Sec 3 · G3 (Express)

Graphs of functions

Practise graphs of functions with step-by-step help. Free account.

1. Cubic, reciprocal and exponential graphs

Six curves: y = x³ rises left to right through the origin; y = −x³ falls; y = 2/x has two branches in the first and third quadrants; y = −2/x in the second and fourth; y = 2/x² has two branches above the x-axis; y = 2ˣ rises, crosses the y-axis at (0, 1) and stays above the x-axis.

Graph	Shape
$y = ax^{3}$	$a > 0$ : rises; $a < 0$ : falls
$y = ax^{3} + bx^{2} + cx + d$	up to two turning points; right-hand end goes up if $a > 0$ , down if $a < 0$
$y = \dfrac{a}{x}$	two branches; never meets either axis
$y = \dfrac{a}{x^{2}}$	two branches, both on the same side of the $x$ -axis
$y = ka^{x}$	crosses the $y$ -axis at $(0, k)$ ; never meets the $x$ -axis

Table of values: substitute each x

\displaystyle y = x^{3} - 3x + 1, \quad x = 2

e.g.

\displaystyle 2^{3} - 3(2) + 1

\displaystyle = 3

Reciprocal

\displaystyle y = \dfrac{6}{x}, \quad x = -4

e.g.

\displaystyle \dfrac{6}{-4}

\displaystyle = -1.5

Exponential

\displaystyle y = 3(2^{x}), \quad x = -2

e.g.

\displaystyle 3(2^{-2})

\displaystyle = \dfrac{3}{4}

\displaystyle = 0.75

A point on the curve: substitute to find k

\displaystyle y = \dfrac{k}{x} \text{ through } (-4, -5)

e.g.

\displaystyle -5 = \dfrac{k}{-4}

\displaystyle k = 20

📖 Worked example

Given that $y = x^{3} - 3x + 1$ , calculate the value of $y$ when $x = -2.5$ , correct to 1 decimal place.

Step 1 of 2

Substitute the given value of x and work it out

y = (-2.5)^{3} - 3(-2.5) + 1

\phantom{y} = -7.125

📖 Worked example

The number of fish in a pond after $t$ months is $N = 200(3^{t})$ . Find the number of fish after 2 months.

Step 1 of 2

Substitute the given value of t

N = 200(3^{2})

3^{2} = 9

📖 Worked example

Which diagram, A, B, C or D, shows the graph of $y = -\dfrac{1}{x}$ ?

Step 1 of 3

What type of graph is it?

y = -\dfrac{1}{x}

is reciprocal (

\dfrac{a}{x}

Water poured in steadily: container	Depth against time
same width all the way up	a straight line
gets wider going up	rises quickly, then more slowly
gets narrower going up	rises slowly, then more quickly
wide part, then narrow part	two straight pieces: gentle, then steep
narrow part, then wide part	two straight pieces: steep, then gentle

📖 Worked example

Water is poured at a constant rate into the empty container shown, which is open at the top. Which graph, A, B, C or D, shows how the depth of the water, $h$ cm, changes with time, $t$ seconds? (Each graph has $h$ going up and $t$ going across.)

Step 1 of 3

How does the container's width change going up?

The container gets narrower going up.

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2. Gradient of a curve

Gradient of the curve at P = gradient of the tangent at P

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

Curve y = x² − 4x + 3 with the tangent at P(3, 0); the tangent also passes through (5, 4); run 2, rise 4.

e.g.

\displaystyle m = \dfrac{4 - 0}{5 - 3}

\displaystyle m = \dfrac{4}{2}

\displaystyle m = 2

two points on the tangent: $\dfrac{\text{rise}}{\text{run}}$

📖 Worked example

The diagram shows the curve $y = -x^{2} + 2x + 3$ and the tangent to the curve at $x = 0$ . Use the two points marked on the tangent to find the gradient of the curve at $x = 0$ .

