# IBDP Maths AI: Topic: SL 2.5: Quadratic models: IB style Questions SL Paper 1

## Question

The following curves are sketches of the graphs of the functions given below, but in a different order. Using your graphic display calculator, match the equations to the curves, writing your answers in the table below.

(the diagrams are not to scale)

## Markscheme

(i) B     (A1)

(ii) D     (A1)

(iii) A     (A1)

(iv) E     (A1)

(v) C     (A1)

(vi) F     (A1)     (C6)[6 marks]

## Question

Factorise the expression $${x^2} – kx$$ .[1]

a.

Hence solve the equation $${x^2} – kx = 0$$ .[1]

b.

The diagram below shows the graph of the function $$f(x) = {x^2} – kx$$ for a particular value of $$k$$.

Write down the value of $$k$$ for this function.[1]

c.

The diagram below shows the graph of the function $$f(x) = {x^2} – kx$$ for a particular value of $$k$$.

Find the minimum value of the function $$y = f(x)$$ .[3]

d.

## Markscheme

$$x(x – k)$$     (A1)     (C1)[1 mark]

a.

$$x = 0$$ or $$x = k$$     (A1)     (C1)

Note: Both correct answers only.[1 mark]

b.

$$k = 3$$     (A1)     (C1)[1 mark]

c.

$${\text{Vertex at }}x = \frac{{ – ( – 3)}}{{2(1)}}$$     (M1)

Note: (M1) for correct substitution in formula.

$$x = 1.5$$     (A1)(ft)

$$y = – 2.25$$     (A1)(ft)

OR

$$f'(x) = 2x – 3$$     (M1)

Note: (M1) for correct differentiation.

$$x = 1.5$$     (A1)(ft)
$$y = – 2.25$$     (A1)(ft)

OR

for finding the midpoint of their 0 and 3     (M1)
$$x = 1.5$$     (A1)(ft)
$$y = – 2.25$$     (A1)(ft)

Note: If final answer is given as $$(1.5{\text{, }}{- 2.25})$$ award a maximum of (M1)(A1)(A0)[3 marks]

d.

## Question

The graph of a quadratic function $$y = f (x)$$ is given below.

Write down the equation of the axis of symmetry.[2]

a.

Write down the coordinates of the minimum point.[2]

b.

Write down the range of $$f (x)$$.[2]

c.

## Markscheme

x = 3     (A1)(A1)     (C2)

Notes: Award (A1) for “ x = ” (A1) for 3.

The mark for x = is not awarded unless a constant is seen on the other side of the equation.[2 marks]

a.

(3, −14)  (Accept x = 3, y = −14)     (A1)(ft)(A1)     (C2)

Note: Award (A1)(A0) for missing coordinate brackets.[2 marks]

b.

y ≥ −14     (A1)(A1)(ft)     (C2)

Notes: Award (A1) for y ≥ , (A1)(ft) for –14.

Accept alternative notation for intervals.[2 marks]

c.

## Question

A quadratic curve with equation y = ax (xb) is shown in the following diagram.

The x-intercepts are at (0, 0) and (6, 0), and the vertex V is at (h, 8).

Find the value of h.[2]

a.

Find the equation of the curve.[4]

b.

## Markscheme

$$\frac{{0 + 6}}{2} = 3$$   $$h = 3$$     (M1)(A1)     (C2)

Note: Award (M1) for any correct method.[2 marks]

a.

$$y = ax(x – 6)$$     (A1)

$$8 = 3a(- 3)$$     (A1)(ft)

$$a = – \frac{8}{9}$$     (A1)(ft)

$$y = – \frac{8}{9} x (x – 6)$$     (A1)(ft)

Notes: Award (A1) for correct substitution of $$b = 6$$ into equation.

Award (A1)(ft) for substitution of their point V into the equation.

OR

$$y = a(x – 3)^2 + 8$$     (A1)(ft)

Note: Award (A1)(ft) for correct substitution of their h into the equation.

$$0 = a(6 – 3)^2 + 8$$ OR $$0 = a(0 – 3)^2 + 8$$     (A1)

Note: Award (A1) for correct substitution of an x intercept.

$$a = – \frac{8}{9}$$     (A1)(ft)

$$y = – \frac{8}{9}(x – 3)^2 + 8$$     (A1)(ft)     (C4)[4 marks]

b.

