Exercises: Drawing graphs of linear functions

1
Draw the graph corresponding to the given table of values.
$𝐱$
$- 8$
$- 4$
$0$
$4$
$8$
$𝐲$
$0$
$1$
$2$
$3$
$4$

For this task you need the following basic knowledge: Lines in coordinate systems
Draw the five given points in a coordinate system and draw a straight line through the points.

$𝐱$
$- 2$
$0$
$2$
$4$
$𝐲$
$- 20$
$- 10$
$0$
$10$

For this task you need the following basic knowledge: Lines in coordinate systems
Draw the four given points in a coordinate system and draw a straight line through the points.
2
Draw the graph of the linear functions in a coordinate system!
$f (x) = - 2 x + 4$

For this task you need the following basic knowledge: Slope/Gradient of a line
Drawing the linear function
First read off the $y$ -intercept and the slope from the function term of the linear function.
In this case:
$f (x) = - 2 x + 4$
You obtain a $y$ -intercept of $t = 4$ and a slope of $m = - 2$ .
First draw the intersection with the $y$ -axis that results from the $y$ -axis intercept. This is at $A (0 / 4)$ .
Then draw a gradient triangle using the gradient. To do this, go one length unit to the right and two length units down. This gives you the point $B = (1 / 2)$ . Now draw the straight line through points $A$ and $B$ .
You obtain the graph $G_{f}$ of $f (x)$ .

$g (x) = \frac{1}{2} x - 2$

For this task you need the following basic knowledge: Slope/Gradient of a line
Drawing the linear function
First read off the $y$ -intercept and the slope from the function term of the linear function.
In this case:
$f (x) = \frac{1}{2} x - 2$
You obtain a $y$ -intercept of $t = - 2$ and a slope of $m = \frac{1}{2}$ .
First draw the intersection with the $y$ -axis that results from the $y$ -axis intercept. This is at $C (0 / - 2)$ .
Then draw a gradient triangle using the gradient. To do this, go one length unit to the right and two length units down. This gives you the point $D = (2 / - 1)$ . Now draw the straight line through points $C$ and $D$ .
You obtain the graph $G_{g}$ of $g (x)$ .

$h (x) = 5$

For this task you need the following basic knowledge: Slope/Gradient of a line
Drawing the linear function
The function $h (x) = 5$ represents a special case of a linear function. The slope of $h (x)$ is $0$ .
This means that the function value does not change regardless of the variable $x$ .
So if you draw the function value $h (x) = 5$ for each $x$ in a coordinate system, you get a straight line that runs parallel to the $x$ axis at the height $y = 5$ .
3
Draw the graphs of the functions with the following equation:
$y = 3 x - 2$

For this task you need the following basic knowledge: Linear function
Determine one point
$y = 3 x - 2$
$- 2$ is the $t$ within the general line equation, or in other words, the $y$ -axis intercept.
$\Rightarrow P (0 | - 2)$
Find the slope
Determine the slope $m$ of the function
$y = 3 x - 2$
$3$ is the $m$ within the general line equation, or in other words, the slope.
$m = 3$
Draw the line
Go from the previously determined point one unit to the right and $3$ upwards, since $m$ is equal to $3$ . Here is a second point of the function.
Then connect the two points to form a straight line.

$y = 2 - x$

For this task you need the following basic knowledge: Linear function
Transform the equation
$y = 2 - x$
First, we do a little transformation, such that we get the form of a general line equation.
$y = - x + 2$
Determine one point
$y = - x + 2$
$+ 2$ is the $t$ within the general line equation, or in other words, the $y$ -axis intercept.
$\Rightarrow P (0 | 2)$
Find the slope
Determine the slope $m$ of the function
$y = - x + 2$
$- 1$ is the $m$ within the general line equation, or in other words, the slope.
$m = - 1$
Draw the line
From the previously determined point, go one unit to the right and $1$ downwards, since $m$ is negative. Here is a second point of the function.
Then connect the two points to form a straight line.

$y = - \frac{3}{4} x - 1$

For this task you need the following basic knowledge: Linear function
Determine one point
$y = - \frac{3}{4} x - 1$
$- 1$ is the $t$ within the general line equation, or in other words, the $y$ -axis intercept.
$\Rightarrow P (0 | - 1)$
Find the slope
Determine the slope $m$ of the function
$y = - \frac{3}{4} x - 1$
$- \frac{3}{4}$ is the $m$ within the general line equation, or in other words, the slope.
$m = - \frac{3}{4}$
Draw the line
From the previously determined point, go one unit to the right and $\frac{3}{4}$ downwards, since $m$ is negative. Here is a second point of the function.
Then connect the two points to form a straight line.