Step 1 of 2

Take the two points on the tangent

A = (-1, 1)

B = (1, 5)

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3. Solving equations graphically

Where the curve meets the line

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

Curve y = x² − 2x − 3 and line y = x + 1 meeting where x = −1 and x = 4.

e.g.

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

\displaystyle x = -1 \text{ or } x = 4

Which line to draw: make the drawn curve one side

\displaystyle x^{2} - 3x - 4 = 0

e.g.

\displaystyle x^{2} - 3x - 4 = 0

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

the curve $y = x^{2} - 2x - 3$ is drawn, so draw $y = x + 1$

📖 Worked example

The diagram shows the graph of $y = x^{2} - 2x$ and the line $y = 3$ . Use the graph to solve the equation $x^{2} - 2x = 3$ .

Step 1 of 3

Each side of the equation is a graph

Left-hand side:

y = x^{2} - 2x

, the curveRight-hand side:

y = 3

, the horizontal line

📖 Worked example

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 $-x^{2} + 5x + 6 = 0$ .

Step 1 of 3

Rearrange so the drawn curve is the left-hand side

-x^{2} + 5x + 6 = 0

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

A range of x: where the curve is below (or above) the line

\displaystyle x^{2} - 2x < x + 4

e.g.

\displaystyle x^{2} - 2x = x + 4

\displaystyle x = -1 \text{ or } x = 4

the curve is below the line between the crossings: $-1 < x < 4$

The line $y = k$ (cubic with two turning points)	Solutions of $f(x) = k$
above the maximum point, or below the minimum point	$1$
through a turning point	$2$
between the two turning points	$3$

📖 Worked example

The diagram shows the graph of $y = x^{3} - 2x + 1$ , and the straight line that must be drawn on the same axes to solve the equation $x^{3} - 2x + 4 = 0$ . Use the graph to solve this equation.

Step 1 of 3

Rearrange so the drawn curve is the left-hand side

x^{3} - 2x + 4 = 0

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

📖 Worked example

The diagram shows the graph of $y = x^{3} - 3x^{2}$ . Find the range of values of $k$ for which the equation $x^{3} - 3x^{2} = k$ has three solutions.

Step 1 of 2

Read the y-coordinates of the turning points

Maximum point:

y = 0

Minimum point:

y = -4

Practise this →

4. Speed–time graphs

Feature	Meaning	Example
gradient	acceleration (m/s²)	$\dfrac{12}{8} = 1.5$ m/s²
gradient going down	deceleration, given as a positive number	$\dfrac{12}{6} = 2$ m/s²
horizontal	constant speed	$12$ m/s from $8$ s to $14$ s
area under the graph	distance travelled (m)	$\dfrac{1}{2} \times (6 + 20) \times 12 = 156$ m
$\dfrac{\text{total distance}}{\text{total time}}$	average speed	$\dfrac{156}{20} = 7.8$ m/s
km/h to m/s	$\times 1000 \div 3600$	$54$ km/h $= 15$ m/s

\displaystyle \text{distance} = \text{area}

📖 Worked example

The diagram shows the speed–time graph of a lorry over 65 seconds. Find the distance, in metres, travelled by the lorry in the first 45 seconds.

Step 1 of 3

Area A

\text{A}: 0 \text{ to } 10 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 10 \times 12 = 60

📖 Worked example

The diagram shows the speed–time graph of a car that starts from rest and comes to rest again. The total distance travelled is 600 m. Find the value of $T$ .

Step 1 of 3

Write its area in terms of T, equal to the distance

\dfrac{1}{2} \times (20 + T) \times 16 = 600

Two vehicles: catching up means equal distances (equal areas)

\displaystyle \text{area under one graph} = \text{area under the other}

e.g.