## Question

The diagram below shows the graph of a quadratic function. The graph passes through the points (6, 0) and (p, 0). The maximum point has coordinates (0.5, 30.25).

Calculate the value of p.[2]

a.

Given that the quadratic function has an equation $$y = -x^2 + bx + c$$ where $$b,{\text{ }}c \in \mathbb{Z}$$, find $$b$$ and $$c$$.[4]

b.

## Markscheme

$$\frac{{(p + 6)}}{2} = 0.5$$     (M1)

$$p = -5$$     (A1)     (C2)[2 marks]

a.

$$\frac{{ – b}}{{2( – 1)}} = 0.5$$     (M1)

$$b = 1$$     (A1)

$$- 0.5^2 + 0.5 + c = 30.25$$     (M1)

$$c = 30$$     (A1)(ft)

Note: Follow through from their value of b.

OR

$$y = (6 – x) (5 + x)$$     (M1)

$$= 30 + x – x^2$$     (A1)

$$b = 1,{\text{ }}c = 30$$     (A1)(A1)(ft)     (C4)

Note: Follow through from their value of p in part (a).[4 marks]

b.

## Question

A quadratic function, $$f(x) = a{x^2} + bx$$, is represented by the mapping diagram below.

Use the mapping diagram to write down two equations in terms of a and b.[2]

a.

Find the value of a.[1]

b.i.

Find the value of b.[1]

b.ii.

Calculate the x-coordinate of the vertex of the graph of f (x).[2]

c.

## Markscheme

4a + 2b = 20

a + b = 8     (A1)

a – b = –4     (A1)     (C2)

Note: Award (A1)(A1) for any two of the given or equivalent equations.[2 marks]

a.

a = 2     (A1)(ft)[1 mark]

b.i.

b = 6     (A1)(ft)     (C2)

Note: Follow through from their (a).[1 mark]

b.ii.

$$x = – \frac{6}{{2(2)}}$$     (M1)

Note: Award (M1) for correct substitution in correct formula.

$$= -1.5$$     (A1)(ft)     (C2) [2 marks]

c.

## Question

The graph of y = 2x2 $$–$$ rx + q is shown for $$– 5 \leqslant x \leqslant 7$$.

The graph cuts the y axis at (0, 4).

Write down the value of q.[1]

a.

The axis of symmetry is x = 2.5.

Find the value of r.[2]

b.

The axis of symmetry is x = 2.5.

Write down the minimum value of y.[1]

c.

The axis of symmetry is x = 2.5.x

Write down the range of y.[2]

d.

## Markscheme

q = 4     (A1)     (C1)[1 mark]

a.

$$2.5 = \frac{r}{4}$$     (M1)

r = 10     (A1)     (C2)[2 marks]

b.

–8.5     (A1)(ft)     (C1)[1 mark]

c.

$$-8.5 \leqslant y \leqslant 104$$     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for their answer to part (c) with correct inequality signs, (A1)(ft) for 104. Follow through from their values of q and r.

Accept 104 ±2 if read from graph.[2 marks]

d.

## Question

The following is the graph of the quadratic function y = f (x).

Write down the solutions to the equation f (x) = 0.[2]

a.

Write down the equation of the axis of symmetry of the graph of f (x).[2]

b.

The equation f (x) = 12 has two solutions. One of these solutions is x = 6. Use the symmetry of the graph to find the other solution.[1]

c.

The minimum value for y is – 4. Write down the range of f (x).[1]

d.

## Markscheme

x = 0, x = 4     (A1)(A1)     (C2)

Notes: Accept 0 and 4.[2 marks]

a.

x = 2     (A1)(A1)     (C2)

Note: Award (A1) for x = constant, (A1) for 2.[2 marks]
b.

x = –2     (A1)     (C1)

Note: Accept –2.[1 mark]

c.

$$y \geqslant -4{\text{ }}(f(x) \geqslant -4)$$     (A1)     (C1)

Notes: Accept alternative notations.

Award (A0) for use of strict inequality.[1 mark]

d.

## Question

The graph of the quadratic function $$f(x) = 3 + 4x – {x^2}$$ intersects the $$y$$-axis at point A and has its vertex at point B .

