Sec 3 G2 Maths Graphs of functions worksheet with answers

A free Sec 3 G2 Maths worksheet on graphs of functions: 7 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 G2 Maths worksheet: Graphs of functions

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The graph shows the distance travelled by a car,

s

m, after

t

s. The tangent at

t = 4

has been drawn. Using the tangent, find the speed, in m/s, when

t = 4

Answerm/s[2]

The graph of

y = -x^{2} + 2x + 3

has been drawn. Find the equation of the straight line that must be drawn on the same axes to solve the equation

-x^{2} + 2 = 0

Answer[2]

The diagram shows the speed–time graph of a bus over 80 seconds. Find the total distance, in metres, travelled by the bus.

Answerm[3]

(a)The curve

y = x^{3} + kx

passes through the point

(1, -2)

. Find the value of

k

Answer (a)[2]

(b)The curve

y = x^{3} + kx

passes through the point

(-2, -12)

. Find the value of

k

Answer (b)[2]

Section B

The diagram shows part of the graph of

y = x^{3} - 3x

, with the tangent to the curve at

x = 1

drawn.

(a)Find the value of

y

when

x = 1.5

, correct to 1 decimal place.

Answer (a)[1]

(b)Use the tangent to find the gradient of the curve at

x = 1

Answer (b)[2]

(c)By drawing a suitable straight line on the graph, the equation

x^{3} - 5x - 2 = 0

can be solved. Find the equation of this straight line.

Answer (c)[2]

The diagram shows part of the graph of

y = x^{3} - 2x + 1

, with the tangent to the curve at

x = 0

drawn.

(a)Find the value of

y

when

x = 1.5

, correct to 1 decimal place.

Answer (a)[1]

(b)Use the tangent to find the gradient of the curve at

x = 0

Answer (b)[2]

(c)By drawing a suitable straight line on the graph, the equation

x^{3} - x + 4 = 0

can be solved. Find the equation of this straight line.

Answer (c)[2]

The table shows some values of

x

and the corresponding values of

y

for

y = -x^{2} - 2x + 5

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$
$y$	$p$	$2$	$5$	$6$	$5$	$2$	$-3$

(a)Find the value of

p

Answer (a)[1]

(b)The grid above shows the graph of

y = -x^{2} - 2x + 5

for

-4 \leq x \leq 2

. Use the graph to solve the equation

-x^{2} - 2x + 5 = 1

, giving your answers correct to 1 decimal place.

Answer (b)[2]

(c)By drawing a tangent, find the gradient of the curve at the point where

x = -2

Answer (c)[2]

(d)By drawing a suitable straight line on the grid, the equation

x^{2} + 3x - 1 = 0

can be solved. Find the equation of this straight line.

Answer (d)[2]

(e)Hence solve the equation

x^{2} + 3x - 1 = 0

, giving your answers correct to 1 decimal place.

Answer (e)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer: Speed

= 2

m/s[2]

A = (3, 2)

B = (5, 6)

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{6 - 2}{5 - 3}

\phantom{m} = 2

Speed

= 2

m/s

Question 2

Answer: Gradient

2

y

-intercept

1

.[2]

-x^{2} + 2 = 0

-x^{2} + 2x + 3 = 2x + 1

The solutions are where the LHS graph

y = -x^{2} + 2x + 3

meets the RHS graph, so draw

y = 2x + 1

.Then read the

x

-coordinates where the curve and the line meet.Gradient

2

y

-intercept

1

Question 3

Answer:

\phantom{\text{Distance}} = 810 \text{ m}

[3]

\text{A}: 0 \text{ to } 30 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 30 \times 6 = 90

\text{B}: 30 \text{ to } 45 \text{ s}, \text{ trapezium } \dfrac{1}{2} \times (6 + 18) \times 15 = 180

\text{C}: 45 \text{ to } 70 \text{ s}, \text{ rectangle } 25 \times 18 = 450

\text{D}: 70 \text{ to } 80 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 10 \times 18 = 90

\text{Distance} = 90 + 180 + 450 + 90

\phantom{\text{Distance}} = 810 \text{ m}

Question 4

(a)Answer:

k = -3

[2]

-2 = (1)^{3} + k(1)

-2 = 1 + k

k = -3

(b)Answer:

k = 2

[2]

-12 = (-2)^{3} + k(-2)

-12 = -8 - 2k

-2k = -4

k = 2

Question 5

(a)Answer:

-1.1

[1]

y = -1.125

= -1.1

(1 d.p.)

(b)Answer:

0

[2]

The tangent passes through

(0, -2)

and

(2, -2)

.Gradient

= \dfrac{0}{2}

= 0

(c)Answer:

y = 2x + 2

[2]

x^{3} - 5x - 2 = 0

x^{3} - 3x = 2x + 2

Draw

y = 2x + 2

Question 6

(a)Answer:

1.4

[1]

y = 1.375

= 1.4

(1 d.p.)

