Linear Functions and Graphs

A pair of coordinates $(x, y)$ pins a single point on a grid. The first number, $x$, says how far across the point sits from the origin; the second number, $y$, says how far up or down. The origin itself is $(0, 0)$, where the two axes cross.

The plane splits into four quadrants, labelled anticlockwise starting at the top right.

A point like $(3, 2)$ sits in the first quadrant, where $x$ and $y$ are both positive. A point like $(-4, 1)$ sits in the second quadrant (negative $x$, positive $y$). The order matters: $(3, 2)$ and $(2, 3)$ are different points. To plot, go along the corridor first, then up the stairs — never the other way around.

To plot a straight-line graph from an equation like $y = 2x + 1$, the friendly recipe is:

1. Make a table of $x$-values, often $x = -2, -1, 0, 1, 2, 3$.
2. Work out each $y$ by substituting into the equation.
3. Plot each point as $(x, y)$ on the grid.
4. Draw one straight line that passes through every point — use a ruler, and extend the line slightly beyond your points.

For $y = 2x + 1$ the table is:

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

So plot $(-2, -3)$, $(-1, -1)$, $(0, 1)$, $(1, 3)$, $(2, 5)$ and rule a line through them.

If a single point looks off the line, treat it as a warning: re-check that one calculation rather than forcing the ruler through it. A straight-line graph should never need "best-fit" — every correct point sits on the line.

For special cases, lines like $y = 4$ are horizontal (every point has $y = 4$), and lines like $x = -2$ are vertical (every point has $x = -2$). These are quick to draw — no table needed.

Every straight line has two key measurements: a gradient (how steep it is) and a y-intercept (where it crosses the y-axis).

The gradient, written $m$, tells you the steepness. Pick any two points on the line and compute

$m = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}$

A positive gradient slopes up to the right; a negative gradient slopes down; a gradient of zero gives a horizontal line.

The y-intercept, written $c$, is the $y$-value of the point where the line crosses the y-axis (so $x = 0$). For $y = 2x + 1$, set $x = 0$ to get $y = 1$, so the y-intercept is $1$ — the line crosses at $(0, 1)$.

Picking two clean points helps. For a line through $(2, 7)$ and $(5, 13)$ the gradient is $\dfrac{13 - 7}{5 - 2} = \dfrac{6}{3} = 2$. Choose points where the line passes exactly through a grid intersection — your numbers will be cleaner and the answer easier to check.

The equation $y = mx + c$ is the single rule that links every straight-line graph to its picture:

• $m$ is the gradient.
• $c$ is the y-intercept.

So $y = 3x - 2$ has gradient $3$ and crosses the y-axis at $-2$. The line $y = -x + 5$ has gradient $-1$ (slopes down) and crosses at $5$.

This makes a lot of questions painless.

• Sketch a line from an equation — mark the y-intercept first, then use the gradient as rise-over-run to step to another point.
• Find the equation from a graph — read $c$ off the y-axis, then use any two clean grid points to find $m$.
• Check whether a point lies on a line — substitute the coordinates into the equation. If both sides match, the point lies on the line.

For example, does $(4, 10)$ lie on $y = 3x - 2$? Substitute: $3(4) - 2 = 10$. Yes! The point lies on the line.

Sometimes the equation is given in a different form, like $2x + y = 7$. Rearrange to make $y$ the subject: $y = -2x + 7$. Gradient $-2$, y-intercept $7$. The form $y = mx + c$ is just one rearrangement away.

Two lines are parallel when they have the same gradient $m$ but different y-intercepts. So $y = 2x + 1$ and $y = 2x - 4$ are parallel — both step up $2$ for every $1$ across, but they cross the y-axis at different heights.

If you need a line parallel to $y = 2x + 1$ passing through $(0, -3)$: keep $m = 2$, change $c$ to $-3$, so $y = 2x - 3$.

A few special lines are worth memorising:

• $y = c$ is horizontal — gradient $0$, crosses the y-axis at $c$.
• $x = a$ is vertical — gradient is undefined, crosses the x-axis at $a$.
• $y = x$ is the diagonal through the origin — gradient $1$, y-intercept $0$.
• $y = -x$ is the other diagonal — gradient $-1$, y-intercept $0$.

For a vertical line, the gradient is undefined, not zero — you would be dividing by zero (no run). That's a small but important distinction worth knowing by heart.

Real-life situations often produce straight-line graphs — and the same skills unlock them. The key questions become:

• What does the gradient mean? It is a rate of change — kilometres per hour, dollars per item, litres per minute.
• What does the y-intercept mean? It is the starting value — distance when $t = 0$, cost when there are zero items, fuel at the start.

For a distance-time graph with $d = 60t + 10$ (km against hours), the gradient $60$ means the speed is $60$ km/h, and the y-intercept $10$ means the journey started $10$ km from the reference point.

For a phone plan that charges \3$plus$$2$per minute, the cost$C$in dollars after$m$minutes is$C = 2m + 3$. A graph of$C$against$m$is a straight line with gradient$2$($$2$per minute) and y-intercept$3$(the$$3$fixed cost). Reading the graph, you can answer questions like *"How many minutes for$$25$?"* by setting$C = 25$:$25 = 2m + 3$, so$m = 11$ minutes.

Always label the axes carefully, use the scale the question provides, and pay attention to the units of the gradient.

Linear functions and graphs are the foundation of so much that follows:

• Simultaneous equations are solved graphically by finding where two lines cross.
• Inequalities like $y \ge 2x + 1$ are shown by shading regions above or below a line.
• Quadratic graphs build on the same coordinate skills but use curves rather than lines.
• In science, almost every "best-fit line" reading uses gradient (rate) and intercept (starting value).

If a later topic feels new, ask whether it is really another reading of $y = mx + c$. Master gradient, intercept and plotting now and the rest of graph work falls into place.