Exponential and quadratic functions problems [78 marks]

1a. [2 marks]

Consider the functions and .

Write down

(i) the -intercept of the graph of ;

(ii) the -intercept of the graph of .

Markscheme(i) (A1)

Note: Accept .

(ii) (A1) (C2)

Note: Accept .

f(x) = x + 1 g(x) = − 23x

x y = f(x)

y y = g(x)

(−1, 0)

−1

(0, − 1)

−1

1b. [2 marks]Solve .

Markscheme (A1)

(A1) (C2)

f(x) = g(x)

(x =) − 2.96 (−2.96135 …)

(x =) 1.34 (1.33508 …)

( ) > ( )

1c. [2 marks]Write down the interval for the values of for which .

Markscheme (A1)(ft)(A1) (C2)

Notes: Award (A1)(ft) for both correct endpoints of the interval, (A1) for correct strict inequalities (or correct openinterval notation).

Follow through from part (b).

x f(x) > g(x)

−2.96 < x < 1.34 OR ]−2.96, 1.34[ OR (−2.96, 1.34)

1

2a. [1 mark]

The axis of symmetry of the graph of a quadratic function has the equation x

.

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 graphof the quadratic function intersects the x-axis.

Markscheme

vertical straight line which may be dotted passing through (A1) (C1)

= − 12

(− , 0)12

2b. [1 mark]Draw the axis of symmetry on the following axes.

Markscheme

vertical straight line which may be dotted passing through (A1) (C1)(− , 0)12

2c. [1 mark]The graph of the quadratic function intersects the -axis at the point . There is a second point, , at whichthe graph of the quadratic function intersects the -axis.

Clearly mark and label point on the axes.

Markschemepoint correctly marked on the -axis (A1)(ft) (C1)

Note: Follow through from part (a).

x N(2, 0) Mx

M

M (−3, 0) x

2d. [4 marks](i) Find the value of and the value of .

(ii) Draw the graph of the function on the axes.

Markscheme(i)

, (A1)(ft)(A1)(ft)

Notes: Follow through from (b).

(ii) smooth parabola passing through and (A1)(ft)

Note: Follow through from their point from part (b).

parabola passing through and symmetrical about (A1)(ft) (C4)

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

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

b c

b = 1 c = −6

M N

M

(0, − 6) x = −0.5

3a. [2 marks]

Consider the quadratic function, , where and are positive integers.

The graph of passes through the point .

Calculate the value of .

f(x) = px(q − x) p q

y = f(x) (6, 0)

q

Markscheme (M1)

(A1)

OR

(M1)

(A1)

OR

(M1)

(A1) (C2)

0 = p(6)(q − 6)

q = 6

f(x) = −p + pqxx2

3 = −pq

−2p

q = 6

f(x) = −p + pqxx2

(x) = pq − 2pxf ′

pq − 2p(3) = 0

q = 6

3b. [2 marks]The vertex of the function is .

Find the value of .

Markscheme (M1)

Note: Award (M1) for correct substitution of the vertex and their into or equivalent or equivalent.

(A1)(ft) (C2)

Note: Follow through from part (a).

(3, 27)

p

27 = p(3)(6 − 3)

(3, 27) q f(x) = px(q − x)

p = 3

(3, 27)

3c. [2 marks]The vertex of the function is .

Write down the range of .

Markscheme (A1)(A1) (C2)

Notes: Award (A1) for , (A1) for as part of an inequality.

Accept alternative notation: .

Award (A0)(A1) for .

Award (A0)(A0) for .

(3, 27)

f

y ⩽ 27 (f(x) ⩽ 27)

y ⩽ (or f(x) ⩽) 27

(−∞, 27], ] − ∞, 27]

[27, − ∞)

(−∞, ∞)

4a. [2 marks]

The number of fish, , in a pond is decreasing according to the model

where and are positive constants, and is the time in months since the number of fish in the pond was first counted.

At the beginning fish were counted.

Find the value of .

Markscheme (M1)

Note: Award (M1) for substituting and equating to .

(A1)(C2)

N

N(t) = a + 40, t ⩾ 0b−t

abt

840

a

a + 40 = 840b0

t = 0 840

a = 800

4 90

4b. [3 marks]After months fish were counted.

Find the value of .

Markscheme (M1)

Note: Award (M1) for correct substitution of their (from part (a)) and in the formula of the function and equating to .

OR OR OR (M1)

Notes: Award second (M1) for correctly rearranging their equation and eliminating the negative index (see aboveexamples).

Accept in place of.

OR

(M1)(M1)

Notes: Award (M1) for a decreasing exponential and a horizontal line that are both in the first quadrant, and (M1) fortheir graphs intersecting.

