Exercises: Linear functions, zeros, axis intercepts

1
Read off the $y$ -axis intercept from the graph.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | 0)$ , so $y = 0$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | 3)$ , so $y = 3$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | - 4)$ , so $y = - 4$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | - 3)$ , so $y = - 3$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | 4)$ , so $y = 4$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | 5)$ , so $y = 5$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | 2)$ , so $y = 2$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | 3)$ , so $y = 3$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | - 1)$ , so $y = - 1$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.

For this task you need the following basic knowledge: Linear function
The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 | - 4)$ , so $y = - 4$ .
You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
2
Read off the zero from the graph.

For this task you need the following basic knowledge: Calculating zeros
The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (0 | 0)$ , so $x = 0$ .
You determine the zero by looking at the intersection of the line with the $x$ -axis.

For this task you need the following basic knowledge: Calculating zeros
The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (1 | 0)$ , so $x = 1$ .
You determine the zero by looking at the intersection of the line with the $x$ -axis.

For this task you need the following basic knowledge: Calculating zeros
The zero is the $x$ -value of the intersection of the line with the $x$ -axis $(2 | 0)$ , so $x = 2$ .
You determine the zero by looking at the intersection of the line with the $x$ -axis.

For this task you need the following basic knowledge: Calculating zeros
The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (3 | 0)$ , so $x = 3$ .
You determine the zero by looking at the intersection of the line with the $x$ -axis.

For this task you need the following basic knowledge: Calculating zeros
The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (3 | 0)$ , so $x = 3$ .
You determine the zero by looking at the intersection of the line with the $x$ -axis.

The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (7 | 0)$ , so $x = 7$ .
You determine the zero by looking at the intersection of the line with the $x$ -axis.
3
Look at the graphs of the functions $a (x)$ and $c (x)$ .
Read off the $y$ -axis intercept and the slope of the lines and enter them in the boxes!
Can you work out the function term from this?
What is the $y$ -axis intercept of $a (x)$ ?

For this task you need the following basic knowledge: Line equation in analytic geometry
Determining the $y$ -axis intercept
You determine the $y$ -axis intercept by looking at the intersection of the straight line with the $y$ -axis. In this case:
The $y$ -axis intercept is the $y$ -value of the intersection point $(0 / 4)$ , i.e. $4$ .
Notice where the straight line $G_{a}$ intersects the $y$ -axis.
Determine the intersection.

What is the slope of $a (x)$ ?

For this task you need the following basic knowledge: Slope/Gradient of a line
Determining the slope
The easiest way to determine the slope of a straight line is to use a gradient triangle.
In the case of $G_{a}$ :
You have to go one length unit to the right and one length unit down.
Therefore, you get the slope $m = - 1$
Pick a gradient triangle (see below).
Then determine the slope.

What is the function term of $a (x)$ ?

For this task you need the following basic knowledge: Linear function
Setting up the function term
The function term of a linear function takes the form:
$y = m x + t$
Here $m$ is the slope and $t$ the $y$ -axis intercept.
If you insert the values from the previous sub-tasks, you get:
$y = - 1 \cdot x + 4$
Simplified, that is:
$y = - x + 4$
The function equation of $G_{a}$ is thus:
$a (x) = - x + 4$
.
Recap the basic knowledge about linear functions.
Put your previous values into the function (see below for a more detailed explanation).

What is the y-axis intercept of c(x)?

For this task you need the following basic knowledge: Line equation in analytic geometry
Determining the $y$ -axis intercept
You determine the $y$ -axis intercept by looking at the intersection of the straight line with the $y$ -axis. In this case:
Therefore, the $y$ -axis intercept of $G_{c}$ is $- 3$ .
Notice where the straight line $G_{c}$ intersects the $y$ -axis.
Determine the intersection.

What is the slope of $c (x)$ ?

For this task you need the following basic knowledge: Slope/Gradient of a line
Determining the slope
The easiest way to determine the slope of a straight line is to use a gradient triangle.
In the case of $G_{c}$ :
You have to go one length unit to the right and two length units up.
Therefore, you get the slope: $m = 2$
Pick a gradient triangle (see below).
Then determine the slope.

What is the function term of $c (x)$ ?

For this task you need the following basic knowledge: Linear function
Setting up the function term
The function term of a linear function takes the form:
$y = m x + t$
Here $m$ is the slope and $t$ the $y$ -axis intercept.
If you insert the values from the previous sub-tasks, you get:
$y = 2 \cdot x - 3$
The functional equation of $G_{c}$ is thus:
$c (x) = 2 x - 3$
Recap the basic knowledge about linear functions.
Put your previous values into the function (see below for a more detailed explanation).
4
Consider the lines $g : y = 2 x - 3$ and $h : y = - 0.5 x + 3$ .
Check whether the points $A (1 | - 1)$ , $B (0.5 | 1.5)$ , $C (- 6 | 5)$ , $D (- 102 | 55)$ and $E (45 | 87)$ are on either of both lines.
Complete the coordinates so that the points lie on $h$ : P $(5 | ?)$ , $Q (- 3.5 | ?)$ , $R (? | 12)$ , $S (? | - 7,5)$ .
Show that $T (2.4 | 1.8)$ lies on both lines. What does this mean?