Find the coordinates of B .[3]

a.

Another point, C , which lies on the graph of $$y = f(x)$$ has the same $$y$$-coordinate as A .
(i)     Plot and label C on the graph above.
(ii)    Find the $$x$$-coordinate of C .[3]

b.

## Markscheme

$$x = – \frac{4}{{ – 2}}$$     (M1)
$$x = 2$$     (A1)

OR

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4 – 2x$$     (M1)
$$x = 2$$     (A1)
$$(2{\text{, }}7)$$ or $$x = 2$$, $$y = 7$$     (A1)     (C3)

Notes: Award (M1)(A1)(A0) for 2, 7 without parentheses.[3 marks]

a.

(i)     C labelled in correct position on graph     (A1)     (C1)

(ii)    $$3 = 3 + 4x – {x^2}$$     (M1)

Note: Award (M1) for correct substitution of $$y = 3$$ into quadratic.

$$(x = )4$$     (A1)     (C2)

OR

Using symmetry of graph $$x = 2 + 2$$ .     (M1)

Note: Follow through from their $$x$$-coordinate of the vertex.

$$(x = )4$$     (A1)(ft)     (C2)[3 marks]

b.

## Question

The $$x$$-coordinate of the minimum point of the quadratic function $$f(x) = 2{x^2} + kx + 4$$ is $$x =1.25$$.

(i) Find the value of $$k$$ .

(ii) Calculate the $$y$$-coordinate of this minimum point.[4]

a.

Sketch the graph of $$y = f(x)$$ for the domain $$– 1 \leqslant x \leqslant 3$$.

[2]

b.

## Markscheme

(i)     $$1.25 = – \frac{k}{{2(2)}}$$     (M1)

OR

$$f'(x) = 4x + k = 0$$     (M1)

Note: Award (M1) for setting the gradient function to zero.

$$k = – 5$$     (A1)     (C2)

(ii)     $$2{(1.25)^2} – 5(1.25) + 4$$     (M1)

$$= 0.875$$     (A1)(ft)     (C2)

Note: Follow through from their $$k$$.[4 marks]

a.

(A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for a curve with correct concavity consistent with their $$k$$ passing through (0, 4).

(A1)(ft) for minimum in approximately the correct place. Follow through from their part (a).[2 marks]

b.

## Question

y = f (x) is a quadratic function. The graph of f (x) intersects the y-axis at the point A(0, 6) and the x-axis at the point B(1, 0). The vertex of the graph is at the point C(2, –2).

Write down the equation of the axis of symmetry.[2]

a.

Sketch the graph of y = f (x) on the axes below for 0 ≤ x ≤ 4 . Mark clearly on the sketch the points A , B , and C.

[3]

b.

The graph of y = f (x) intersects the x-axis for a second time at point D.

Write down the x-coordinate of point D.[1]

c.

## Markscheme

x = 2     (A1)(A1)     (C2)

Notes: Award (A1)(A0) for “ x = constant” (other than 2). Award (A0)(A1) for y = 2. Award (A0)(A0) for only seeing 2. Award (A0)(A0) for 2 = –b / 2a.[2 marks]

a.

(A1) for correctly plotting and labelling A, B and C

(A1) for a smooth curve passing through the three given points

(A1) for completing the symmetry of the curve over the domain given.     (A3)     (C3)

Notes: For A marks to be awarded for the curve, each segment must be a reasonable attempt at a continuous curve. If straight line segments are used, penalise once only in the last two marks.[3 marks]

b.

3     (A1)(ft)     (C1)

Notes: (A0) for coordinates. Accept x = 3 or D = 3 .[1 mark]

c.

## Question

Part of the graph of the quadratic function f is given in the diagram below.

On this graph one of the x-intercepts is the point (5, 0). The x-coordinate of the maximum point is 3.

The function f is given by $$f (x) = -x^2 + bx + c$$, where $$b,c \in \mathbb{Z}$$

Find the value of

(i) b ;

(ii) c .[3]

a.

The domain of f is  0 ≤ x ≤ 6.

Find the range of f .[3]

b.

## Markscheme

(i) $$3 = \frac{{-b}}{{-2}}$$     (M1)

Note: Award (M1) for correct substitution in formula.

