Basic Graphs – GCSE Maths

A coordinate grid is formed by two perpendicular number lines called axes.

The x-axis runs horizontally (left ↔ right).
The y-axis runs vertically (up ↕ down).
They cross at the origin, labelled $O$, which has coordinates $(0, 0)$.

The four quadrants and how to read coordinates

Reading & Plotting Coordinates

A coordinate is written as $(x,\ y)$ — always across first, then up/down.

Memory trick: think of coordinates like reading a map — go along the corridor before climbing the stairs.
Across (x) first → Up/down (y) second.

Point	Coordinates	How to plot
A	$(3,\ 2)$	Go 3 right, then 2 up
B	$(-2,\ -3)$	Go 2 left, then 3 down
C	$(-4,\ 1)$	Go 4 left, then 1 up
D	$(0,\ -2)$	Stay on y-axis, go 2 down

The gradient of a line measures its steepness and direction. It tells you how much the $y$-value changes for every 1 unit increase in $x$.

Gradient Formula

$$m = \frac{\text{rise}}{\text{run}} = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

Positive gradient

$m > 0$ — slopes up ↗

Negative gradient

$m < 0$ — slopes down ↘

Zero gradient

$m = 0$ — horizontal →

Undefined

vertical — no gradient

💡 Tip: A steeper line has a larger absolute gradient. A line going up from left to right is positive; down from left to right is negative.

Worked Example — Finding gradient from two points

Find the gradient of the line through $(1,\ 3)$ and $(5,\ 11)$.

①Label the points: $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (5, 11)$

②Substitute: $m = \dfrac{11 - 3}{5 - 1} = \dfrac{8}{4} = 2$

③The gradient is $\mathbf{2}$. The line slopes upward (positive).

Worked Example — Finding gradient from two points (negative)

Find the gradient of the line through $(2,\ 7)$ and $(6,\ 3)$.

①$(x_1, y_1) = (2, 7)$, $(x_2, y_2) = (6, 3)$

②$m = \dfrac{3 - 7}{6 - 2} = \dfrac{-4}{4} = -1$

③Gradient is $-1$. The line slopes downward.

Horizontal Lines: $y = k$

Every point on a horizontal line has the same $y$-value. The equation is simply:

$y = k$ (where $k$ is any number)

Gradient = $0$
Parallel to the $x$-axis
Crosses the $y$-axis at $y = k$

Vertical Lines: $x = k$

Every point on a vertical line has the same $x$-value. The equation is:

$x = k$ (where $k$ is any number)

Gradient is undefined
Parallel to the $y$-axis
Crosses the $x$-axis at $x = k$

💡 Remember: $y = k$ is horizontal (the $y$-value is fixed). $x = k$ is vertical (the $x$-value is fixed). Don't mix them up!

The general equation of a straight line is:

$$y = mx + c$$

$m$ = gradient | $c$ = y-intercept (where the line crosses the y-axis)

Positive gradient ($m > 0$) — lines that slope upward

Negative gradient ($m < 0$) — lines that slope downward

How to draw a straight line from its equation

Identify $m$ (gradient) and $c$ (y-intercept) from $y = mx + c$.
Plot the y-intercept: place a point at $(0,\ c)$.
Use the gradient to find a second point: from the intercept, go 1 across and $m$ up (or down if $m$ is negative).
Draw a straight line through both points and extend it.

Worked Example — Drawing the line $y = 2x - 1$

①$m = 2$, $c = -1$

②Plot y-intercept at $(0,\ -1)$

③Gradient = 2, so from $(0,-1)$ go 1 right and 2 up → new point at $(1,\ 1)$

④Repeat: from $(1,1)$ go 1 right, 2 up → $(2,\ 3)$. Draw the line.

Check: Substitute $x=3$: $y = 2(3)-1 = 5$. Does the point $(3, 5)$ lie on your line?

Worked Example — Drawing $y = -\frac{1}{2}x + 3$

①$m = -\tfrac{1}{2}$, $c = 3$

②Plot y-intercept: $(0,\ 3)$

③Gradient $= -\tfrac{1}{2}$: go 2 right and 1 down → new point $(2,\ 2)$

④Go 2 right, 1 down again → $(4,\ 1)$. Draw the line through all three points.

A quadratic contains an $x^2$ term. Its graph is a smooth curve called a parabola.

Key shapes:
$y = x^2$ → U-shaped parabola (opens upward)
$y = -x^2$ → ∩-shaped parabola (opens downward)
$y = x^2 + c$ → same U-shape, shifted up or down by $c$

Plotting $y = x^2$ — building a table of values

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$x^2$	$9$	$4$	$1$	$0$	$1$	$4$	$9$
$y$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

Three parabolas: $y=x^2$ (green), $y=-x^2$ (red), $y=x^2+2$ (purple dashed)

Key facts about parabolas:
• They are symmetrical about the y-axis (for $y = x^2 + c$)
• The lowest (or highest) point is called the vertex
• $y = x^2$ has its vertex at the origin $(0,0)$
• $y = x^2 + c$ shifts the graph up by $c$ (or down if $c$ is negative)

Worked Example — Plotting $y = x^2 - 3$

①Build a table for $x = -3$ to $3$

$x$	-3	-2	-1	0	1	2	3
$x^2$	9	4	1	0	1	4	9
$x^2-3$	6	1	-2	-3	-2	1	6

②Plot each pair $(x, y)$ on the grid.

