Date | November 2019 | Marks available | 1 | Reference code | 19N.2.SL.TZ0.T_4 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Draw | Question number | T_4 | Adapted from | N/A |
Question
The graph of the quadratic function f(x)=12(x-2)(x+8) intersects the y-axis at (0, c).
The vertex of the function is (-3, -12.5).
The equation f(x)=12 has two solutions. The first solution is x=-10.
Let T be the tangent at x=-3.
Find the value of c.
Write down the equation for the axis of symmetry of the graph.
Use the symmetry of the graph to show that the second solution is x=4.
Write down the x-intercepts of the graph.
On graph paper, draw the graph of y=f(x) for -10≤x≤4 and -14≤y≤14. Use a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis.
Write down the equation of T.
Draw the tangent T on your graph.
Given f(a)=5.5 and f'(a)=-6, state whether the function, f, is increasing or decreasing at x=a. Give a reason for your answer.
Markscheme
12(0-2)(0+8) OR 12(02+6(0)-16) (or equivalent) (M1)
Note: Award (M1) for evaluating f(0).
(c=) -8 (A1)(G2)
Note: Award (G2) if -8 or (0, -8) seen.
[2 marks]
x=-3 (A1)(A1)
Note: Award (A1) for “x= constant”, (A1) for the constant being -3. The answer must be an equation.
[2 marks]
(-3--10)+-3 (M1)
OR
(-8--10)+2 (M1)
OR
-10+x2=-3 (M1)
OR
diagram showing axis of symmetry and given points (x-values labels, -10, -3 and 4, are sufficient) and an indication that the horizontal distances between the axis of symmetry and the given points are 7. (M1)
Note: Award (M1) for correct working using the symmetry between x=-10 and x=-3. Award (M0) if candidate has used x=-10 and x=4 to show the axis of symmetry is x=-3. Award (M0) if candidate solved f(x)=12 or evaluated f(-10) and f(4).
(x=) 4 (AG)
[1 mark]
-8 and 2 (A1)(A1)
Note: Accept x=-8, y=0 and x=2, y=0 or (-8, 0) and (2, 0), award at most (A0)(A1) if parentheses are omitted.
[2 marks]
(A1)(A1)(A1)(A1)(ft)
Note: Award (A1) for labelled axes with correct scale, correct window. Award (A1) for the vertex, (-3, -12.5), in correct location.
Award (A1) for a smooth continuous curve symmetric about their vertex. Award (A1)(ft) for the curve passing through their x and y intercepts in correct location. Follow through from their parts (a) and (d).
If graph paper is not used:
Award at most (A0)(A0)(A1)(A1)(ft). Their graph should go through their -8 and 2 for the last (A1)(ft) to be awarded.
[4 marks]
y=-12.5 OR y=0x-12.5 (A1)(A1)
Note: Award (A1) for "y= constant", (A1) for the constant being -12.5. The answer must be an equation.
[2 marks]
tangent to the graph drawn at x=-3 (A1)(ft)
Note: Award (A1) for a horizontal straight-line tangent to curve at approximately x=-3. Award (A0) if a ruler is not used. Follow through from their part (e).
[1 mark]
decreasing (A1)
gradient (of tangent line) is negative (at x=a) OR f'(a)<0 (R1)
Note: Do not accept "gradient (of tangent line) is -6". Do not award (A1)(R0).
[2 marks]