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Maths · Graphs

Linear graphs

Straight-line graphs from a table of values, the gradient as "change in y over change in x", and reading the gradient and y-intercept straight from \(y = mx + c\).

  • 5 key terms
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Before We Start

Answer each one, then check.

  1. 1

    Work out \(3 \times (-2) + 1\).

    Show answerHide answer

    \(-5\)

  2. 2

    Which axis is horizontal?

    Show answerHide answer

    The \(x\)-axis

  3. 3

    Write down the coordinates of the origin.

    Show answerHide answer

    \((0, 0)\)

  4. 4

    Rearrange \(y - 4 = 2x\) to make \(y\) the subject.

    Show answerHide answer

    \(y = 2x + 4\)

Learning Objectives

  1. 1Draw a straight-line graph from a table of values.
  2. 2Recognise lines of the form \(x = a\) and \(y = b\).
  3. 3Work out the gradient of a line.
  4. 4Use \(y = mx + c\) to find the gradient and \(y\)-intercept.

Tables of Values

To draw \(y = 2x + 1\), work out \(y\) for a few values of \(x\), plot the points and join them with a ruler.

  • Substitute

    When \(x = -2\), \(y = 2 \times (-2) + 1 = -3\).

  • Three points at least

    Two points fix a line; a third catches a mistake.

  • Straight line

    If the points are not in a straight line, check your table.

  • Across the grid

    Draw the line through all the points, from edge to edge of the given range.

A Table of Values for y = 2x + 1

  • \(-2\)

    Working: \(2 \times (-2) + 1\). y: \(-3\)

  • \(-1\)

    Working: \(2 \times (-1) + 1\). y: \(-1\)

  • 0

    Working: \(2 \times 0 + 1\). y: 1

  • 1

    Working: \(2 \times 1 + 1\). y: 3

  • 2

    Working: \(2 \times 2 + 1\). y: 5

  • 3

    Working: \(2 \times 3 + 1\). y: 7

Horizontal and Vertical Lines

Some lines have only one letter in their equation.

  • \(y = b\)

    A horizontal line: every point on it has \(y\)-coordinate \(b\). The \(x\)-axis is \(y = 0\).

  • \(x = a\)

    A vertical line: every point on it has \(x\)-coordinate \(a\). The \(y\)-axis is \(x = 0\).

  • \(y = x\)

    A diagonal line through the origin, where both coordinates are equal.

  • \(y = -x\)

    The other diagonal through the origin.

Gradient

Gradient \(= \dfrac{\text{change in } y}{\text{change in } x}\).

  • Positive

    The line goes up from left to right.

  • Negative

    The line goes down from left to right.

  • Zero

    A horizontal line has gradient 0.

  • From two points

    Between \((x_1, y_1)\) and \((x_2, y_2)\), gradient \(= \dfrac{y_2 - y_1}{x_2 - x_1}\).

Gradient from Two Points

Work out the gradient of the line through \((1, 3)\) and \((4, 12)\).

Show the solutionHide the solution
  1. 1 Change in \(y\) \(12 - 3 = 9\)
  2. 2 Change in \(x\) \(4 - 1 = 3\)
  3. 3 Divide \(9 \div 3 = 3\)

AnswerGradient 3

y = mx + c

When the equation is written as \(y = mx + c\), you can read the line straight off it.

  • \(m\)

    The gradient - the number in front of \(x\).

  • \(c\)

    The \(y\)-intercept: the line crosses the \(y\)-axis at \((0, c)\).

  • Rearrange first

    \(y\) must be on its own: \(2y = 4x + 6\) becomes \(y = 2x + 3\).

  • Order doesn't matter

    \(y = 5 - 3x\) has gradient \(-3\) and \(y\)-intercept 5.

Finding the Equation from a Graph

A straight line crosses the \(y\)-axis at \((0, -2)\) and passes through \((3, 4)\). Find its equation.

Show the solutionHide the solution
  1. 1 \(y\)-intercept \(c = -2\)
  2. 2 Gradient \(\dfrac{4 - (-2)}{3 - 0} = \dfrac{6}{3} = 2\)
  3. 3 Write \(y = mx + c\) \(y = 2x - 2\)

Answer\(y = 2x - 2\)

Line Match-Up

Sort these six equations into pairs that share something - same gradient, same intercept or both: \(y = 3x + 2\), \(y = 2 - x\), \(y = 3x - 5\), \(2y = 6x + 4\), \(y = -x + 7\), \(y = 2x + 2\). Then sketch each line.