OR

$$-1 + b + c = 0$$

$$-25+5b + c = 0$$

$$-24 + 4b = 0$$     (M1)

Note: Award (M1) for setting up 2 correct simultaneous equations.

OR

$$-2x + b = 0$$     (M1)

Note: Award (M1) for correct derivative of $$f (x)$$ equated to zero.

$$b = 6$$     (A1)     (C2)

(ii) $$0 = -(5)^2 + 6 \times 5 + c$$

$$c =-5$$     (A1)(ft)     (C1)

Note: Follow through from their value for b.

Note: Alternatively candidates may answer part (a) using the method below, and not as two separate parts.

$$(x – 5)(-x +1)$$     (M1)

$$-x^2 +6x – 5$$     (A1)

$$b = 6{\text{ }} c = -5$$     (A1)     (C3)[3 marks]

a.

–5 ≤ y ≤ 4     (A1)(ft)(A1)(ft)(A1)     (C3)

Notes: Accept [–5, 4]. Award (A1)(ft) for –5, (A1)(ft) for 4. (A1) for inequalities in the correct direction or brackets with values in the correct order or a clear word statement of the range. Follow through from their part (a).[3 marks]

b.

## Question

Consider the quadratic function y = f (x) , where f (x) = 5 − x + ax2.

It is given that f (2) = −5 . Find the value of a .[2]

a.

Find the equation of the axis of symmetry of the graph of y = f (x) .[2]

b.

Write down the range of this quadratic function.[2]

c.

## Markscheme

−5 = 5 − (2) + a(2)2     (M1)

Note: Award (M1) for correct substitution in equation.

(a =) −2     (A1)     (C2)[2 marks]

a.

$$x = – \frac{1}{4}$$     (–0.25)     (A1)(A1)(ft)     (C2)

Notes: Follow through from their part (a). Award (A1)(A0)(ft) for “ x = constant”. Award (A0)(A1)(ft) for $$y = – \frac{1}{4}$$.[2 marks]

b.

f (x) ≤ 5.125     (A1)(A1)(ft)     (C2)

Notes: Award (A1) for f (x) ≤ (accept y). Do not accept strict inequality. Award (A1)(ft) for 5.125 (accept 5.13). Accept other correct notation, for example, (−∞, 5.125]. Follow through from their answer to part (b).[2 marks]

c.

## Question

The graph of the quadratic function $$f (x) = c + bx − x^2$$ intersects the y-axis at point A(0, 5) and has its vertex at point B(2, 9).

Write down the value of c.[1]

a.

Find the value of b.[2]

b.

Find the x-intercepts of the graph of f .[2]

c.

Write down $$f (x)$$ in the form $$f (x) = −(x − p) (x + q)$$.[1]

d.

## Markscheme

5     (A1)     (C1)

a.

$$\frac{{ – b}}{{2( – 1)}} = 2$$     (M1)

Note: Award (M1) for correct substitution in axis of symmetry formula.

OR

$$y = 5 + bx – x^2$$

$$9 = 5 + b (2) – (2)^2$$     (M1)

Note: Award (M1) for correct substitution of 9 and 2 into their quadratic equation.

$$(b =) 4$$     (A1)(ft)     (C2)

Note: Follow through from part (a).

b.

5, −1     (A1)(ft)(A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b), irrespective of working shown.

c.

$$f (x) = -(x – 5)(x + 1)$$     (A1)(ft)     (C1)

Notes: Follow through from part (c).

d.

## Question

The graph of the quadratic function $$f(x) = a{x^2} + bx + c$$ intersects the y-axis at point A (0, 5) and has its vertex at point B (4, 13).

Write down the value of $$c$$.[1]

a.

By using the coordinates of the vertex, B, or otherwise, write down two equations in $$a$$ and $$b$$.[3]

b.

Find the value of $$a$$ and of $$b$$.[2]

c.

## Markscheme

5     (A1)     (C1)[1 mark]

a.

at least one of the following equations required

$$a{(4)^2} + 4b + 5 = 13$$

$$4 = – \frac{b}{{2a}}$$

$$a{(8)^2} + 8b + 5 = 5$$     (A2)(A1)     (C3)

Note: Award (A2)(A0) for one correct equation, or its equivalent, and (C3) for any two correct equations.

