IB Applications and Interpretation SL · Functions · Section 2 of 7

Linear functions and linear models

3 worked examples · By Álvaro López

Key idea

A linear function $f (x) = m x + c$ has a straight line as its graph: $m$ is the gradient (how much $y$ changes for each unit increase in $x$ ) and $c$ is the value at $x = 0$ . It is used to model any situation with a constant rate of change, such as a cost with a fixed charge or a tank that empties at a steady rate.

Gradient and y-intercept

The gradient measures the steepness of a line. If $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two points on the line, the gradient is

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

A positive gradient indicates an increasing line, a negative one a decreasing line, and $m = 0$ gives a horizontal line. A vertical line has no gradient and is not the graph of a function. The $y$ -intercept is the value of $y$ when $x = 0$ , that is, the point where the line cuts the $y$ -axis.

Line y = 2x + 1 with y-intercept 1, points A(2; 5) and B(6; 13), an auxiliary triangle with Δx = 4 and Δy = 8, and gradient m = 2

Forms of the equation of a line

To convert from one form to another, simply rearrange. For example, the line with gradient $- 3$ passing through $(1, 4)$ is $y - 4 = - 3 (x - 1)$ , that is, $y = - 3 x + 7$ . In general form it is $3 x + y - 7 = 0$ . If you are asked for the equation of the line through two points, first calculate $m$ and then substitute one of the points into $y - y_{1} = m (x - x_{1})$ .

Parallel and perpendicular lines

Parallel lines have the same steepness and so never meet. Perpendicular lines form a right angle: the gradient of one is the negative reciprocal of the other. If $m_{1} = \frac{2}{5}$ , a perpendicular has $m_{2} = - \frac{5}{2}$ .

Piecewise functions

Sometimes a situation follows one rule over one interval and a different rule over the next. It is described by a piecewise function, which has a formula for each interval. A car park may charge $1.50$ € per hour for the first four hours and $3$ € for each hour after that:

$C (t) = {1.5 t 6 + 3 (t - 4) if 0 \leq t \leq 4 if t > 4$

To calculate $C (5)$ choose the piece to which $t = 5$ belongs: $C (5) = 6 + 3 \cdot 1 = 9$ €. To draw it, plot each formula only on its interval. On the GDC you do this by entering each formula with its interval of $x$ .

Linear models

A linear model is a function $f (x) = m x + c$ that describes a real situation. It is solved in three steps: translate the question into a function, solve using mathematics and interpret the result in context. In the model:

$m$ is the rate of change: how much the dependent variable increases (if $m > 0$ ) or decreases (if $m < 0$ ) for each unit of the independent variable. Its unit must be given, for example "euros per kilogram".
$c$ is the initial value: the value of the variable when $x = 0$ , such as a fixed charge.

If $c = 0$ , the relationship is one of direct proportion and $m = \frac{y}{x}$ is the constant of proportionality.

Point of intersection of two lines

Two models are set equal where their graphs cross. To find that point, equate the two expressions and solve for $x$ , or draw the two lines on the GDC and find the intersection point. Then substitute $x$ into either function to obtain the $y$ -coordinate. In a supply and demand context, that point is the equilibrium of the market: the price at which the quantity people want to buy equals the quantity people want to sell.

Worked examples

Example 1 Basic

Line through two points

A line passes through the points $(2, 5)$ and $(6, 13)$ .

a) Find its gradient.

b) Write its equation in the form $y = m x + c$ .

Solution

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

b) Using the point–gradient form and the point $(2, 5)$ : $y - 5 = 2 (x - 2)$ , so $y = 2 x + 1$ . Check with the other point: $2 \cdot 6 + 1 = 13$ . Correct.

Example 2 Medium

Parallel and perpendicular through a point

The line $L$ has equation $y = - \frac{2}{3} x + 4$ . Consider the point $P (6, 1)$ .

a) Find the equation of the line parallel to $L$ that passes through $P$ .

b) Find the equation of the line perpendicular to $L$ that passes through $P$ .

Solution

a) A parallel line has the same gradient, $m = - \frac{2}{3}$ . Using the point–gradient form: $y - 1 = - \frac{2}{3} (x - 6) \Rightarrow y = - \frac{2}{3} x + 4 + 1 = - \frac{2}{3} x + 5$

b) The gradient of a perpendicular satisfies $m \cdot (- \frac{2}{3}) = - 1$ , so $m = \frac{3}{2}$ . Then $y - 1 = \frac{3}{2} (x - 6) \Rightarrow y = \frac{3}{2} x - 9 + 1 = \frac{3}{2} x - 8$

Check of b): at $x = 6$ , $\frac{3}{2} \cdot 6 - 8 = 1$ , so it passes through $P$ .

Example 3 Hard

Supply and demand for oranges

In a market, the demand for oranges (kilograms people want to buy per day) depends on the price $p$ , in euros per kilogram, according to $D (p) = 420 - 60 p$ . The supply (kilograms put on sale per day) is $S (p) = 40 p + 60$ . Both are valid for prices between $0$ and $7$ €/kg.

a) Calculate how many kilograms are demanded if the price is $2$ €/kg.

b) Find the equilibrium price and the quantity sold at that price.

c) State the gradient of $D$ and interpret it.

Supply and demand: decreasing demand line D(p) = 420 − 60p and increasing supply line S(p) = 40p + 60, which meet at the equilibrium (3.6, 204)

Solution

a) $D (2) = 420 - 60 \cdot 2 = 300$ . 300 kg are demanded per day.

b) At equilibrium supply and demand are equal: $D (p) = S (p)$ . $420 - 60 p = 40 p + 60 \Rightarrow 360 = 100 p \Rightarrow p = 3.6$ The quantity is $S (3.6) = 40 \cdot 3.6 + 60 = 204$ . Check with demand: $D (3.6) = 420 - 216 = 204$ . Using the GDC, the intersection point of the two lines is $(3.6; 204)$ . The equilibrium price is $3.60$ €/kg and 204 kg are sold per day.

c) The gradient of $D$ is $- 60$ . It means that for every $1$ € increase in the price per kilogram, demand falls by 60 kg per day.

In the exam: give the price and the quantity with their units, and write the equation $D (p) = S (p)$ before solving it. If you interpret a gradient, state the quantity, the sign (rises or falls), the unit and which variable changes for each unit of which other variable.

Common mistakes

Summary

Gradient: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ . It is the rate of change; $c$ is the value at $x = 0$ .
Forms of the line: $y = m x + c$ , $y - y_{1} = m (x - x_{1})$ and $a x + b y + d = 0$ .
Parallel: $m_{1} = m_{2}$ . Perpendicular: $m_{1} m_{2} = - 1$ .
A piecewise function uses one formula per interval; it is evaluated by choosing the piece that applies.
A linear model is interpreted: $m$ with its unit and $c$ as the initial value.
The intersection of two lines is found by equating the functions or with the GDC.

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