\displaystyle \dfrac{1}{2} \times 8 \times 14 + 14(x - 8) = 12x

\displaystyle x = 28

equal speeds is only where the two graphs cross

📖 Worked example

A car starts from rest and accelerates uniformly to a speed of $14$ m/s in 8 s, then travels at that speed. As it starts, a lorry travelling at a constant speed of 12 m/s passes it, in the same direction. The diagram shows their speed–time graphs. The car catches up with the lorry after $T$ seconds. Find the value of $T$ .

Step 1 of 4

Distance of the car after T s: the area under its graph (a triangle, then a rectangle)

\dfrac{1}{2} \times 8 \times 14 + 14(T - 8)

= 14T - 56

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Sec 3 · G3 (Express)

Graphs of functions

Practise graphs of functions with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Cubic, reciprocal and exponential graphs

Graph	Shape
$y = ax^{3}$	$a > 0$ : rises; $a < 0$ : falls
$y = ax^{3} + bx^{2} + cx + d$	up to two turning points; right-hand end goes up if $a > 0$ , down if $a < 0$
$y = \dfrac{a}{x}$	two branches; never meets either axis
$y = \dfrac{a}{x^{2}}$	two branches, both on the same side of the $x$ -axis
$y = ka^{x}$	crosses the $y$ -axis at $(0, k)$ ; never meets the $x$ -axis

Table of values: substitute each x

\displaystyle y = x^{3} - 3x + 1, \quad x = 2

e.g.

\displaystyle 2^{3} - 3(2) + 1

\displaystyle = 3

Reciprocal

\displaystyle y = \dfrac{6}{x}, \quad x = -4

e.g.

\displaystyle \dfrac{6}{-4}

\displaystyle = -1.5

Exponential

\displaystyle y = 3(2^{x}), \quad x = -2

e.g.

\displaystyle 3(2^{-2})

\displaystyle = \dfrac{3}{4}

\displaystyle = 0.75

A point on the curve: substitute to find k

\displaystyle y = \dfrac{k}{x} \text{ through } (-4, -5)

e.g.

\displaystyle -5 = \dfrac{k}{-4}

\displaystyle k = 20

📖 Worked example

Given that $y = x^{3} - 3x + 1$ , calculate the value of $y$ when $x = -2.5$ , correct to 1 decimal place.

Step 1 of 2

Substitute the given value of x and work it out

y = (-2.5)^{3} - 3(-2.5) + 1

\phantom{y} = -7.125

📖 Worked example

The number of fish in a pond after $t$ months is $N = 200(3^{t})$ . Find the number of fish after 2 months.

Step 1 of 2

Substitute the given value of t

N = 200(3^{2})

3^{2} = 9

📖 Worked example

Which diagram, A, B, C or D, shows the graph of $y = -\dfrac{1}{x}$ ?

Step 1 of 3

What type of graph is it?

y = -\dfrac{1}{x}

is reciprocal (

\dfrac{a}{x}

Water poured in steadily: container	Depth against time
same width all the way up	a straight line
gets wider going up	rises quickly, then more slowly
gets narrower going up	rises slowly, then more quickly
wide part, then narrow part	two straight pieces: gentle, then steep
narrow part, then wide part	two straight pieces: steep, then gentle

📖 Worked example

Step 1 of 3

How does the container's width change going up?

The container gets narrower going up.

Practise this →

2. Gradient of a curve

Gradient of the curve at P = gradient of the tangent at P

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

e.g.

\displaystyle m = \dfrac{4 - 0}{5 - 3}

\displaystyle m = \dfrac{4}{2}

\displaystyle m = 2

two points on the tangent: $\dfrac{\text{rise}}{\text{run}}$

📖 Worked example

The diagram shows the curve $y = -x^{2} + 2x + 3$ and the tangent to the curve at $x = 0$ . Use the two points marked on the tangent to find the gradient of the curve at $x = 0$ .

Step 1 of 2

Take the two points on the tangent

A = (-1, 1)

B = (1, 5)

Practise this →

3. Solving equations graphically

Where the curve meets the line

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

e.g.