$y = - \frac{1}{2} x + 2$

For this task you need the following basic knowledge: Linear function
Determine one point
$y = - \frac{1}{2} x + 2$
$+ 2$ is the $t$ within the general line equation, or in other words, the $y$ -axis intercept.
$\Rightarrow P (0 | 2)$
Find the slope
Determine the slope $m$ of the function
$y = - \frac{1}{2} x + 2$
$- \frac{1}{2}$ is the $m$ within the general line equation, or in other words, the slope.
$m = - \frac{1}{2}$
Draw the line
From the previously determined point, go one unit to the right and $\frac{1}{2}$ downwards, since $m$ is negative. Here is a second point of the function.
Then connect the two points to form a straight line.

$y = \frac{3}{4} x + 1$

For this task you need the following basic knowledge: Linear function
Determine one point
$y = \frac{3}{4} x + 1$
$1$ is the $t$ within the general line equation, or in other words, the $y$ -axis intercept.
$\Rightarrow P (0 | 1)$
Find the slope
Determine the slope $m$ of the function
$y = \frac{3}{4} x + 1$
$\frac{3}{4}$ is the $m$ within the general line equation, or in other words, the slope.
$m = \frac{3}{4}$
Draw the line
From the previously determined point go one unit to the right and $\frac{3}{4}$ upwards, since $m$ is positive. Here is a second point of the function.
Then connect the two points to form a straight line.
4
Draw the graphs of the following lines with the point of intersection with the $y$ -axis and the gradient triangle. Calculate the point of intersection with the $x$ -axis and check the result using the graph.
$f (x) = 2 x - 5$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = 2 x - 5$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 5)$
$\Rightarrow m_{f} = 2$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$2 x - 5$ $=$ $0$ $+ 5$
$2 x$ $=$ $5$ $: 2$
$x_{0}$ $=$ $2.5$
$\Rightarrow P_{x} (2.5 | 0)$

$f (x) = - x - 3$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - x - 3$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 3)$
$\Rightarrow m_{f} = - 1$
$- x - 3$ $=$ $0$ $+ x$
$- 3$ $=$ $x_{0}$
$\Rightarrow P_{x} (- 3 | 0)$

$f (x) = \frac{1}{2} x + 1$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{1}{2} x + 1$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | 1)$
$\Rightarrow m_{f} = \frac{1}{2}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$\frac{1}{2} x + 1$ $=$ $0$ $- 1$
$\frac{1}{2} x$ $=$ $- 1$ $\cdot 2$
$x_{0}$ $=$ $- 2$
$\Rightarrow P_{x} (- 2 | 0)$

$f (x) = - \frac{1}{2} x - 2$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - \frac{1}{2} x - 2$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 2)$
$\Rightarrow m_{f} = - \frac{1}{2}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0..
$- \frac{1}{2} x - 2$ $=$ $0$ $+ 2$
$- \frac{1}{2} x$ $=$ $2$ $\cdot (- 2)$
$x_{0}$ $=$ $- 4$
$\Rightarrow P_{x} (- 4 | 0)$

$f (x) = \frac{1}{3} x - \frac{1}{2}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{1}{3} x - \frac{1}{2}$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - \frac{1}{2})$
$\Rightarrow m_{f} = \frac{1}{3}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$\frac{1}{3} x - \frac{1}{2}$ $=$ $0$ $+ \frac{1}{2}$
$\frac{1}{3} x$ $=$ $\frac{1}{2}$ $\cdot 3$
$x_{0}$ $=$ $\frac{3}{2}$
$\Rightarrow P_{x} (\frac{3}{2} | 0)$

$f (x) = - \frac{1}{4} x + \frac{3}{2}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - \frac{1}{4} x + \frac{3}{2}$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | \frac{3}{2})$
$\Rightarrow m_{f} = - \frac{1}{4}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$- \frac{1}{4} x + \frac{3}{2}$ $=$ $0$ $- \frac{3}{2}$
$- \frac{1}{4} x$ $=$ $- \frac{3}{2}$ $\cdot (- 4)$
$x_{0}$ $=$ $6$
$\Rightarrow P_{x} (6 | 0)$

$f (x) = \frac{2}{3} x + 2$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{2}{3} x + 2$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | 2)$
$\Rightarrow m_{f} = \frac{2}{3}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$\frac{2}{3} x + 2$ $=$ $0$ $- 2$
$\frac{2}{3} x$ $=$ $- 2$ $: \frac{2}{3}$
$x_{0}$ $=$ $- 3$
$\Rightarrow P_{x} (- 3 | 0)$