The equation $$a{(0)^2} + b(0) = 5$$ earns no marks.[3 marks]

b.

$$a = – \frac{1}{2},{\text{ }}b = 4$$     (A1)(ft)(A1)(ft)     (C2)

Note: Follow through from their equations in part (b), but only if their equations lead to unique solutions for $$a$$ and $$b$$.[2 marks]

c.

## Question

The axis of symmetry of the graph of a quadratic function has the equation x $$= – \frac{1}{2}$$

Draw the axis of symmetry on the following axes.

The graph of the quadratic function intersects the x-axis at the point N(2, 0) . There is a second point, M, at which the graph of the quadratic function intersects the x-axis.[1]

a.

Draw the axis of symmetry on the following axes.

[1]

a.

The graph of the quadratic function intersects the $$x$$-axis at the point $${\text{N}}(2, 0)$$. There is a second point, $${\text{M}}$$, at which the graph of the quadratic function intersects the $$x$$-axis.

Clearly mark and label point $${\text{M}}$$ on the axes.[1]

b.

(i)     Find the value of $$b$$ and the value of $$c$$.

(ii)     Draw the graph of the function on the axes.[4]

c.

## Markscheme

vertical straight line which may be dotted passing through $$\left( { – \frac{1}{2},{\text{ }}0} \right)$$     (A1)     (C1)

a.

vertical straight line which may be dotted passing through $$\left( { – \frac{1}{2},{\text{ }}0} \right)$$     (A1)     (C1)

a.

point $${\text{M }}( – 3,{\text{ }}0)$$ correctly marked on the $$x$$-axis     (A1)(ft)     (C1)

Note: Follow through from part (a).

b.

(i)     $$b = 1$$, $$c = – 6$$     (A1)(ft)(A1)(ft)

(ii)     smooth parabola passing through $${\text{M}}$$ and $${\text{N}}$$     (A1)(ft)

Note: Follow through from their point $${\text{M}}$$ from part (b).

parabola passing through $$(0,{\text{ }} – 6)$$ and symmetrical about $$x = – 0.5$$     (A1)(ft) (C4)

Note: Follow through from part (c)(i).

If parabola is not smooth and not concave up award at most (A1)(A0).

c.

## Question

Consider the quadratic function, $$f(x) = px(q – x)$$, where $$p$$ and $$q$$ are positive integers.

The graph of $$y = f(x)$$ passes through the point $$(6,{\text{ }}0)$$.

Calculate the value of $$q$$.[2]

a.

The vertex of the function is $$(3,{\text{ }}27)$$.

Find the value of $$p$$.[2]

b.

The vertex of the function is $$(3,{\text{ }}27)$$.

Write down the range of $$f$$.[2]

c.

## Markscheme

$$0 = p(6)(q – 6)$$      (M1)

$$q = 6$$     (A1)

OR

$$f(x) = – p{x^2} + pqx$$

$$3 = \frac{{ – pq}}{{ – 2p}}$$     (M1)

$$q = 6$$     (A1)

OR

$$f(x) = – p{x^2} + pqx$$

$$f'(x) = pq – 2px$$     (M1)

$$pq – 2p(3) = 0$$

$$q = 6$$     (A1)     (C2)

a.

$$27 = p(3)(6 – 3)$$     (M1)

Note: Award (M1) for correct substitution of the vertex $$(3,{\text{ }}27)$$ and their $$q$$ into or equivalent $$f(x) = px(q – x)$$ or equivalent.

$$p = 3$$     (A1)(ft)     (C2)

Note: Follow through from part (a).

b.

$$y \leqslant 27\;\;\;\left( {f(x) \leqslant 27} \right)$$     (A1)(A1)     (C2)

Notes: Award (A1) for $$y \leqslant {\text{ }}\left( {{\text{or }}f(x) \leqslant } \right)$$, (A1) for $$27$$ as part of an inequality.

Accept alternative notation: $$( – \infty ,{\text{ }}27],{\text{ }}] – \infty ,{\text{ }}27]$$.

Award (A0)(A1) for $$[27,{\text{ }}- \infty )$$.