(b)Answer:

-2

[2]

The tangent passes through

(-1, 3)

and

(1, -1)

.Gradient

= \dfrac{-4}{2}

= -2

(c)Answer:

y = -x - 3

[2]

x^{3} - x + 4 = 0

x^{3} - 2x + 1 = -x - 3

Draw

y = -x - 3

Question 7

(a)Answer:

p = -3

[1]

p = -(-4)^{2} - 2(-4) + 5

\phantom{p} = -3

(b)Answer:

x = -3.2

x = 1.2

[2]

Draw the line

y = 1

on the grid.It meets the curve where

x = -3.2

x = 1.2

(c)Answer:

2

(accept

1.8

2.2

)[2]

Draw the tangent at

(-2, 5)

. It passes through

(-3, 3)

and

(-1, 7)

.Gradient

= \dfrac{7 - 3}{-1 - (-3)}

= 2

(d)Answer:

y = x + 4

[2]

x^{2} + 3x - 1 = 0

-x^{2} - 3x + 1 = 0

(multiply by

-1

)

-x^{2} - 2x + 5 = x + 4

Draw the line

y = x + 4

(e)Answer:

x = -3.3

x = 0.3

[2]

Draw

y = x + 4

through

(0, 4)

and

(2, 6)

.It meets the curve where

x = -3.3

x = 0.3

Sec 3 G2 Maths Graphs of functions worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Graphs of functions notes. Or practise it online with step-by-step help when you get one wrong.

Section B

The diagram shows part of the graph of

y = x^{3} - 3x

, with the tangent to the curve at

x = 1

drawn.

(a)Find the value of

y

when

x = 1.5

, correct to 1 decimal place.

Answer (a)[1]

(b)Use the tangent to find the gradient of the curve at

x = 1

Answer (b)[2]

(c)By drawing a suitable straight line on the graph, the equation

x^{3} - 5x - 2 = 0

can be solved. Find the equation of this straight line.

Answer (c)[2]

x

-4

-3

-2

-1

0

1

2

y

p

2

5

6

5

2

-3

Answers and worked solutions

Question 1

Answer: Speed

= 2

m/s[2]

A = (3, 2)

B = (5, 6)

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{6 - 2}{5 - 3}

\phantom{m} = 2

Speed

= 2

m/s

Question 2

Answer: Gradient

2

y

-intercept

1

.[2]

-x^{2} + 2 = 0

-x^{2} + 2x + 3 = 2x + 1

The solutions are where the LHS graph

y = -x^{2} + 2x + 3

meets the RHS graph, so draw

y = 2x + 1

.Then read the

x

-coordinates where the curve and the line meet.Gradient

2

y

-intercept

1

Question 3

Answer:

\phantom{\text{Distance}} = 810 \text{ m}

[3]

\text{A}: 0 \text{ to } 30 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 30 \times 6 = 90

\text{B}: 30 \text{ to } 45 \text{ s}, \text{ trapezium } \dfrac{1}{2} \times (6 + 18) \times 15 = 180

\text{C}: 45 \text{ to } 70 \text{ s}, \text{ rectangle } 25 \times 18 = 450

\text{D}: 70 \text{ to } 80 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 10 \times 18 = 90

\text{Distance} = 90 + 180 + 450 + 90

\phantom{\text{Distance}} = 810 \text{ m}

Question 4

(a)Answer:

k = -3

[2]

-2 = (1)^{3} + k(1)

-2 = 1 + k

k = -3

(b)Answer:

k = 2

[2]

-12 = (-2)^{3} + k(-2)

-12 = -8 - 2k

-2k = -4

k = 2

Question 5

(a)Answer:

-1.1

[1]

y = -1.125

= -1.1

(1 d.p.)

(b)Answer:

0

[2]

The tangent passes through

(0, -2)

and

(2, -2)

.Gradient

= \dfrac{0}{2}

= 0

(c)Answer:

y = 2x + 2

[2]

x^{3} - 5x - 2 = 0

x^{3} - 3x = 2x + 2

Draw

y = 2x + 2

Question 6

(a)Answer:

1.4

[1]

y = 1.375

= 1.4

(1 d.p.)

(b)Answer:

-2

[2]

The tangent passes through

(-1, 3)

and

(1, -1)

.Gradient

= \dfrac{-4}{2}

= -2

(c)Answer:

y = -x - 3

[2]

x^{3} - x + 4 = 0

x^{3} - 2x + 1 = -x - 3

Draw

y = -x - 3

Question 7

(a)Answer:

p = -3

[1]

p = -(-4)^{2} - 2(-4) + 5

\phantom{p} = -3

(b)Answer:

x = -3.2

x = 1.2

[2]

Draw the line

y = 1

on the grid.It meets the curve where

x = -3.2

x = 1.2

(c)Answer:

2

(accept

1.8

2.2

)[2]

Draw the tangent at

(-2, 5)

. It passes through

(-3, 3)

and

(-1, 7)

.Gradient

= \dfrac{7 - 3}{-1 - (-3)}

= 2

(d)Answer:

y = x + 4

[2]

x^{2} + 3x - 1 = 0

-x^{2} - 3x + 1 = 0

(multiply by

-1

)

-x^{2} - 2x + 5 = x + 4

Draw the line

y = x + 4

(e)Answer:

x = -3.3

x = 0.3

[2]

Draw

y = x + 4

through

(0, 4)

and

(2, 6)

.It meets the curve where

x = -3.3

x = 0.3