For graphs drawn in both first and second quadrants award at most (M1)(M0).

(A1)(ft) (C3)

Note: Follow through from their answer to part (a) only if is positive.

4 90

b

800 + 40 = 90b−4

a4 90

= 16b4 =1b4

116

b = 16−−√4 b = 1614

80050

16

b = 2

a

4c. [1 mark]The number of fish in the pond will not decrease below .

Write down the value of .

Markscheme (A1) (C1)

p

p

40

5a. [3 marks]

George leaves a cup of hot coffee to cool and measures its temperature every minute. His results are shown in the table below.

Write down the decrease in the temperature of the coffee

(i) during the first minute (between t = 0 and t =1) ;

(ii) during the second minute;

(iii) during the third minute.

Markscheme(i) 40

(ii) 20

(iii) 10 (A3)

Notes: Award (A0)(A1)(ft)(A1)(ft) for −40, −20, −10.

Award (A1)(A0)(A1)(ft) for 40, 60, 70 seen.

Award (A0)(A0)(A1)(ft) for −40, −60, −70 seen.

5b. [2 marks]Assuming the pattern in the answers to part (a) continues, show that.

Markscheme or equivalent (A1)(M1)

Note: Award (A1) for 5 seen, (M1) for difference from 24 indicated.

(AG)

Note: If 19 is not seen award at most (A1)(M0).

k = 19

24 − k = 5

k = 19

5c. [4 marks]Use the seven results in the table to draw a graph that shows how the temperature of the coffee changes during the first sixminutes.

Use a scale of 2 cm to represent 1 minute on the horizontal axis and 1 cm to represent 10 °C on the vertical axis.

Markscheme

(A1)(A1)(A1)(A1)

Note: Award (A1) for scales and labelled axes (t or “time” and y or “temperature”).

Accept the use of x on the horizontal axis only if “time” is also seen as the label.

Award (A2) for all seven points accurately plotted, award (A1) for 5 or 6 points accurately plotted, award (A0) for 4 points orfewer accurately plotted.

Award (A1) for smooth curve that passes through all points on domain [0, 6].

If graph paper is not used or one or more scales is missing, award a maximum of (A0)(A0)(A0)(A1).

5d. [2 marks]The function that models the change in temperature of the coffee is y = p (2 )+ q.

(i) Use the values t = 0 and y = 94 to form an equation in p and q.

(ii) Use the values t =1 and y = 54 to form a second equation in p and q.

−t

Markscheme(i)

(A1)

(ii) (A1)

Note: The equations need not be simplified; accept, for example.

94 = p + q

54 = 0.5p + q

94 = p( ) + q2−0

[2 marks]5e. Solve the equations found in part (d) to find the value of p and the value of q.

Markschemep = 80, q = 14 (G1)(G1)(ft)

Note: If the equations have been incorrectly simplified, follow through even if no working is shown.

5f. [2 marks]The graph of this function has a horizontal asymptote.

Write down the equation of this asymptote.

Markschemey = 14 (A1)(A1)(ft)

Note: Award (A1) for y = a constant, (A1) for their 14. Follow through from part (e) only if their q lies between 0 and 15.25inclusive.

5g. [4 marks]George decides to model the change in temperature of the coffee with a linear function using correlation and linear regression.

Use the seven results in the table to write down

(i) the correlation coefficient;

(ii) the equation of the regression line y on t.

Markscheme(i) –0.878 (–0.87787...) (G2)

Note: Award (G1) if –0.877 seen only. If negative sign omitted award a maximum of (A1)(A0).

(ii) y = –11.7t + 71.6 (y = –11.6517...t + 71.6336...) (G1)(G1)

Note: Award (G1) for –11.7t, (G1) for 71.6.

If y = is omitted award at most (G0)(G1).

If the use of x in part (c) has not been penalized (the axis has been labelled “time”) then award at most (G0)(G1).

[2 marks]5h. Use the equation of the regression line to estimate the temperature of the coffee at t = 3.

Markscheme−11.6517...(3) + 71.6339... (M1)

Note: Award (M1) for correct substitution in their part (g)(ii).

= 36.7 (36.6785...) (A1)(ft)(G2)

Note: Follow through from part (g). Accept 36.5 for use of the 3sf answers from part (g).

[2 marks]5i. Find the percentage error in this estimate of the temperature of the coffee at t = 3.

Markscheme (M1)

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

= 52.8% (52.82738...) (A1)(ft)(G2)

Note: Follow through from part (h). Accept 52.1% for use of 36.5.

