⚙️ Function Kingdom · Functions

Linear Functions

Work with straight lines: slope as a rate of change, intercepts, and the equation $y =$ mx $+ b$ .

In short

Slope is rise over run — the constant rate of change of a linear function.
In $y =$ mx $+ b, m$ is the slope and b is the y-intercept (the value when $x = 0)$ .
For the y-intercept set $x = 0$ ; for the x-intercept set $y = 0$ .
A fixed start plus a constant change per step is always a linear model, with a negative slope when something decreases.

Constant rate of change

A function is linear when it changes by the same amount for every step of the input. Its graph is a straight line, and the size of that constant step is the slope.

$m =$ (change in $y) /$ (change in $x) =$ rise $/$ run

Slope is a rate: metres per second, gold per quest, degrees per hour. A slope of 3 means "up 3 for every 1 across"; a slope of -2 means "down 2 for every 1 across". A horizontal line has slope 0, and a vertical line has no slope at all, because its run is 0 and division by zero is not allowed.

For two points $(x1, y1)$ and $(x2, y2)$ :

$m = (y2$ - $y1) / (x2$ - x1)

Start with the same point on the top and the bottom. Mixing the order flips the sign.

Slope-intercept form

The most useful way to write a line is

$y = m x + b$

where m is the slope and b is the y-intercept — the height at which the line crosses the y-axis, that is the value of y when $x = 0$ .

Reading a line from a graph is then a two-step routine: find where it crosses the y-axis (that is b), then count the rise over a convenient run (that is m).

The x-intercept is where the line crosses the x-axis, and every point there has $y = 0$ . So set $y = 0$ and solve:

$y = 3x$ - $12 -> 3x$ - $12 = 0 -> x = 4$

Do not confuse the two intercepts. For the y-intercept set $x = 0$ ; for the x-intercept set $y = 0$ .

Building the equation of a line

From a point and a slope. You already know m, so only b is missing. Substitute the point and solve:

slope 4 through $(2, 11): 11 = 4(2) + b$ , so $b = 3$ , and $y = 4x + 3$

From two points. Find the slope first, then use either point to find b:

$(1, 5)$ and $(3, 11): m = (11$ - $5)/(3$ - $1) = 3 5 = 3(1) + b$ , so $b = 2$ , and $y = 3x + 2$

Always check your equation against the point that you did not use — here $3(3) + 2 = 11$ . Correct.

An equivalent shortcut is point-slope form, y - $y1 =$ m(x - x1), which rearranges to the same line.

Linear models in the world

Whenever there is a fixed starting amount plus a constant change per step, the situation is linear.

"Join the guild for 30 gold, then pay 5 gold per quest" becomes

$y = 5x + 30$

Here $b = 30$ is the joining fee (the cost before any quest) and $m = 5$ is the cost per quest.

When something is being used up, the rate is negative: "start with 80 arrows, use 6 per hour" gives $y = -6x + 80$ .

The four representations from function notation all reappear here. The table shows a constant step; the graph shows a straight line; the words describe a start and a rate; the equation lets you compute. Being able to switch between them is what makes linear functions so useful.

Worked examples

Example 1

Find the slope of the line through $(-2, 7)$ and $(4, -5)$ .

Use $m = (y2$ - $y1) / (x2$ - x1), keeping the same point first on top and bottom.
Change in $y: -5$ - $7 = -12$ .
Change in $x: 4$ - $(-2) = 6$ .
$m = -12 / 6 = -2$ . The negative sign means the line falls as x increases.

Example 2

Write the equation of the line with slope -3 passing through $(5, 4)$ .

Start from $y =$ mx $+ b$ with $m = -3$ , so $y = -3x + b$ .
Substitute the point: $4 = -3(5) + b$ .
$4 = -15 + b$ , so $b = 19$ .
The line is $y = -3x + 19$ . Check: $-3(5) + 19 = 4$ . Correct.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

A line passes through $(2, 7)$ and $(5, 9)$ . What is its slope? Give the answer as a fraction in lowest terms (a whole number if it simplifies to one).

Answer: $2/3$

$m = (y2$ - $y1) / (x2$ - x1)
$m = (9$ - $(7)) / (5$ - (2))
$m = 2 / 3$
$m = 2/3$

Problem 2

Difficulty 3 of 5

A straight line crosses the y-axis at $(0, 8)$ and also passes through $(2, 16)$ . Write its equation as an expression in x (for example 2x - 5).

Answer: $4x + 8$

y-intercept: the line meets the y-axis at $(0, 8)$ , so $b = 8$ .
Slope: rise $/$ run $= 8 / 2 = 4$ .
$y = 4x + 8$

Problem 3

Difficulty 4 of 5

A line has slope -6 and passes through the point $(0, -12)$ . Write its equation as an expression in x (for example 2x - 5).

Answer: -6x - 12

$y =$ mx $+ b$ with $m = -6$ .
Substitute $(0, -12): -12 = -6(0) + b$
$-12 = 0 + b$ , so $b = -12$
$y = -6x$ - 12

Common mistakes

Computing run over rise instead of rise over run.
Subtracting the coordinates in opposite orders on the top and bottom, which flips the sign.
Swapping m and b when reading or writing $y =$ mx $+ b$ .
Forgetting the starting amount, so the model becomes rate x time with no intercept.

What you should be able to do

Compute the slope between two points.
Write the equation of a line from a slope and a point.
Interpret slope and intercept in a real-world context.
Identify parallel and perpendicular slopes.

Where this fits in the curriculum

Common Core

8.EE.B.6
Grade 8 — Use similar triangles to explain slope, and derive y = mx + b for a line through the origin and elsewhere.
8.F.B.4
Grade 8 — Construct a function to model a linear relationship and interpret its rate of change and initial value.
HSG-GPE.B.5
High school — Prove and use the slope criteria for parallel and perpendicular lines.

Ontario

MTH1W.C3.2
Grade 9 de-streamed — Represent linear relations using tables of values, graphs and equations, and identify the rate of change and the initial value.
MTH1W.C4.4
Grade 9 de-streamed — Determine the equations of lines by connecting rates of change with slopes and initial values with y-intercepts.

SAT

Heart of Algebra
Slope, intercepts, and the equation and graph of a line.