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Date	November 2019	Marks available	2	Reference code	19N.2.SL.TZ0.T_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Write down	Question number	T_4	Adapted from	N/A

Question

The graph of the quadratic function $f (x) = \frac{1}{2} (x - 2) (x + 8)$ $f (x) = \frac{1}{2} (x - 2) (x + 8)$ intersects the $y$ $y$ -axis at $(0, c)$ $(0, c)$ .

The vertex of the function is $(- 3, - 12.5)$ $(- 3, - 12.5)$ .

The equation $f (x) = 12$ $f (x) = 12$ has two solutions. The first solution is $x = - 10$ $x = - 10$ .

Let $T$ $T$ be the tangent at $x = - 3$ $x = - 3$ .

Find the value of $c$ $c$ .

[2]

Write down the equation for the axis of symmetry of the graph.

[2]

Use the symmetry of the graph to show that the second solution is $x = 4$ $x = 4$ .

[1]

Write down the $x$ $x$ -intercepts of the graph.

[2]

On graph paper, draw the graph of $y = f (x)$ $y = f (x)$ for $- 10 \leq x \leq 4$ $- 10 \leq x \leq 4$ and $- 14 \leq y \leq 14$ $- 14 \leq y \leq 14$ . Use a scale of $1 cm$ $1 cm$ to represent $1$ $1$ unit on the $x$ $x$ -axis and $1 cm$ $1 cm$ to represent $2$ $2$ units on the $y$ $y$ -axis.

[4]

Write down the equation of $T$ $T$ .

[2]

f.i.

Draw the tangent $T$ $T$ on your graph.

[1]

f.ii.

Given $f (a) = 5.5$ $f (a) = 5.5$ and $f' (a) = - 6$ $f' (a) = - 6$ , state whether the function, $f$ $f$ , is increasing or decreasing at $x = a$ $x = a$ . Give a reason for your answer.

[2]

Markscheme

$\frac{1}{2} (0 - 2) (0 + 8)$ $\frac{1}{2} (0 - 2) (0 + 8)$ OR $\frac{1}{2} (0^{2} + 6 (0) - 16)$ $\frac{1}{2} (0^{2} + 6 (0) - 16)$ (or equivalent) (M1)

Note: Award (M1) for evaluating $f (0)$ $f (0)$ .

$(c =) - 8$ $(c =) - 8$ (A1)(G2)

Note: Award (G2) if $- 8$ $- 8$ or $(0, - 8)$ $(0, - 8)$ seen.

[2 marks]

$x = - 3$ $x = - 3$ (A1)(A1)

Note: Award (A1) for “ $x =$ $x =$ constant”, (A1) for the constant being $- 3$ $- 3$ . The answer must be an equation.

[2 marks]

$(- 3 - - 10) + - 3$ $(- 3 - - 10) + - 3$ (M1)

$(- 8 - - 10) + 2$ $(- 8 - - 10) + 2$ (M1)

$\frac{- 10 + x}{2} = - 3$ $\frac{- 10 + x}{2} = - 3$ (M1)

diagram showing axis of symmetry and given points ( $x$ $x$ -values labels, $- 10$ $- 10$ , $- 3$ $- 3$ and $4$ $4$ , are sufficient) and an indication that the horizontal distances between the axis of symmetry and the given points are $7$ $7$ . (M1)

Note: Award (M1) for correct working using the symmetry between $x = - 10$ $x = - 10$ and $x = - 3$ $x = - 3$ . Award (M0) if candidate has used $x = - 10$ $x = - 10$ and $x = 4$ $x = 4$ to show the axis of symmetry is $x = - 3$ $x = - 3$ . Award (M0) if candidate solved $f (x) = 12$ $f (x) = 12$ or evaluated $f (- 10)$ $f (- 10)$ and $f (4)$ $f (4)$ .

$(x =) 4$ $(x =) 4$ (AG)

[1 mark]

$- 8$ $- 8$ and $2$ $2$ (A1)(A1)

Note: Accept $x = - 8, y = 0$ $x = - 8, y = 0$ and $x = 2, y = 0$ $x = 2, y = 0$ or $(- 8, 0)$ $(- 8, 0)$ and $(2, 0)$ $(2, 0)$ , 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, $(- 3, - 12.5)$ $(- 3, - 12.5)$ , in correct location.
Award (A1) for a smooth continuous curve symmetric about their vertex. Award (A1)(ft) for the curve passing through their $x$ $x$ and $y$ $y$ 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 $- 8$ $- 8$ and $2$ $2$ for the last (A1)(ft) to be awarded.

[4 marks]

$y = - 12.5$ $y = - 12.5$ OR $y = 0 x - 12.5$ $y = 0 x - 12.5$ (A1)(A1)

Note: Award (A1) for " $y =$ $y =$ constant", (A1) for the constant being $- 12.5$ $- 12.5$ . The answer must be an equation.

[2 marks]

f.i.

tangent to the graph drawn at $x = - 3$ $x = - 3$ (A1)(ft)

Note: Award (A1) for a horizontal straight-line tangent to curve at approximately $x = - 3$ $x = - 3$ . Award (A0) if a ruler is not used. Follow through from their part (e).