$f (x) = - \frac{3}{4} x - 1$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - \frac{3}{4} x - 1$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 1)$
$\Rightarrow m_{f} = - \frac{3}{4}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$- \frac{3}{4} x - 1$ $=$ $0$ $+ 1$
$- \frac{3}{4} x$ $=$ $1$ $: (- \frac{3}{4})$
$x_{0}$ $=$ $- \frac{4}{3}$
$\Rightarrow P_{x} (- \frac{4}{3} | 0)$

$f (x) = - 3 x + \frac{5}{10}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - 3 x + \frac{5}{10}$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | \frac{5}{10})$
$\Rightarrow m_{f} = - 3$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$- 3 x + \frac{5}{10}$ $=$ $0$ $- \frac{5}{10}$
$- 3 x$ $=$ $- \frac{1}{2}$ $: (- 3)$
$x_{0}$ $=$ $- \frac{1}{6}$
$\Rightarrow P_{x} (\frac{1}{6} | 0)$

$f (x) = \frac{5}{7} x - \frac{12}{4}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{5}{7} x - \frac{12}{4} = \frac{5}{7} x - 3$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 3)$
$\Rightarrow m_{f} = \frac{5}{7}$
Calculate the point of intersection with the $x$ -axis. Set the function term equal to 0.
$\frac{5}{7} x - 3$ $=$ $0$ $+ 3$
$\frac{5}{7} x$ $=$ $3$ $: \frac{5}{7}$
$x_{0}$ $=$ $\frac{21}{5}$
$\Rightarrow P_{x} (\frac{21}{5} | 0)$
5
Draw the graphs of each of the following functions in a coordinate system.
$f (x) = - \frac{2}{3} x + 2$

For this task you need the following basic knowledge: Lines in coordinate systems
Draw the $y$ -axis intercept as a point (here $A (0 | 2)$ ).
Go according to the slope 3 to the right and 2 down and draw the point (here $B (3 | 0)$ ).
Connect the two points to form a straight line.

$f (x) = 2 x - 4$

For this task you need the following basic knowledge: Lines in coordinate systems
Draw the $y$ -axis intercept as a point (here $A (0 | - 4)$ ).
Go according to the slope 1 to the right and 2 up and draw the point (here $B (1 | - 2)$ ).
Connect the two points to form a straight line.

$f (x) = - \frac{5}{4} x + 1$

For this task you need the following basic knowledge: Lines in coordinate systems
Draw the $y$ -axis intercept as a point (here $A (0 | 1)$ ).
Go according to the slope 4 to the right and 5 down and draw the point (here $B (4 | - 4)$ ).
Connect the two points to form a straight line.

$f (x) = - 4 x + 5$

For this task you need the following basic knowledge: Lines in coordinate systems
Draw the $y$ -axis intercept as a point (here $A (0 | 5)$ ).
Go according to the slope 1 to the right and 4 down and draw the point (here $B (1 | 1)$ ).
Connect the two points to form a straight line.

$f (x) = - 0.3 x$

For this task you need the following basic knowledge: Lines in coordinate systems
The comparison with the general form of the straight line equation , results in: Intercept t=0 and slope $m = - \frac{3}{10}$
From the value of the $y$ -axis intercept $t = 0$ it follows that our line is a line passes through the origin: $A = (0 | 0)$ .
Write the slope as a fraction: $- 0,3 = - \frac{3}{10}$ . Go 10 to the right and 3 down according to the slope. There you find a second point on the line $B (10 | - 3)$ .
The line is drawn through the two points $A$ and $B$ .

$f (x) = 2.5$

For this task you need the following basic knowledge: Lines in coordinate systems
The $y$ -value of the straight line is always 2.5. Therefore, the straight line is parallel to the $x$ -axis.
6
Consider the function $f$ with $y = 0.3 \cdot x$ .
Compute a table of values for $f$ with $x \in [- 3,3]$ where the step length is $Δ x = 1$ .

For this task you need the following basic knowledge: Linear function
The table of values for $y = 0.3 \cdot x$ with step length $Δ x = 1$ :
x
-3
-2
-1
0
1
2
3
y
-0.9
-0.6
-0.3
0
0.3
0.6
0.9

Plot the points of the function $f$ in a coordinate system and draw the graph of the function $f$ .

