Question
Let \(f\) be a quadratic function. Part of the graph of \(f\) is shown below.
The vertex is at P(\(4\), \(2\)) and the y-intercept is at Q(\(0\), \(6\)) .
Write down the equation of the axis of symmetry.
The function f can be written in the form \(f(x) = a{(x – h)^2} + k\) .
Write down the value of h and of k .
The function f can be written in the form \(f(x) = a{(x – h)^2} + k\) .
Find a .
Answer/Explanation
Markscheme
\(x = 4\) (must be an equation) A1 N1
[1 mark]
\(h = 4\) , \(k = 2\) A1A1 N2
[2 marks]
attempt to substitute coordinates of any point on the graph into f (M1)
e.g. \(f(0) = 6\) , \(6 = a{(0 – 4)^2} + 2\) , \(f(4) = 2\)
correct equation (do not accept an equation that results from \(f(4) = 2\) ) (A1)
e.g. \(6 = a{( – 4)^2} + 2\) , \(6 = 16a + 2\)
\(a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)\) A1 N2
[3 marks]
Question
Let \(f(x) = \sin x + \frac{1}{2}{x^2} – 2x\) , for \(0 \le x \le \pi \) .
Let \(g\) be a quadratic function such that \(g(0) = 5\) . The line \(x = 2\) is the axis of symmetry of the graph of \(g\) .
The function \(g\) can be expressed in the form \(g(x) = a{(x – h)^2} + 3\) .
Find \(f'(x)\) .
Find \(g(4)\) .
(i) Write down the value of \(h\) .
(ii) Find the value of \(a\) .
Find the value of \(x\) for which the tangent to the graph of \(f\) is parallel to the tangent to the graph of \(g\) .
Answer/Explanation
Markscheme
\(f'(x) = \cos x + x – 2\) A1A1A1 N3
Note: Award A1 for each term.
[3 marks]
recognizing \(g(0) = 5\) gives the point (\(0\), \(5\)) (R1)
recognize symmetry (M1)
eg vertex, sketch 
\(g(4) = 5\) A1 N3
[3 marks]
(i) \(h = 2\) A1 N1
(ii) substituting into \(g(x) = a{(x – 2)^2} + 3\) (not the vertex) (M1)
eg \(5 = a{(0 – 2)^2} + 3\) , \(5 = a{(4 – 2)^2} + 3\)
working towards solution (A1)
eg \(5 = 4a + 3\) , \(4a = 2\)
\(a = \frac{1}{2}\) A1 N2
[4 marks]
\(g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5\)
correct derivative of \(g\) A1A1
eg \(2 \times \frac{1}{2}(x – 2)\) , \(x – 2\)
evidence of equating both derivatives (M1)
eg \(f’ = g’\)
correct equation (A1)
eg \(\cos x + x – 2 = x – 2\)
working towards a solution (A1)
eg \(\cos x = 0\) , combining like terms
\(x = \frac{\pi }{2}\) A1 N0
Note: Do not award final A1 if additional values are given.
[6 marks]
Question
Let \(f(x) = a{(x – h)^2} + k\). The vertex of the graph of \(f\) is at \((2, 3)\) and the graph passes through \((1, 7)\).
Write down the value of \(h\) and of \(k\).
Find the value of \(a\).
Answer/Explanation
Markscheme
\(h = 2,{\text{ }}k = 3\) A1A1 N2
[2 marks]
attempt to substitute \((1,7)\) in any order into their \(f(x)\) (M1)
eg \(7 = a{(1 – 2)^2} + 3{\text{, }}7 = a{(1 – 3)^2} + 2{\text{, }}1 = a{(7 – 2)^2} + 3\)
correct equation (A1)
eg \(7 = a + 3\)
a = 4 A1 N2
[3 marks]
Examiners report
[N/A]
[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.
Hence, show that \(p = 3\).
The graph of \(f\)has its vertex on the \(x\)-axis.
Find the coordinates of the vertex of the graph of \(f\).
The graph of \(f\) has its vertex on the \(x\)-axis.
Write down the solution of \(f(x) = 0\).
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 \(a\).
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 \(h\).
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 \(k\).
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 \(g(x) = A{x^2} + Bx + C\).
Answer/Explanation
Markscheme
correct value \(0\), or \(36 – 12p\) A2 N2
[2 marks]
correct equation which clearly leads to \(p = 3\) A1
eg \(36 – 12p = 0,{\text{ }}36 = 12p\)
\(p = 3\) AG N0
[1 mark]
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]
\(x = 1\) A1 N1
[1 mark]
\(a = 3\) A1 N1
[1 mark]
\(h = 1\) A1 N1
[1 mark]
\(k = 0\) A1 N1
[1 mark]
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]
Question
Consider \(f(x) = {x^2} + qx + r\). The graph of \(f\) has a minimum value when \(x = – 1.5\).
The distance between the two zeros of \(f\) is 9.
Show that the two zeros are 3 and \( – 6\).
Find the value of \(q\) and of \(r\).