Award (A0)(A0) for $$( – \infty ,{\text{ }}\infty )$$.

c.

## Question

A building company has many rectangular construction sites, of varying widths, along a road.

The area, $$A$$, of each site is given by the function

$A(x) = x(200 – x)$

where $$x$$ is the width of the site in metres and $$20 \leqslant x \leqslant 180$$.

Site S has a width of $$20$$ m. Write down the area of S.[1]

a.

Site T has the same area as site S, but a different width. Find the width of T.[2]

b.

When the width of the construction site is $$b$$ metres, the site has a maximum area.

(i)     Write down the value of $$b$$.

(ii)     Write down the maximum area.[2]

c.

The range of $$A(x)$$ is $$m \leqslant A(x) \leqslant n$$.

Hence write down the value of $$m$$ and of $$n$$.[1]

d.

## Markscheme

$$3600{\text{ (}}{{\text{m}}^2})$$     (A1)(C1)

a.

$$x(200 – x) = 3600$$     (M1)

Note: Award (M1) for setting up an equation, equating to their $$3600$$.

$$180{\text{ (m)}}$$     (A1)(ft)     (C2)

b.

(i)     $$100{\text{ (m)}}$$     (A1)     (C1)

(ii)     $$10\,000{\text{ (}}{{\text{m}}^2})$$     (A1)(ft)(C1)

c.

$$m = 3600\;\;\;$$and$$\;\;\;n = 10\,000$$     (A1)(ft)     (C1)

Notes: Follow through from part (a) and part (c)(ii), but only if their $$m$$ is less than their $$n$$. Accept the answer $$3600 \leqslant A \leqslant 10\,000$$.

d.

## Question

Consider the function $$f(x) = a{x^2} + c$$.

Find $$f'(x)$$[1]

a.

Point $${\text{A}}( – 2,\,5)$$  lies on the graph of $$y = f(x)$$ . The gradient of the tangent to this graph at $${\text{A}}$$ is $$– 6$$ .

Find the value of $$a$$ .[3]

b.

Find the value of $$c$$ .[2]

c.

## Markscheme

$$2ax$$      (A1)   (C1)

Note: Award (A1) for $$2ax$$.  Award (A0) if other terms are seen.

a.

$$2a( – 2) = – 6$$       (M1)(M1)

Note: Award (M1) for correct substitution of $$x = – 2$$  in their gradient function, (M1) for equating their gradient function to $$– 6$$ . Follow through from part (a).

$$(a = )1.5\,\,\,\left( {\frac{3}{2}} \right)$$       (A1)(ft) (C3)

b.

$${\text{their }}1.5 \times {( – 2)^2} + c = 5$$         (M1)

Note: Award (M1) for correct substitution of their $$a$$ and point $${\text{A}}$$. Follow through from part (b).

$$(c = ) – 1$$         (A1)(ft) (C2)

c.

## Question

The graph of a quadratic function has $$y$$-intercept 10 and one of its $$x$$-intercepts is 1.

The $$x$$-coordinate of the vertex of the graph is 3.

The equation of the quadratic function is in the form $$y = a{x^2} + bx + c$$.

Write down the value of $$c$$.[1]

a.

Find the value of $$a$$ and of $$b$$.[4]

b.

Write down the second $$x$$-intercept of the function.[1]

c.

## Markscheme

10     (A1)     (C1)

Note:     Accept $$(0,{\text{ }}10)$$.[1 mark]

a.

$$3 = \frac{{ – b}}{{2a}}$$

$$0 = a{(1)^2} + b(1) + c$$

$$10 = a{(6)^2} + b(6) + c$$

$$0 = a{(5)^2} + b(5) + c$$     (M1)(M1)

Note:     Award (M1) for each of the above equations, provided they are not equivalent, up to a maximum of (M1)(M1). Accept equations that substitute their 10 for $$c$$.

OR

sketch graph showing given information: intercepts $$(1,{\text{ }}0)$$ and $$(0,{\text{ }}10)$$ and line $$x = 3$$     (M1)

$$y = a(x – 1)(x – 5)$$     (M1)

Note:     Award (M1) for $$(x – 1)(x – 5)$$ seen.

$$a = 2$$     (A1)(ft)

$$b = – 12$$     (A1)(ft)     (C4)

Note:     Follow through from part (a).

