Question 12 [Maximum mark: 7]
Consider the function f (x) = ax 2 + bx + c . The graph of y = f (x) is shown in the diagram. The vertex of the graph has coordinates (0.5, −12.5) . The graph intersects the x-axis at two points, (−2, 0) and ( p, 0).
a. Find the value of p . [1]
b. Find the value of
(i)a .
(ii)b .
(iii) c . [5]
c. Write down the equation of the axis of symmetry of the graph. [1]
Answer/Explanation
(a) 3
(b) METHOD 1
\(0 = 4a -2b+c\),
\(0 = 9a+ 3b+ c\)
\(- \frac{25}{2}= \frac{1}{4}a + \frac{1}{2}\) b + c\)
on solving we get a , b , c
(i) 2 (ii) – 2(iii) -12
METHOD 2 \(−12.5 = a (0.5+ 2)(0.5- 2)\) (i) \(a=2\)
\(0 = 2x(3)2+ 3b + c \)
\(0 = 2x(-2)2+ (-2b)+c\)
(ii) b = −2(iii) c = −12
(c) x = 0.5
Question
The following curves are sketches of the graphs of the functions given below, but in a different order. Using your graphic display calculator, match the equations to the curves, writing your answers in the table below.
(the diagrams are not to scale)
Answer/Explanation
Markscheme
(i) B (A1)
(ii) D (A1)
(iii) A (A1)
(iv) E (A1)
(v) C (A1)
(vi) F (A1) (C6)[6 marks]
Question
Factorise the expression \({x^2} – kx\) .[1]
Hence solve the equation \({x^2} – kx = 0\) .[1]
The diagram below shows the graph of the function \(f(x) = {x^2} – kx\) for a particular value of \(k\).
Write down the value of \(k\) for this function.[1]
The diagram below shows the graph of the function \(f(x) = {x^2} – kx\) for a particular value of \(k\).
Find the minimum value of the function \(y = f(x)\) .[3]
Answer/Explanation
Markscheme
\(x(x – k)\) (A1) (C1)[1 mark]
\(x = 0\) or \(x = k\) (A1) (C1)
Note: Both correct answers only.[1 mark]
\(k = 3\) (A1) (C1)[1 mark]
\({\text{Vertex at }}x = \frac{{ – ( – 3)}}{{2(1)}}\) (M1)
Note: (M1) for correct substitution in formula.
\(x = 1.5\) (A1)(ft)
\(y = – 2.25\) (A1)(ft)
OR
\(f'(x) = 2x – 3\) (M1)
Note: (M1) for correct differentiation.
\(x = 1.5\) (A1)(ft)
\(y = – 2.25\) (A1)(ft)
OR
for finding the midpoint of their 0 and 3 (M1)
\(x = 1.5\) (A1)(ft)
\(y = – 2.25\) (A1)(ft)
Note: If final answer is given as \((1.5{\text{, }}{- 2.25})\) award a maximum of (M1)(A1)(A0)[3 marks]
Question
The graph of a quadratic function \(y = f (x)\) is given below.
Write down the equation of the axis of symmetry.[2]
Write down the coordinates of the minimum point.[2]
Write down the range of \(f (x)\).[2]
Answer/Explanation
Markscheme
x = 3 (A1)(A1) (C2)
Notes: Award (A1) for “ x = ” (A1) for 3.
The mark for x = is not awarded unless a constant is seen on the other side of the equation.[2 marks]
(3, −14) (Accept x = 3, y = −14) (A1)(ft)(A1) (C2)
Note: Award (A1)(A0) for missing coordinate brackets.[2 marks]
y ≥ −14 (A1)(A1)(ft) (C2)
Notes: Award (A1) for y ≥ , (A1)(ft) for –14.
Accept alternative notation for intervals.[2 marks]
Question
A quadratic curve with equation y = ax (x − b) is shown in the following diagram.
The x-intercepts are at (0, 0) and (6, 0), and the vertex V is at (h, 8).
Find the value of h.[2]
Find the equation of the curve.[4]
Answer/Explanation
Markscheme
\(\frac{{0 + 6}}{2} = 3\) \(h = 3\) (M1)(A1) (C2)
Note: Award (M1) for any correct method.[2 marks]
\(y = ax(x – 6)\) (A1)
\(8 = 3a(- 3)\) (A1)(ft)
\(a = – \frac{8}{9}\) (A1)(ft)
\(y = – \frac{8}{9} x (x – 6)\) (A1)(ft)
Notes: Award (A1) for correct substitution of \(b = 6\) into equation.
Award (A1)(ft) for substitution of their point V into the equation.
OR
\(y = a(x – 3)^2 + 8\) (A1)(ft)
Note: Award (A1)(ft) for correct substitution of their h into the equation.
\(0 = a(6 – 3)^2 + 8\) OR \(0 = a(0 – 3)^2 + 8\) (A1)
Note: Award (A1) for correct substitution of an x intercept.
\(a = – \frac{8}{9}\) (A1)(ft)
\(y = – \frac{8}{9}(x – 3)^2 + 8\) (A1)(ft) (C4)[4 marks]