Drawing Linear Graphs: Practise Online

Either paper 2 marks

A straight line is drawn on the grid.

(a)
Use the graph to write down the value of \(y\) when \(x = 4\).

(b)
Use the graph to write down the value of \(x\) when \(y = -7\).
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(a) \(-6\)
(b) \(5\)
Find \(4\) on the \(x\)-axis and go straight down until you meet the line.

Now read across to the \(y\)-axis:

\(y = -6\)
This time start on the \(y\)-axis at \(-7\) and go across to the right until you meet the line.

Then read straight down (or up) to the \(x\)-axis:

\(x = 5\)
Either paper 2 marks

Each grid has \(x\) and \(y\) from \(-6\) to \(6\).

(a)
On this grid, draw the line with equation \(y = -4\)

(b)
On this grid, draw the line with equation \(x = 5\)

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(a) A horizontal line through \((0, -4)\), \((1, -4)\) and every other point with \(y = -4\)
(b) A vertical line through \((5, 0)\), \((5, 1)\) and every other point with \(x = 5\)
(a)

(b)

\(y = -4\) means every point on the line has \(y\)-coordinate \(-4\), whatever \(x\) is.

So the line goes through \((0, -4)\) and \((3, -4)\): rule a flat line through them, right across the grid.
\(x = 5\) means every point on the line has \(x\)-coordinate \(5\), whatever \(y\) is.

So the line goes through \((5, 0)\) and \((5, 3)\): rule an upright line through them, from the bottom of the grid to the top.
- (a) B1the horizontal line through \(y = -4\)
- (b) B1the vertical line through \(x = 5\), right across the grid
Either paper 3 marks
The table shows some values of \(y = 2x - 1\)

x -3 -2 -1 0 1

y -5 1

Three of the \(y\) values are missing.
(a)
Work out the value of \(y\) when \(x = 0\).

(b)
Work out the value of \(y\) when \(x = -3\).

(c)
Work out the value of \(y\) when \(x = -1\).
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(a) \(-1\)
(b) \(-7\)
(c) \(-3\)
Put \(x = 0\) into the equation:

\(y = 2 \times 0 - 1 = -1\)

Multiplying by zero leaves nothing, so the answer is always the number on the end — that is why the graph crosses the \(y\)-axis at \(-1\).
\(y = 2 \times (-3) - 1 = -6 - 1 = -7\)
\(y = 2 \times (-1) - 1 = -2 - 1 = -3\)

Check it against the row: going one step to the right along the table always adds \(2\) to \(y\).
Either paper 2 marks

Here are four points.

\((-2, -4) \qquad (2, 8) \qquad (4, 12) \qquad (3, 8)\)

One of them lies on the line \(y = 3x - 1\)

Write down the coordinates of that point.
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\((3, 8)\)
A point lies on the line when putting its \(x\)-coordinate into \(y = 3x - 1\) gives back its \(y\)-coordinate.

Try \(x = 3\):

\(y = 3 \times (3) - 1 = 8\)

That is the \(y\)-coordinate written beside it, so \((3, 8)\) is the point on the line. The other three give a different \(y\), so they are off the line.
- M1substitutes an \(x\)-value from the list into \(y = 3x - 1\)
- A1point \((3, 8)\) identified
Either paper 3 marks

Each grid below has one straight line drawn on it.

(a)
Write down the equation of the line drawn on this grid.

(b)
Write down the equation of the line drawn on this grid.

(c)
Suppose both lines were drawn on the same grid.

Write down the coordinates of the point where they would cross.
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(a) \(y = 2\)
(b) \(x = 3\)
(c) \((3, 2)\)
The line is flat, so every point on it has the same \(y\)-coordinate, namely \(2\). It does not matter what \(x\) is.

The equation is \(y = 2\)
This line is upright, so every point on it has the same \(x\)-coordinate, namely \(3\). It does not matter what \(y\) is.

The equation is \(x = 3\)

A common slip is to write \(y = 3\) here — but \(y\) takes every value on this line, so \(y\) cannot be what the equation fixes.
The crossing point is on both lines, so it has \(x = 3\) (from the upright line) and \(y = 2\) (from the flat one).

