Introduction to Graphs

A graph turns a relationship between two quantities into a picture. Once you can read that picture — where a line crosses, how steeply it climbs — you can answer questions about the relationship without doing any algebra at all. This topic sets up the machinery you will use for the rest of the course.

The Cartesian plane

Two number lines crossing at right angles make the Cartesian plane. The horizontal line is the $x$-axis, the vertical line is the $y$-axis, and they meet at the origin, the point $(0,0)$.

Every point is named by an ordered pair $(x, y)$: go across first, then up. The order matters — $(3, 5)$ and $(5, 3)$ are different points.

The axes cut the plane into four quadrants, numbered anticlockwise starting from the top right:

A point sitting on an axis is in no quadrant at all: $(0, 5)$ is on the $y$-axis, $(-2, 0)$ is on the $x$-axis.

Choosing a scale

When you draw a graph yourself, the scale is your decision, and marks are given for a sensible one.

• Use the grid you are given. If the paper offers 20 squares and your $x$ values run from $0$ to $10$, use 2 squares per unit — not 1, which wastes half the page.
• Keep each axis uniform. Squares must not jump from 1 unit to 5 units partway along.
• The two axes need not match each other. $x$ in steps of $1$ and $y$ in steps of $10$ is perfectly acceptable if the numbers call for it.
• Label both axes and state the scale.

The equation of a straight line

Every straight line that is not vertical can be written

$$y = mx + c$$

where $m$ is the gradient and $c$ is the $y$-intercept.

The gradient tells you how steep the line is: it is the change in $y$ for every $1$ unit increase in $x$. The $y$-intercept tells you where the line crosses the $y$-axis, at the point $(0, c)$.

For $y = 5x + 2$: the gradient is $5$ and the line crosses the $y$-axis at $(0, 2)$. Every step of $1$ to the right lifts the line $5$ units.

For $y = -2x + 7$: the gradient is $-2$. A negative gradient means the line falls as you move right. It crosses the $y$-axis at $(0, 7)$.

Sign and steepness at a glance

The size of $m$ is the steepness, ignoring direction. A line with gradient $-4$ is steeper than one with gradient $2$, even though it is falling.

Gradient from two points

If you know two points on a line, $(x_1, y_1)$ and $(x_2, y_2)$, the gradient is

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Rise over run: the change in $y$ divided by the change in $x$.

Find the gradient of the line through $(1, 2)$ and $(5, 10)$.

$$m = \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2$$

The line rises $2$ units for every $1$ across.

It does not matter which point you call first, as long as you are consistent. Taking them the other way round:

$$m = \frac{2 - 10}{1 - 5} = \frac{-8}{-4} = 2$$

Same answer.

Is a point on the line?

A point lies on a line exactly when its coordinates satisfy the equation. Substitute and check.

Does $(2, 7)$ lie on $y = 3x + 1$?

Substitute $x = 2$:

$$y = 3(2) + 1 = 7$$

The equation gives $7$ and the point has $y = 7$, so yes, $(2, 7)$ is on the line.

Does $(4, 10)$ lie on $y = 3x + 1$?

$$y = 3(4) + 1 = 13 \ne 10$$

So no. The point $(4, 13)$ is on the line; $(4, 10)$ sits $3$ units below it.

Finding the equation of a line

There are two standard situations.

Given the gradient and one point

Put the gradient into $y = mx + c$, substitute the point to find $c$, then write the equation out.

Find the equation of the line with gradient $2$ passing through $(3, 1)$.

Start with $m = 2$:

$$y = 2x + c$$

The line goes through $(3, 1)$, so $x = 3$ and $y = 1$ must fit:

$$1 = 2(3) + c$$ $$1 = 6 + c$$ $$c = -5$$

So the equation is

$$y = 2x - 5$$

Check: at $x = 3$, $y = 2(3) - 5 = 1$. Correct.

Given two points

Find the gradient first, then use either point exactly as above.

Find the equation of the line through $(1, 4)$ and $(3, 10)$.

$$m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$$

So $y = 3x + c$. Using $(1, 4)$:

$$4 = 3(1) + c \quad \Rightarrow \quad c = 1$$

The equation is $y = 3x + 1$.

Check with the other point, which you have not used yet: at $x = 3$, $y = 3(3) + 1 = 10$. Correct — and this check is worth doing every time, because it catches an arithmetic slip in the gradient.

Special cases

Horizontal and vertical lines

A horizontal line has gradient $0$. Its equation is $y = k$ — every point on it has the same $y$-coordinate. The line through $(1, 3)$ and $(7, 3)$ is $y = 3$.

A vertical line has undefined gradient, because the run is zero and you cannot divide by zero. Its equation is $x = k$. The line through $(4, 1)$ and $(4, 9)$ is $x = 4$.

Parallel lines

Two lines are parallel exactly when their gradients are equal. $y = 3x + 1$ and $y = 3x - 8$ are parallel: same steepness, different intercepts, so they never meet.

Find the line parallel to $y = 4x - 1$ passing through $(2, 3)$.

Parallel means the same gradient, $m = 4$:

$$3 = 4(2) + c \quad \Rightarrow \quad c = 3 - 8 = -5$$

So $y = 4x - 5$.

Intercepts

The $y$-intercept is found by setting $x = 0$; the $x$-intercept by setting $y = 0$.

For $y = 2x - 6$:

• $x = 0$ gives $y = -6$, so the $y$-intercept is $(0, -6)$.
• $y = 0$ gives $0 = 2x - 6$, so $x = 3$ and the $x$-intercept is $(3, 0)$.

Those two points are enough to draw the whole line.