + 6x - 3 + 6\)

\(g(x) =  - 3{x^2} + 6x + 3\)     A1     N3

[4 marks]

 

e.

Examiners report

[N/A]
a(i).
[N/A]
a(ii).
[N/A]
b.
[N/A]
c.
[N/A]
d(i).
[N/A]
d(ii).
[N/A]
d(iii).
[N/A]
e.

Syllabus sections

Topic 2 - Functions and equations » 2.4 » The quadratic function \(x \mapsto a{x^2} + bx + c\) : its graph, \(y\)-intercept \((0, c)\). Axis of symmetry.
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