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Date	May 2022	Marks available	2	Reference code	22M.1.SL.TZ2.7
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Write down	Question number	7	Adapted from	N/A

Question

The following diagram shows part of the graph of a quadratic function $f$ .

The graph of $f$ has its vertex at $(3, 4)$ , and it passes through point $Q$ as shown.

The function can be written in the form $f (x) = a {(x - h)}^{2} + k$ .

The line $L$ is tangent to the graph of $f$ at $Q$ .

Now consider another function $y = g (x)$ . The derivative of $g$ is given by $g' (x) = f (x) - d$ , where $d \in ℝ$ .

Write down the equation of the axis of symmetry.

[1]

Write down the values of $h$ and $k$ .

[2]

b.i.

Point $Q$ has coordinates $(5, 12)$ . Find the value of $a$ .

[2]

b.ii.

Find the equation of $L$ .

[4]

Find the values of $d$ for which $g$ is an increasing function.

[3]

Find the values of $x$ for which the graph of $g$ is concave-up.

[3]

Markscheme

$x = 3$ A1

Note: Must be an equation in the form “ $x =$ ”. Do not accept $3$ or $\frac{- b}{2 a} = 3$ .

[1 mark]

$h = 3, k = 4$ (accept $a {(x - 3)}^{2} + 4$ ) A1A1

[2 marks]

b.i.

attempt to substitute coordinates of $Q$ (M1)

$12 = a {(5 - 3)}^{2} + 4, 4 a + 4 = 12$

$a = 2$ A1

[2 marks]

b.ii.

recognize need to find derivative of $f$ (M1)

$f' (x) = 4 (x - 3)$ or $f' (x) = 4 x - 12$ A1

$f' (5) = 8$ (may be seen as gradient in their equation) (A1)

$y - 12 = 8 (x - 5)$ or $y = 8 x - 28$ A1

Note: Award A0 for $L = 8 x - 28$ .

[4 marks]

METHOD 1

Recognizing that for $g$ to be increasing, $f (x) - d > 0$ , or $g' > 0$ (M1)

The vertex must be above the $x$ -axis, $4 - d > 0, d - 4 < 0$ (R1)

$d < 4$ A1

METHOD 2

attempting to find discriminant of $g'$ (M1)

${(- 12)}^{2} - 4 (2) (22 - d)$

recognizing discriminant must be negative (R1)

$- 32 + 8 d < 0$ OR $Δ < 0$

$d < 4$ A1

[3 marks]

recognizing that for $g$ to be concave up, $g'' > 0$ (M1)

$g'' > 0$ when $f' > 0, 4 x - 12 > 0, x - 3 > 0$ (R1)

$x > 3$ A1

[3 marks]

Examiners report

In parts (a) and (b) of this question, a majority of candidates recognized the connection between the coordinates of the vertex and the axis of symmetry and the values of $h$ and $k$ , and most candidates were able to successfully substitute the coordinates of point Q to find the value of $a$ . In part (c), the candidates who recognized the need to use the derivative to find the gradient of the tangent were generally successful in finding the equation of the line, although many did not give their equation in the proper form in terms of $x$ and $y$ , and instead wrote $L = 8 x - 28$ , thus losing the final mark. Parts (d) and (e) were much more challenging for candidates. Although a good number of candidates recognized that $g' (x) > 0$ in part (d), and $g'' (x) > 0$ in part (e), very few were able to proceed beyond this point and find the correct inequalities for their final answers.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.6—Quadratic function

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18M.2.AHL.TZ2.H_10a.iii:
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21N.1.SL.TZ0.1a.i:
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18M.2.AHL.TZ2.H_10a.ii:
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18M.2.AHL.TZ2.H_10a.iv:
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18M.2.AHL.TZ2.H_10a.i:
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21M.1.SL.TZ1.7d:
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22M.1.SL.TZ1.7b.ii:
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22M.1.SL.TZ1.7c:
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EXN.1.SL.TZ0.3:
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22M.1.SL.TZ2.7b.ii:
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22M.1.AHL.TZ2.11c:
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16N.1.SL.TZ0.S_1b:
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21M.2.SL.TZ2.3b:
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21M.2.SL.TZ2.5a:
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21M.2.SL.TZ2.5b:
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19N.1.SL.TZ0.S_3b:
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