Answer/Explanation
Markscheme
recognition that the \(x\)-coordinate of the vertex is \( – 1.5\) (seen anywhere) (M1)
eg\(\,\,\,\,\,\)axis of symmetry is \( – 1.5\), sketch, \(f'( – 1.5) = 0\)
correct working to find the zeroes A1
eg\(\,\,\,\,\,\)\( – 1.5 \pm 4.5\)
\(x = – 6\) and \(x = 3\) AG N0
[2 marks]
METHOD 1 (using factors)
attempt to write factors (M1)
eg\(\,\,\,\,\,\)\((x – 6)(x + 3)\)
correct factors A1
eg\(\,\,\,\,\,\)\((x – 3)(x + 6)\)
\(q = 3,{\text{ }}r = – 18\) A1A1 N3
METHOD 2 (using derivative or vertex)
valid approach to find \(q\) (M1)
eg\(\,\,\,\,\,\)\(f'( – 1.5) = 0,{\text{ }} – \frac{q}{{2a}} = – 1.5\)
\(q = 3\) A1
correct substitution A1
eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0\)
\(r = – 18\) A1
\(q = 3,{\text{ }}r = – 18\) N3
METHOD 3 (solving simultaneously)
valid approach setting up system of two equations (M1)
eg\(\,\,\,\,\,\)\(9 + 3q + r = 0,{\text{ }}36 – 6q + r = 0\)
one correct value
eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r = – 18\) A1
correct substitution A1
eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0,{\text{ }}{3^2} + 3q – 18 = 0,{\text{ }}36 – 6q – 18 = 0\)
second correct value A1
eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r = – 18\)
\(q = 3,{\text{ }}r = – 18\) N3
[4 marks]
Question
Let f(x) = ax2 − 4x − c. A horizontal line, L , intersects the graph of f at x = −1 and x = 3.
The equation of the axis of symmetry is x = p. Find p.
Hence, show that a = 2.
The equation of L is y = 5 . Find the value of c.
Answer/Explanation
Markscheme
METHOD 1 (using symmetry to find p)
valid approach (M1)
eg \(\frac{{ – 1 + 3}}{2}\), 
p = 1 A1 N2
Note: Award no marks if they work backwards by substituting a = 2 into \( – \frac{b}{{2a}}\) to find p.
Do not accept \(p = \frac{2}{a}\).
METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a M1
eg a + 4 − c = a(32) − 4(3) − c, f(−1) = f(3)
correct working A1
eg 8a = 16
a = 2 AG N0
valid approach to find p (M1)
eg \( – \frac{b}{{2a}},\,\,\,\frac{4}{{2\left( 2 \right)}}\)
p = 1 A1 N2
[2 marks]
METHOD 1
valid approach M1
eg \( – \frac{b}{{2a}},\,\,\,\frac{4}{{2a}}\) (might be seen in (i)), f’ (1) = 0
correct equation A1
eg \(\frac{4}{{2a}}\) = 1, 2a(1) − 4 = 0
a = 2 AG N0
METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a M1
eg a + 4 − c = a(32) − 4(3) − c, f(−1) = f(3)
correct working A1
eg 8a = 16
a = 2 AG N0
[2 marks]
valid approach (M1)
eg f(−1) = 5, f(3) =5
correct working (A1)
eg 2 + 4 − c = 5, 18 − 12 − c = 5
c = 1 A1 N2
[3 marks]
Question
Let \(f\left( x \right) = p{x^2} + qx – 4p\), where p ≠ 0. Find Find the number of roots for the equation \(f\left( x \right) = 0\).
Justify your answer.
Answer/Explanation
Markscheme
METHOD 1
evidence of discriminant (M1)
eg \({b^2} – 4ac,\,\,\Delta \)
correct substitution into discriminant (A1)
eg \({q^2} – 4p\left( { – 4p} \right)\)
correct discriminant A1
eg \({q^2} + 16{p^2}\)
\(16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)\) A1
\({q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)\) A1
\({q^2} + 16{p^2} > 0\) A1
\(f\) has 2 roots A1 N0
METHOD 2
y-intercept = −4p (seen anywhere) A1
if p is positive, then the y-intercept will be negative A1
an upward-opening parabola with a negative y-intercept R1
eg sketch that must indicate p > 0.
if p is negative, then the y-intercept will be positive A1
a downward-opening parabola with a positive y-intercept R1
eg sketch that must indicate p > 0.
\(f\) has 2 roots A2 N0
[7 marks]
Question
Let \(f(x) = 3{(x + 1)^2} – 12\) .
Show that \(f(x) = 3{x^2} + 6x – 9\) .
For the graph of f
(i) write down the coordinates of the vertex;
(ii) write down the y-intercept;
(iii) find both x-intercepts.
Hence sketch the graph of f .
Let \(g(x) = {x^2}\) . The graph of f may be obtained from the graph of g by the following two transformations
a stretch of scale factor t in the y-direction,
followed by a translation of \(\left( \begin{array}{l}
p\\
q
\end{array} \right)\) .
Write down \(\left( \begin{array}{l}
p\\
q
\end{array} \right)\) and the value of t .
Answer/Explanation
Markscheme
\(f(x) = 3({x^2} + 2x + 1) – 12\) A1
\( = 3{x^2} + 6x + 3 – 12\) A1
\( = 3{x^2} + 6x – 9\) AG N0
[2 marks]
(i) vertex is \(( – 1, – 12)\) A1A1 N2
(ii) \(y = – 9\) , or \((0, – 9)\) A1 N1
(iii) evidence of solving \(f(x) = 0\) M1
e.g. factorizing, formula
correct working A1
e.g. \(3(x + 3)(x – 1) = 0\) , \(x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}\)
\(x = – 3\) , \(x = 1\) , or \(( – 3{\text{, }}0){\text{, }}(1{\text{, }}0)\) A1A1 N2
[7 marks]
A1A1A1 N3
Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximately correct positions. Scale and labelling not required.
[3 marks]
\(\left( \begin{array}{l}
p\\
q
\end{array} \right) = \left( \begin{array}{l}
– 1\\
– 12
\end{array} \right)\) , \(t = 3\) A1A1A1 N3
[3 marks]