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Date	May 2017	Marks available	3	Reference code	17M.1.SL.TZ1.S_9
Level	Standard Level	Paper	Paper 1 (with calculator from previous syllabus)	Time zone	Time zone 1
Command term	Find	Question number	S_9	Adapted from	N/A

Question

A quadratic function $f$ can be written in the form $f(x) = a(x - p)(x - 3)$ . The graph of $f$ has axis of symmetry $x = 2.5$ and $y$ -intercept at $(0,{\text{ }} - 6)$

Find the value of $p$ .

[3]

Find the value of $a$ .

[3]

The line $y = kx - 5$ is a tangent to the curve of $f$ . Find the values of $k$ .

[8]

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an $x$ -intercept (M1)

eg $\,\,\,\,\,$ $x - 3 = 0$ , M17/5/MATME/SP1/ENG/TZ1/09.a/M

valid approach (M1)

eg $\,\,\,\,\,$ $3 - 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5$

$p = 2$ A1 N2

METHOD 2 (expanding f (x))

correct expansion (accept absence of $a$ ) (A1)

eg $\,\,\,\,\,$ $a{x^2} - a(3 + p)x + 3ap,{\text{ }}{x^2} - (3 + p)x + 3p$

valid approach involving equation of axis of symmetry (M1)

eg $\,\,\,\,\,$ $\frac{{ - b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}$

$p = 2$ A1 N2

METHOD 3 (using derivative)

correct derivative (accept absence of $a$ ) (A1)

eg $\,\,\,\,\,$ $a(2x - 3 - p),{\text{ }}2x - 3 - p$

valid approach (M1)

eg $\,\,\,\,\,$ $f’(2.5) = 0$

$p = 2$ A1 N2

[3 marks]

attempt to substitute $(0,{\text{ }} - 6)$ (M1)

eg $\,\,\,\,\,$ $- 6 = a(0 - 2)(0 - 3),{\text{ }}0 = a( - 8)( - 9),{\text{ }}a{(0)^2} - 5a(0) + 6a = - 6$

correct working (A1)

eg $\,\,\,\,\,$ $- 6 = 6a$

$a = - 1$ A1 N2

[3 marks]

METHOD 1 (using discriminant)

recognizing tangent intersects curve once (M1)

recognizing one solution when discriminant = 0 M1

attempt to set up equation (M1)

eg $\,\,\,\,\,$ $g = f,{\text{ }}kx - 5 = - {x^2} + 5x - 6$

rearranging their equation to equal zero (M1)

eg $\,\,\,\,\,$ ${x^2} - 5x + kx + 1 = 0$

correct discriminant (if seen explicitly, not just in quadratic formula) A1

eg $\,\,\,\,\,$ ${(k - 5)^2} - 4,{\text{ }}25 - 10k + {k^2} - 4$

correct working (A1)

eg $\,\,\,\,\,$ $k - 5 = \pm 2,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}$

$k = 3,{\text{ }}7$ A1A1 N0

METHOD 2 (using derivatives)

attempt to set up equation (M1)

eg $\,\,\,\,\,$ $g = f,{\text{ }}kx - 5 = - {x^2} + 5x - 6$

recognizing derivative/slope are equal (M1)

eg $\,\,\,\,\,$ $f’ = {m_T},{\text{ }}f' = k$

correct derivative of $f$ (A1)

eg $\,\,\,\,\,$ $- 2x + 5$

attempt to set up equation in terms of either $x$ or $k$ M1

eg $\,\,\,\,\,$ $( - 2x + 5)x - 5 = - {x^2} + 5x - 6,{\text{ }}k\left( {\frac{{5 - k}}{2}} \right) - 5 = - {\left( {\frac{{5 - k}}{2}} \right)^2} + 5\left( {\frac{{5 - k}}{2}} \right) - 6$

rearranging their equation to equal zero (M1)

eg $\,\,\,\,\,$ ${x^2} - 1 = 0,{\text{ }}{k^2} - 10k + 21 = 0$

correct working (A1)

eg $\,\,\,\,\,$ $x = \pm 1,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}$

$k = 3,{\text{ }}7$ A1A1 N0

[8 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.6—Quadratic function

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