If it is not clear which is $$a$$ and which is $$b$$ award at most (A0)(A1)(ft).[4 marks]

b.

5     (A1)     (C1)[1 mark]

c.

## Question

Consider the following graphs of quadratic functions.

The equation of each of the quadratic functions can be written in the form $$y = a{x^2} + bx + c$$, where $$a \ne 0$$.

Each of the sets of conditions for the constants $$a$$, $$b$$ and $$c$$, in the table below, corresponds to one of the graphs above.

Write down the number of the corresponding graph next to each set of conditions.

## Markscheme

(A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note:     Award (A1) for each correct entry.[6 marks]

## Question

A quadratic function $$f$$ is given by $$f(x) = a{x^2} + bx + c$$. The points $$(0,{\text{ }}5)$$ and $$( – 4,{\text{ }}5)$$ lie on the graph of $$y = f(x)$$.

The $$y$$-coordinate of the minimum of the graph is 3.

Find the equation of the axis of symmetry of the graph of $$y = f(x)$$.[2]

a.

Write down the value of $$c$$.[1]

b.

Find the value of $$a$$ and of $$b$$.[3]

c.

## Markscheme

$$x = – 2$$     (A1)(A1)     (C2)

Note:     Award (A1) for $$x =$$ (a constant) and (A1) for $$– 2$$.[2 marks]

a.

$$(c = ){\text{ }}5$$     (A1)     (C1)[1 mark]

b.

$$– \frac{b}{{2a}} = – 2$$

$$a{( – 2)^2} – 2b + 5 = 3$$ or equivalent

$$a{( – 4)^2} – 4b + 5 = 5$$ or equivalent

$$2a( – 2) + b = 0$$ or equivalent     (M1)

Note:     Award (M1) for two of the above equations.

$$a = 0.5$$     (A1)(ft)

$$b = 2$$     (A1)(ft)     (C3)

Note:     Award at most (M1)(A1)(ft)(A0) if the answers are reversed.

Follow through from parts (a) and (b).[3 marks]

c.

## Question

Maria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, $$Q$$, is given by

$$Q = 882 – 45p$$,

where $$p$$ is the price of a kilogram of cheese in euros (EUR).

Maria earns $$(p – 6.80){\text{ EUR}}$$ for each kilogram of cheese sold.

To calculate her weekly profit $$W$$, in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.

Write down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.[1]

a.

Find how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.[2]

b.

Write down an expression for $$W$$ in terms of $$p$$.[1]

c.

Find the price, $$p$$, that will give Maria the highest weekly profit.[2]

d.

## Markscheme

522 (kg)     (A1)     (C1)[1 mark]

a.

$$522(8 – 6.80)$$ or equivalent     (M1)

Note:     Award (M1) for multiplying their answer to part (a) by $$(8 – 6.80)$$.

626 (EUR) (626.40)     (A1)(ft)     (C2)

Note:     Follow through from part (a).[2 marks]

b.

$$(W = ){\text{ }}(882 – 45p)(p – 6.80)$$     (A1)

OR

$$(W = ) – 45{p^2} + 1188p – 5997.6$$     (A1)     (C1)[1 mark]

c.

sketch of $$W$$ with some indication of the maximum     (M1)

OR

$$– 90p + 1188 = 0$$     (M1)

Note:     Award (M1) for equating the correct derivative of their part (c) to zero.

OR

$$(p = ){\text{ }}\frac{{ – 1188}}{{2 \times ( – 45)}}$$     (M1)

Note:     Award (M1) for correct substitution into the formula for axis of symmetry.

$$(p = ){\text{ }}13.2{\text{ (EUR)}}$$     (A1)(ft)     (C2)

Note:     Follow through from their part (c), if the value of $$p$$ is such that $$6.80 < p < 19.6$$.[2 marks]

d.

## Question

Consider the quadratic function $$f\left( x \right) = a{x^2} + bx + 22$$.

The equation of the line of symmetry of the graph $$y = f\left( x \right){\text{ is }}x = 1.75$$.

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.[1]

a.

Using this information, write down a second equation in terms of a and b.[1]

b.

Hence find the value of a and of b.[2]

c.

The graph intersects the x-axis at a second point, P.

