Linear function – Scitura

What is it about?

A linear function has the equation

f(x) = m \cdot x + c

Its graph is always a straight line.

The y-intercept $c$ tells you where the line crosses the y-axis: at the point $(0, c)$ .
The slope $m$ tells you how steep the line is: go 1 to the right and the line goes $m$ up – or down if $m$ is negative. This is what the slope triangle shows.

If you know two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ on the line, the slope is

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}

The zero (x-intercept) is where the line crosses the x-axis. Set $f(x) = 0$ : $x_0 = -\frac{c}{m}$ (for $m \neq 0$ ). For $m = 0$ the line is horizontal: if $c \neq 0$ there is no zero, if $c = 0$ the line lies on the x-axis.

A vertical line is not the graph of a function, because one $x$ -value belongs to infinitely many $y$ -values. That is why you cannot set one up in the simulation.

Explore

Change only $c$ . What happens to the line – and what stays the same? Load example
Change only $m$ and watch the slope triangle. What happens for $m = 0$ ? Load example
Drag the highlighted points directly in the coordinate system: the point on the y-axis and the tip of the slope triangle.
Turn on the second line $g$ . When do two lines not intersect? Parallel lines

Exercises

Exercise 1: Read off the equation

Load exercise Read the slope and the y-intercept from the graph and write down the equation. The results on the right are hidden – reveal them only to check.

Show solution

The line crosses the y-axis at 2, so $c = 2$ . Going 1 to the right, it goes 1.5 down, so $m = -1.5$ :

$f(x) = -1.5x + 2$

Exercise 2: Line through two points

Find the equation of the line through $A(-2, -1)$ and $B(2, 5)$ . Then set it up in the simulation and check that it passes through both points.

Show solution

$m = \frac{5 - (-1)}{2 - (-2)} = \frac{6}{4} = 1.5$

Substituting $A$ : $-1 = 1.5 \cdot (-2) + c$ , so $c = 2$ and $f(x) = 1.5x + 2$ . Show in simulation

Exercise 3: Find the zero

Calculate the zero of $f(x) = 0.5x - 2$ and check it in the simulation.

Show solution

$0 = 0.5x - 2 \;\Rightarrow\; x = 4$ . The line crosses the x-axis at $N(4, 0)$ . Show in simulation

Exercise 4: Perpendicular lines

Set up $f(x) = 2x - 1$ and turn on the second line. Find a line $g$ that crosses $f$ at a right angle. What do you notice about the two slopes?

Show solution

For example $g(x) = -0.5x + 2$ . For perpendicular lines $m \cdot m_2 = -1$ , so $m_2 = -\frac{1}{2}$ . The y-intercept of $g$ can be anything. Show in simulation

Exercise 5: Phone tariff (modelling)

A phone tariff costs a basic fee of €5 per month plus €2 per gigabyte used.

Write down the equation for the monthly cost ( $x$ : gigabytes, $f(x)$ : cost in €).
How much does a month with 3 GB cost?
A second tariff costs €8 basic fee plus €1 per GB. From how many GB on is it cheaper?

Show solution

$f(x) = 2x + 5$
$f(3) = 2 \cdot 3 + 5 = 11$ , so €11.
$g(x) = x + 8$ . Intersection: $2x + 5 = x + 8 \Rightarrow x = 3$ . At 3 GB both tariffs cost €11. For more than 3 GB the second tariff is cheaper. Show both tariffs

Ideas for the classroom

Use: introduction in grade 7/8 or revision before quadratic functions. On the projector, let the class predict what will happen before moving a slider.
Notation: The simulation writes $f(x) = mx + c$ (the German version writes $f(x) = mx + t$ ). If your textbook uses $b$ for the y-intercept, point this out briefly.
Common stumbling blocks:
- Slope and y-intercept are confused.
- For a negative slope, the slope triangle is drawn “upwards”.
- The constant function ( $m = 0$ ) is not recognised as a linear function.
- A graph that looks flatter is taken to have a smaller slope, although only the axes are scaled differently. The simulation uses equal units on both axes.
Exercise mode: Share → Hide results creates a link or QR code with the equation, zero and intersection hidden – ideal for reading-off exercises. Lock controls prevents accidental changes.
Differentiation: Faster learners use the second line to explore parallel and perpendicular lines or solve systems of linear equations graphically.