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

\displaystyle x = -1 \text{ or } x = 4

Which line to draw: make the drawn curve one side

\displaystyle x^{2} - 3x - 4 = 0

e.g.

\displaystyle x^{2} - 3x - 4 = 0

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

the curve $y = x^{2} - 2x - 3$ is drawn, so draw $y = x + 1$

📖 Worked example

The diagram shows the graph of $y = x^{2} - 2x$ and the line $y = 3$ . Use the graph to solve the equation $x^{2} - 2x = 3$ .

Step 1 of 3

Each side of the equation is a graph

Left-hand side:

y = x^{2} - 2x

, the curveRight-hand side:

y = 3

, the horizontal line

📖 Worked example

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 $-x^{2} + 5x + 6 = 0$ .

Step 1 of 3

Rearrange so the drawn curve is the left-hand side

-x^{2} + 5x + 6 = 0

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

A range of x: where the curve is below (or above) the line

\displaystyle x^{2} - 2x < x + 4

e.g.

\displaystyle x^{2} - 2x = x + 4

\displaystyle x = -1 \text{ or } x = 4

the curve is below the line between the crossings: $-1 < x < 4$

The line $y = k$ (cubic with two turning points)	Solutions of $f(x) = k$
above the maximum point, or below the minimum point	$1$
through a turning point	$2$
between the two turning points	$3$

📖 Worked example

The diagram shows the graph of $y = x^{3} - 2x + 1$ , and the straight line that must be drawn on the same axes to solve the equation $x^{3} - 2x + 4 = 0$ . Use the graph to solve this equation.

Step 1 of 3

Rearrange so the drawn curve is the left-hand side

x^{3} - 2x + 4 = 0

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

📖 Worked example

The diagram shows the graph of $y = x^{3} - 3x^{2}$ . Find the range of values of $k$ for which the equation $x^{3} - 3x^{2} = k$ has three solutions.

Step 1 of 2

Read the y-coordinates of the turning points

Maximum point:

y = 0

Minimum point:

y = -4

Practise this →

4. Speed–time graphs

Feature	Meaning	Example
gradient	acceleration (m/s²)	$\dfrac{12}{8} = 1.5$ m/s²
gradient going down	deceleration, given as a positive number	$\dfrac{12}{6} = 2$ m/s²
horizontal	constant speed	$12$ m/s from $8$ s to $14$ s
area under the graph	distance travelled (m)	$\dfrac{1}{2} \times (6 + 20) \times 12 = 156$ m
$\dfrac{\text{total distance}}{\text{total time}}$	average speed	$\dfrac{156}{20} = 7.8$ m/s
km/h to m/s	$\times 1000 \div 3600$	$54$ km/h $= 15$ m/s

\displaystyle \text{distance} = \text{area}

📖 Worked example

The diagram shows the speed–time graph of a lorry over 65 seconds. Find the distance, in metres, travelled by the lorry in the first 45 seconds.

Step 1 of 3

Area A

\text{A}: 0 \text{ to } 10 \text{ s}, \text{ triangle } \dfrac{1}{2} \times 10 \times 12 = 60

📖 Worked example

The diagram shows the speed–time graph of a car that starts from rest and comes to rest again. The total distance travelled is 600 m. Find the value of $T$ .

Step 1 of 3

Write its area in terms of T, equal to the distance

\dfrac{1}{2} \times (20 + T) \times 16 = 600

Two vehicles: catching up means equal distances (equal areas)

\displaystyle \text{area under one graph} = \text{area under the other}

e.g.

\displaystyle \dfrac{1}{2} \times 8 \times 14 + 14(x - 8) = 12x

\displaystyle x = 28

equal speeds is only where the two graphs cross

📖 Worked example

Step 1 of 4

Distance of the car after T s: the area under its graph (a triangle, then a rectangle)

\dfrac{1}{2} \times 8 \times 14 + 14(T - 8)

= 14T - 56

Practise this →

Try a question: Graphs of functions

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Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

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