Find the x-coordinate of P.[2]

d.

## Markscheme

$$1.75 = \frac{{ – b}}{{2a}}$$ (or equivalent)      (A1) (C1)

Note: Award (A1) for $$f\left( x \right) = {\left( {1.75} \right)^2}a + 1.75b$$ or for $$y = {\left( {1.75} \right)^2}a + 1.75b + 22$$ or for $$f\left( {1.75} \right) = {\left( {1.75} \right)^2}a + 1.75b + 22$$.[1 mark]

a.

$${\left( { – 2} \right)^2} \times a + \left( { – 2} \right) \times b + 22 = 0$$ (or equivalent)      (A1) (C1)

Note: Award (A1) for $${\left( { – 2} \right)^2} \times a + \left( { – 2} \right) \times b + 22 = 0$$ seen.

Award (A0) for $$y = {\left( { – 2} \right)^2} \times a + \left( { – 2} \right) \times b + 22$$.[1 mark]

b.

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]

c.

−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).

(=) 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.

(=) 5.5     (A1) (C2)[2 marks]

d.

## Question

A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by

C(x) = (x − 75)2 + 100.

The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.

Find the cost of producing 70 shirts.[2]

a.

Find the value of s.[2]

b.

Find the number of shirts produced when the cost of production is lowest.[2]

c.

## Markscheme

(70 − 75)2 + 100     (M1)

Note: Award (M1) for substituting in x = 70.

125     (A1) (C2)[2 marks]

a.

(s − 75)2 + 100 = 500     (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

OR

(M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95    (A1) (C2)[2 marks]

b.

(M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

OR

$$\frac{{95 + 55}}{2}$$     (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

OR

(C’(x) =) 2x − 150 = 0     (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75     (A1) (C2)[2 marks]

c.

## Question

Consider the function $$f\left( x \right) = \frac{{{x^4}}}{4}$$.

Find f’(x)[1]

a.

Find the gradient of the graph of f at $$x = – \frac{1}{2}$$.[2]

b.

Find the x-coordinate of the point at which the normal to the graph of f has gradient $${ – \frac{1}{8}}$$.[3]

c.

## Markscheme

x3     (A1) (C1)

Note: Award (A0) for $$\frac{{4{x^3}}}{4}$$ and not simplified to x3.[1 mark]

a.

$${\left( { – \frac{1}{2}} \right)^3}$$     (M1)

Note: Award (M1) for correct substitution of $${ – \frac{1}{2}}$$ into their derivative.

$${ – \frac{1}{8}}$$  (−0.125)     (A1)(ft) (C2)

Note: Follow through from their part (a).[2 marks]

b.

x3 = 8     (A1)(M1)

Note: Award (A1) for 8 seen maybe seen as part of an equation y = 8x + c(M1) for equating their derivative to 8.

(x =) 2     (A1) (C3)

Note: Do not accept (2, 4).[3 marks]

c.

## Question

A small manufacturing company makes and sells $$x$$ machines each month. The monthly cost $$C$$ , in dollars, of making $$x$$ machines is given by
$C(x) = 2600 + 0.4{x^2}{\text{.}}$The monthly income $$I$$ , in dollars, obtained by selling $$x$$ machines is given by
$I(x) = 150x – 0.6{x^2}{\text{.}}$$$P(x)$$ is the monthly profit obtained by selling $$x$$ machines.

Find $$P(x)$$ .[2]

a.

Find the number of machines that should be made and sold each month to maximize $$P(x)$$ .[2]

b.

Use your answer to part (b) to find the selling price of each machine in order to maximize $$P(x)$$ .[2]

c.

## Markscheme

$$P(x) = I(x) – C(x)$$     (M1)
$$= – {x^2} + 150x – 2600$$     (A1)    (C2)

a.

$$– 2x + 150 = 0$$     (M1)

Note: Award (M1) for setting $$P'(x) = 0$$ .

OR

Award (M1) for sketch of $$P(x)$$ and maximum point identified.     (M1)
$$x = 75$$     (A1)(ft)     (C2)

$$\frac{{7875}}{{75}}$$     (M1)
Note: Award (M1) for $$7875$$ seen.
$$= 105$$ (A1)(ft)     (C2)