Accept 52.9 % for use of 36.7. If partial working ( omitted) is followed by their correct answer award (M1)(A1). If partial working is followed by an incorrect answer award

(M0)(A0). The percentage sign is not required.

× 10036.6785...−2424

×100

[2 marks]6a.

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). Thevertex of the graph is at the point C(2, –2).

Write down the equation of the axis of symmetry.

Markschemex = 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]

6b. [3 marks]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.

Markscheme

(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 linesegments are used, penalise once only in the last two marks.

[3 marks]

6c. [1 mark]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.

Markscheme3 (A1)(ft) (C1)

Notes: (A0) for coordinates. Accept x = 3 or D = 3 .

[1 mark]

7a. [4 marks]

The diagram shows part of the graph of, and its horizontal asymptote. The graph passes through the points (0, a) and (b, 3.5).

Find the value of

(i) a ;

(ii) b .

Markscheme(i) 2 + 3 (M1)

Note: Award (M1) for correct substitution.

= 4 (A1) (C2)

(ii) 3.5 = 2 + 3 (M1)

Note: Award (M1) for correct substitution.

b = 1 (A1) (C2)

[4 marks]

y = + 32−x

0

−b

[2 marks]7b. Write down the equation of the horizontal asymptote to this graph.

Markschemey = 3 (A1)(A1) (C2)

Notes: y = constant (other than 3) award (A1)(A0).

[2 marks]

[1 mark]8a.

The graph of the quadratic function intersects the y-axis at point A(0, 5) and has its vertex at point B(2, 9).

Write down the value of c.

Markscheme5 (A1) (C1)

f(x) = c + bx − x2

[2 marks]8b. Find the value of b.

Markscheme (M1)

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

OR

(M1)

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

(A1)(ft) (C2)

Note: Follow through from part (a).

= 2−b

2(−1)

y = 5 + bx − x2

9 = 5 + b(2) − (2)2

(b =)4

[2 marks]8c. Find the x-intercepts of the graph of f .

Markscheme5, −1 (A1)(ft)(A1)(ft) (C2)

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

8d. [1 mark]Write down in the form

.f(x)f(x) = −(x − p)(x + q)

Markscheme (A1)(ft) (C1)

Notes: Follow through from part (c).

f(x) = −(x − 5)(x + 1)

9a. [1 mark]

The number of bacteria in a colony is modelled by the function

,

where is the number of bacteria and

is the time in hours.

Write down the number of bacteria in the colony at time.

Markscheme800 (A1) (C1)

N(t) = 800 × , t ⩾ 030.5t

N

t

t = 0

[3 marks]9b. Calculate the number of bacteria present at 2 hours and 30 minutes. Give your answer correct to the nearest hundred bacteria.

Markscheme (M1)

Note: Award (M1) for correctly substituted formula.

... (A1)

(A1) (C3)

Notes: Final (A1) is given for correctly rounding their answer. This may be awarded regardless of a preceding (A0).

800 × 3(0.5×2.5)

= 3158.57

= 3200

[2 marks]9c. Calculate the time, in hours, for the number of bacteria to reach 5500.

Markscheme (M1)

Notes: Award (M1) for equating function to 5500. Accept correct alternative methods.

(3.50968...) (A1) (C2)

5500 = 800 × 3(0.5×t)

= 3.51 hours

10a. [1 mark]

In a trial for a new drug, scientists found that the amount of the drug in the bloodstream decreased over time, according to the model

where

is the amount of the drug in the bloodstream in mg per litre

and

is the time in hours.

Write down the amount of the drug in the bloodstream at

.

Markscheme (A1) (C1)

[1 mark]

D(t) = 1.2 × (0.87 , t ⩾ 0)t

D

(mg )l−1

t

t = 0

1.2 (mg )l−1

[2 marks]10b. Calculate the amount of the drug in the bloodstream after 3 hours.

Markscheme (M1)

Note: Award (M1) for correct substitution into given formula.

(A1) (C2)

[2 marks]

1.2 × (0.87)3

= 0.790 (mg ) (0.790203 … )l−1

10c. [3 marks]Use your graphic display calculator to determine the time it takes for the amount of the drug in the bloodstream to decrease to

.0.333 mg1−1

Printed for Washington-Lee High School

© International Baccalaureate Organization 2017 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme (M1)

Note: Award (M1) for setting up the equation.

(M1) Notes: Some indication of scale is to be shown, for example the window used on the calculator.

Accept alternative methods.

(hours) (

, 9 hours 12 minutes, 9:12) (A1) (C3)[3 marks]

1.2 × = 0.3330.87t

9.21

9.20519…