③Join with a smooth curve. The vertex is at $(0, -3)$.

④Compare to $y = x^2$: the graph has shifted 3 units downward.

Coordinates $(x, y)$

Across first ($x$), then up/down ($y$). Origin = $(0,0)$. Four quadrants.

Gradient $m$

$m = \dfrac{y_2-y_1}{x_2-x_1}$. Positive ↗, negative ↘, zero → horizontal.

Horizontal: $y = k$

Fixed $y$-value, $m = 0$, parallel to x-axis.

Vertical: $x = k$

Fixed $x$-value, gradient undefined, parallel to y-axis.

Straight line: $y=mx+c$

$m$ = gradient, $c$ = y-intercept. Plot intercept then use gradient.

Quadratic: $y=x^2$

U-shaped parabola, vertex at origin. $y=-x^2$ opens down. $+c$ shifts up.

Question 1

Write down the coordinates of the following points on a grid:

A is 3 right and 4 up from the origin. B is 2 left and 1 down from the origin. C is on the y-axis, 5 up.

▶ Show solution

A = $(3,\ 4)$

B = $(-2,\ -1)$

C = $(0,\ 5)$ (on the y-axis means $x=0$)

Question 2

Find the gradient of the line passing through the points $(2,\ 5)$ and $(6,\ 13)$.

▶ Show solution

$$m = \frac{13-5}{6-2} = \frac{8}{4} = 2$$

The gradient is $\mathbf{2}$ (positive — line slopes upward).

Question 3

Find the gradient of the line passing through $(-1,\ 6)$ and $(3,\ -2)$.

▶ Show solution

$$m = \frac{-2-6}{3-(-1)} = \frac{-8}{4} = -2$$

The gradient is $-2$ (negative — line slopes downward).

Question 4

Write the equation of a horizontal line passing through the point $(3,\ -4)$.

▶ Show solution

A horizontal line has a fixed $y$-value. The $y$-value of the point is $-4$.

Equation: $y = -4$

Question 5

Write the equation of a vertical line passing through the point $(-2,\ 7)$.

▶ Show solution

A vertical line has a fixed $x$-value. The $x$-value of the point is $-2$.

Equation: $x = -2$

Question 6

For the line $y = 3x - 2$, state the gradient and the y-intercept, then find the coordinates of the point where it crosses the x-axis.

▶ Show solution

Comparing with $y = mx + c$: gradient $m = 3$, y-intercept $c = -2$.

The line crosses the x-axis when $y = 0$:

$0 = 3x - 2 \Rightarrow 3x = 2 \Rightarrow x = \dfrac{2}{3}$

The line crosses the x-axis at $\left(\dfrac{2}{3},\ 0\right)$.

Question 7

Draw the line $y = -2x + 4$ by finding the y-intercept and one more point.

▶ Show solution

$m = -2$, $c = 4$.

Y-intercept: plot $(0,\ 4)$.

Gradient $= -2$: from $(0,4)$ go 1 right, 2 down → $(1,\ 2)$.

Go 1 right, 2 down again → $(2,\ 0)$ (this is the x-intercept).

Draw a straight line through $(0, 4)$, $(1, 2)$, $(2, 0)$.

Question 8

Complete the table of values for $y = x^2 + 1$, then describe what the graph looks like.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	?	?	?	?	?	?	?

▶ Show solution

$y = x^2 + 1$:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$10$	$5$	$2$	$1$	$2$	$5$	$10$

The graph is a U-shaped parabola, symmetrical about the y-axis, with its vertex (minimum point) at $(0,\ 1)$ — which is $y=x^2$ shifted 1 unit upward.

Question 9

A line has gradient $-3$ and passes through the point $(0,\ 5)$. Write its equation and find the value of $y$ when $x = 4$.

▶ Show solution

The line passes through $(0,5)$ so the y-intercept is $c = 5$. Gradient $m = -3$.

Equation: $y = -3x + 5$

When $x = 4$: $y = -3(4) + 5 = -12 + 5 = -7$

Question 10

The graph of $y = -x^2 + 4$ is a parabola.

(a) What is the maximum value of $y$, and at what value of $x$ does it occur?

(b) At what values of $x$ does the graph cross the x-axis?

(c) Is the parabola U-shaped or ∩-shaped?

▶ Show solution

(a) $y = -x^2 + 4$ is maximised when $x^2$ is smallest, i.e. when $x = 0$: $y_{\max} = 4$. The vertex is at $(0,\ 4)$.

(b) Set $y = 0$: $-x^2 + 4 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. The graph crosses the x-axis at $(-2, 0)$ and $(2, 0)$.

(c) Because the coefficient of $x^2$ is negative ($-1$), the parabola opens downward — it is ∩-shaped.