For this task you need the following basic knowledge: Linear function
Points of the function $f$ in the coordinate system and graph of the function $f$ :

Zu welcher besonderen Art von Geraden gehört der Graph der Funktion $f$ ?

For this task you need the following basic knowledge: Linear function
The graph of the function $f$ is a line that passes through the origin $(0 | 0)$ of the coordinate system.

Check by calculation whether the points $P (5 | 1.4)$ and $Q (8 | 2.4)$ lie on the graph of the function $f$ .

For this task you need the following basic knowledge: Linear function
Insert the coordinates of $P$ and $Q$ into the function equation:
Point $P (5 | 1.4)$ :
$y$ $=$ $0.3 \cdot x$
↓
plug in coordinates
$1.4$ $=$ $0.3 \cdot 5$
$1.4$ $=$ $1.5$
↓
false statement
$=$
Answer: The point $P$ does not lie on the graph of the function $f$ .
Point $Q (8 | 2.4)$ :
$y$ $=$ $0.3 \cdot x$
↓
plug in coordinates
$2.4$ $=$ $0.3 \cdot 8$
$2.4$ $=$ $2.4$
↓
true statement
$=$
Answer: The point $Q$ lies on the graph of the function $f$ .
Additional illustration of the graph with the two points $P$ and $Q$
(Ths is not required by the problem setting, but just provided for better visualization.)
$P (5 | 1.4) \notin G_{f}$
$Q (8 | 2.4) \in G_{f}$

$\frac{1}{2} x + 1$	$=$	$0$	$- 1$
$\frac{1}{2} x$	$=$	$- 1$	$\cdot 2$
$x_{0}$	$=$	$- 2$

$- \frac{1}{2} x - 2$	$=$	$0$	$+ 2$
$- \frac{1}{2} x$	$=$	$2$	$\cdot (- 2)$
$x_{0}$	$=$	$- 4$

$\frac{1}{3} x - \frac{1}{2}$	$=$	$0$	$+ \frac{1}{2}$
$\frac{1}{3} x$	$=$	$\frac{1}{2}$	$\cdot 3$
$x_{0}$	$=$	$\frac{3}{2}$

$- \frac{1}{4} x + \frac{3}{2}$	$=$	$0$	$- \frac{3}{2}$
$- \frac{1}{4} x$	$=$	$- \frac{3}{2}$	$\cdot (- 4)$
$x_{0}$	$=$	$6$

$\frac{2}{3} x + 2$	$=$	$0$	$- 2$
$\frac{2}{3} x$	$=$	$- 2$	$: \frac{2}{3}$
$x_{0}$	$=$	$- 3$

$- \frac{3}{4} x - 1$	$=$	$0$	$+ 1$
$- \frac{3}{4} x$	$=$	$1$	$: (- \frac{3}{4})$
$x_{0}$	$=$	$- \frac{4}{3}$

$- 3 x + \frac{5}{10}$	$=$	$0$	$- \frac{5}{10}$
$- 3 x$	$=$	$- \frac{1}{2}$	$: (- 3)$
$x_{0}$	$=$	$- \frac{1}{6}$

$\frac{5}{7} x - 3$	$=$	$0$	$+ 3$
$\frac{5}{7} x$	$=$	$3$	$: \frac{5}{7}$
$x_{0}$	$=$	$\frac{21}{5}$

$y$	$=$	$0.3 \cdot x$
	↓	plug in coordinates
$1.4$	$=$	$0.3 \cdot 5$
$1.4$	$=$	$1.5$
	↓	false statement
	$=$

$y$	$=$	$0.3 \cdot x$
	↓	plug in coordinates
$2.4$	$=$	$0.3 \cdot 8$
$2.4$	$=$	$2.4$
	↓	true statement
	$=$

$𝐱$	$- 8$	$- 4$	$0$	$4$	$8$
$𝐲$	$0$	$1$	$2$	$3$	$4$

$𝐱$	$- 2$	$0$	$2$	$4$
$𝐲$	$- 20$	$- 10$	$0$	$10$

Exercises: Drawing graphs of linear functions

Drawing the linear function

Drawing the linear function

Drawing the linear function

Determine one point

Find the slope

Draw the line

Transform the equation

Determine one point

Find the slope

Draw the line

Determine one point

Find the slope

Draw the line

Determine one point

Find the slope

Draw the line

Determine one point

Find the slope

Draw the line

Insert the coordinates of P and Q into the function equation:

Additional illustration of the graph with the two points P and Q

Insert the coordinates of $P$ and $Q$ into the function equation:

Additional illustration of the graph with the two points $P$ and $Q$