Date | May 2018 | Marks available | 2 | Reference code | 18M.1.sl.TZ1.12 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 12 | Adapted from | N/A |
Question
Consider the quadratic function f(x)=ax2+bx+22.
The equation of the line of symmetry of the graph y=f(x) is x=1.75.
The graph intersects the x-axis at the point (−2 , 0).
Using only this information, write down an equation in terms of a and b.
Using this information, write down a second equation in terms of a and b.
Hence find the value of a and of b.
The graph intersects the x-axis at a second point, P.
Find the x-coordinate of P.
Markscheme
1.75=−b2a (or equivalent) (A1) (C1)
Note: Award (A1) for f(x)=(1.75)2a+1.75b or for y=(1.75)2a+1.75b+22 or for f(1.75)=(1.75)2a+1.75b+22.
[1 mark]
(−2)2×a+(−2)×b+22=0 (or equivalent) (A1) (C1)
Note: Award (A1) for (−2)2×a+(−2)×b+22=0 seen.
Award (A0) for y=(−2)2×a+(−2)×b+22.
[1 mark]
a = −2, b = 7 (A1)(ft)(A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
Accept answers(s) embedded as a coordinate pair.
[2 marks]
−2x2 + 7x + 22 = 0 (M1)
Note: Award (M1) for correct substitution of a and b into equation and setting to zero. Follow through from part (c).
(x =) 5.5 (A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
OR
x-coordinate = 1.75 + (1.75 − (−2)) (M1)
Note: Award (M1) for correct use of axis of symmetry and given intercept.
(x =) 5.5 (A1) (C2)
[2 marks]