1. Rearrange into \(y = mx + c\) first.

2. Write down \(m\) and \(c\) for each.

3. Group them.

A good answer shows: \(y = 3x + 2\) and \(2y = 6x + 4\) are the same line (gradient 3, intercept 2). \(y = 3x - 5\) is parallel to them. \(y = 2 - x\) and \(y = -x + 7\) are parallel (gradient \(-1\)). \(y = 2x + 2\), \(y = 3x + 2\) and \(y = 2 - x\) share intercept 2.

Can I...?

  1. 1Complete a table of values.
  2. 2Plot and draw a straight-line graph.
  3. 3Recognise \(x = a\), \(y = b\), \(y = x\) and \(y = -x\).
  4. 4Find the gradient from a graph.
  5. 5Find the gradient between two points.
  6. 6Read \(m\) and \(c\) from \(y = mx + c\).
  7. 7Rearrange an equation into \(y = mx + c\).
  8. 8Find the equation of a line from its graph.

Summary & Exam Focus

  • Gradient \(= \dfrac{\text{change in } y}{\text{change in } x}\).
  • \(y = mx + c\): \(m\) is the gradient, \(c\) the \(y\)-intercept.
  • \(x = a\) is vertical; \(y = b\) is horizontal.

Exam focus

The graph shows a straight line. Find the equation of the line. (3 marks) (3 marks)

For the gradient, choose two points where the line crosses grid lines exactly, and check the sign: a line going down to the right has a negative gradient.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Linear
Making a straight line when plotted.
Gradient
The steepness of a line: change in \(y\) divided by change in \(x\).
\(y\)-intercept
Where a line crosses the \(y\)-axis.
Coordinates
A pair \((x, y)\) giving a point's position.
Origin
The point \((0, 0)\) where the axes cross.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 2 marks

    Complete the table of values for \(y = 3x - 2\) for \(x = -1, 0, 1, 2, 3\).

    Show answerHide answer

    Model answer

    \(y = -5, -2, 1, 4, 7\)

    Mark scheme

    • At least 3 correct values — B1
    • All 5 correct — B1
  2. Question 2 Non-calculator 2 marks

    Work out the gradient of the straight line that passes through the points \((-1, 2)\) and \((3, 10)\).

    Show answerHide answer

    Model answer

    \(\dfrac{10 - 2}{3 - (-1)} = \dfrac{8}{4} = 2\)

    Mark scheme

    • \(\dfrac{10 - 2}{3 - (-1)}\) — M1
    • 2 — A1
  3. Question 3 Non-calculator 3 marks

    The graph shows a straight line. Find the equation of the line.

    A straight line on a coordinate grid crossing the y-axis at −2 and passing through (3, 4).
    Show answerHide answer

    Model answer

    The line crosses the \(y\)-axis at \(-2\). Gradient \(= \dfrac{4 - (-2)}{3 - 0} = 2\). \(y = 2x - 2\)

    Mark scheme

    • Gradient 2 — M1
    • \(y\)-intercept \(-2\) — M1
    • \(y = 2x - 2\) — A1
  4. Question 4 Non-calculator 2 marks

    Does the point \((3, 7)\) lie on the line \(y = 4x - 5\)? Give a reason for your answer.

    Show answerHide answer

    Model answer

    When \(x = 3\), \(y = 4 \times 3 - 5 = 7\). Yes, the point lies on the line.

    Mark scheme

    • \(4 \times 3 - 5\) — M1
    • Yes, with 7 shown — C1

Quick check

  1. What is the gradient of \(y = 5x - 3\)?

    1. A\(-3\)
    2. B5
    3. C\(-\frac{3}{5}\)
    4. D2
    Show answerHide answer

    B: 5

    In \(y = mx + c\), the gradient is \(m\), the number in front of \(x\).

  2. Which line is vertical?

    1. A\(x = 4\)
    2. B\(y = 4\)
    3. C\(y = 4x\)
    4. D\(y = x + 4\)
    Show answerHide answer

    A: \(x = 4\)

    Every point on \(x = 4\) has \(x\)-coordinate 4, so it is a vertical line.

  3. What is the \(y\)-intercept of \(3y = 6x - 9\)?

    1. A\(-9\)
    2. B6
    3. C2
    4. D\(-3\)
    Show answerHide answer

    D: \(-3\)

    Divide by 3: \(y = 2x - 3\).

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