The point is \((3, 2)\).
Either paper 2 marks

You are going to draw the graph of \(y = -x - 2\) on this grid, for values of \(x\) from \(-4\) to \(4\).

The quickest way to draw a straight line is to work out the two end points, plot them and join them up.

(a)
Find the coordinates of the point you would plot for \(x = -4\).

(b)
Find the coordinates of the point you would plot for \(x = 4\).
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(a) \((-4, 2)\)
(b) \((4, -6)\)
Substitute \(x = -4\) into \(y = -x - 2\):

\(y = 4 - 2 = 2\)

The point is \((-4, 2)\).
Substitute \(x = 4\) into \(y = -x - 2\):

\(y = -4 - 2 = -6\)

The point is \((4, -6)\).

Plot a third point in the middle as a check — if it does not sit on the straight line through the other two, one of the three is wrong.
Either paper 3 marks

On the grid, draw the graph of \(y = -2x + 3\) for values of \(x\) from \(-4\) to \(4\).

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A straight line through \((0, 3)\) and \((1, 1)\)

Work out two points on the line, and a third as a check.

When \(x = 0\), \(y = 3\). When \(x = 1\), \(y = -2 \times 1 + 3 = 1\). When \(x = 2\), \(y = -1\).

So the line goes through \((0, 3)\), \((1, 1)\) and \((2, -1)\): it crosses the \(y\)-axis at \(3\) and goes 2 down for every square across.

Plot the points and rule one straight line through them, right across the grid.
- B1a straight line through \((0, 3)\)
- B1a straight line with gradient \(-2\)
- B1the correct line
Either paper 4 marks
The table gives some values of \(y = -x - 4\)

x -1 0 1 2 3

y -3 -5 -6

Two of the \(y\) values are missing.
(a)
Work out the value of \(y\) when \(x = 0\).

(b)
Work out the value of \(y\) when \(x = 3\).

(c)
On the grid, plot the five points from the table and join them with a straight line, for values of \(x\) from \(-1\) to \(3\).

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(a) \(-4\)
(b) \(-7\)
(c) Points at \((-1, -3)\), \((0, -4)\), \((1, -5)\), \((2, -6)\) and \((3, -7)\), joined by one straight line
(c)

\(y = -1 \times (0) - 4 = 0 - 4 = -4\)
\(y = -1 \times (3) - 4 = -3 - 4 = -7\)
The completed table gives the points \((-1, -3)\), \((0, -4)\), \((1, -5)\), \((2, -6)\) and \((3, -7)\).

Plot each one, then join them with a ruler. They lie on a straight line that crosses the \(y\)-axis at \(-4\).
- (c) B1at least four of the five points plotted correctly
- (c) B1all five points plotted correctly and joined by a straight line
Either paper 3 marks

A straight line is drawn on the grid. Both axes are numbered in twos, so each square is worth \(2\).

The dashed lines show one reading being taken.

(a)
Use the graph to write down the value of \(y\) when \(x = 6\).

(b)
Use the graph to work out the value of \(x\) when \(y = -6\).
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(a) \(4\)
(b) \(-4\)
The dashed lines meet the \(x\)-axis at \(6\) and the \(y\)-axis at the value being read.

Counting in twos along the \(y\)-axis gives

\(y = 4\)
Do the same thing the other way round. Find \(-6\) on the \(y\)-axis — remember each square is \(2\) — and go across to the line.

From there go straight up to the \(x\)-axis:

\(x = -4\)
- (b) M1reads across from \(-6\) on the y-axis to the line, then straight up to the x-axis
- (b) A1x = \(-4\)
Either paper 2 marks
A student has filled in a table of values for \(y = 2x + 5\)

x -3 -2 -1 0 1

y -1 1 3 5 5

One of the \(y\) values is wrong.
(a)
Write down the value of \(x\) for the entry that is wrong.

(b)
Write down the correct value of \(y\) for that entry.
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(a) \(1\)
(b) \(7\)
Every step of \(1\) to the right should add \(2\) to \(y\), because the gradient is \(2\). One step in the table does not.

The entry that breaks the pattern is the one at \(x = 1\), which should read \(7\) but reads \(5\).
\(y = 2 \times (1) + 5 = 2 + 5 = 7\)