For this task you need the following basic knowledge: Line equation
Part 1
Drawing a sketch
Select any point on the straight line, e.g. the $y$ -axis intercepts $(0 | - 3)$ and $(0 | 3)$ . Go from there 1 unit to the right and, corresponding to the slopes, $m_{g} = 2$ upwards and $m_{h} = - 0.5$ downwards. Connect the two points to form a straight line.
If you look at the straight line and, for example, the position of point $A$ , you will see that it will hardly lie on the straight line $h$ , but probably on $g$ . Similarly, you can decide for other points whether a mathematical check is worthwhile: Point $D (- 102 / 55)$ can only lie on $h$ , for example.

Check by computation
$A (1 | - 1)$ on $g$
Put the coordinates of the points into the equation in question. So set $y = - 1$ and $x = 1$ .
$- 1 = 2 \cdot 1 - 3$
Check if the equation is a true statement.
$- 1 = - 1$
Thi is a true statement.
$\Rightarrow$ $A$ lies on $g$ .
$B$ does not lie on any of the straight lines. This can be clearly seen in the sketch.
$C$ is plugged into $h$ : $5 = - 0,5 \cdot (- 6) + 3 \Rightarrow 5 = 3 + 3$ This statement is false, so $C$ does not lie on $h$ .
$D$ is plugged into $h$ : $55 = - 0.5 \cdot (- 102) + 3 \Rightarrow 55 = 51 + 3$ This statement is false, so $D$ does not lie on $h$ .
$E$ is plugged into $g$ : $87 = 2 \cdot 45 - 3$ This statement is true, so $E$ lies on $g$ .
Part 2
Completing coordinates
$h : y = - 0.5 x + 3$ ; $P (5 | ?)$
The given coordinate of the point (the $x$ -coordinate) is inserted into the function equation and the missing $y$ -coordinate is calculated from it.
$y = - 0.5 \cdot 5 + 3 = - 2.5 + 3 = 0.5$
$\Rightarrow$ $P (5 | 0,5)$
Q: $y = - 0.5 \cdot (- 3.5) + 3 = 1.75 + 3 = 4.75$ $\Rightarrow$ $Q (- 3.5 | 4.75)$
R: $12 = - 0.5 x + 3 \Rightarrow 0.5 x = - 9 \Rightarrow x = - 18$ $\Rightarrow$ $R (- 18 | 12)$
S: $- 7.5 = - 0.5 x + 3 \Rightarrow 0.5 x = 10.5 \Rightarrow x = 21$ $\Rightarrow$ $S (21 | - 7.5)$
Part 3
Proof for $T$
$T (2.4 | 1.8)$
Insert the coordinates of $T$ into both line equations. If the statements are true, $T$ lies on the lines.
on $g$ : $1.8 = 2 \cdot 2.4 - 3$
$1.8 = 4.8 - 3$
$1.8 = 1.8$
on $h$ : $1.8 = - 0.5 \cdot 2.4 + 3$
$1.8 = - 1.2 + 3$
$1.8 = 1.8$
$\Rightarrow$ Both equations give correct statements, so the point $T$ lies on both straight lines.
$\Rightarrow$ $T$ Is the intersection point of the lines.
5
Draw the graphs of the following lines including the point of intersection with the $y$ -axis and a 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 intersection point with the $x$ -axis. This is done by setting 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.
$\Rightarrow P_{y} (0 | - 3)$
$\Rightarrow m_{f} = - 1$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$- x - 3$ $=$ $0$ $+ x$
$- 3$ $=$ $x_{0}$
$\Rightarrow P_{x} (- 3 | 0)$

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

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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting 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 intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\frac{5}{7} x - 3$ $=$ $0$ $+ 3$
$\frac{5}{7} x$ $=$ $3$ $: \frac{5}{7}$
$x_{0}$ $=$ $\frac{21}{5}$
$\Rightarrow P_{x} (\frac{21}{5} | 0)$
6
Draw the lines $y = 3 x - 2$ and $y = - \frac{3}{4} x + 1$ into a coordinate system. Determine the zeros and the point of intersection.

For this task you need the following basic knowledge: Linear function
Draw the graphs
Determining the zeros
$y = 3 x - 2$
Set $y = 0$ to determine the zero point. This is the point where the line intersects the $x$ -axis.
$3 x - 2$ $=$ $0$ $+ 2$
$3 x$ $=$ $2$ $: 3$
$x_{N1}$ $=$ $\frac{2}{3}$
Proceed in the same way for the second line.
$y$ $=$ $- \frac{3}{4} x + 1$
↓
Set y=0 to determine the zero.
$- \frac{3}{4} x + 1$ $=$ $0$ $- 1$
$- \frac{3}{4} x$ $=$ $- 1$ $: (- \frac{3}{4})$
$x$ $=$ $\frac{1}{\frac{3}{4}}$
↓
You divide by a fraction -> multiply by the reciprocal value
$x_{N2}$ $=$ $\frac{4}{3}$
Determining of the intersection point
Set the two function equations equal. The straight lines intersect where both have the same $y$ -value at the same $x$ -position.
$3 x - 2$ $=$ $- \frac{3}{4} x + 1$ $+ \frac{3}{4} x + 2$
$3 x + \frac{3}{4} x$ $=$ $1 + 2$
$3.75 x$ $=$ $3$ $: 3.75$
$x$ $=$ $\frac{3}{3.75}$
$x_{S}$ $=$ $0.8$
↓
Plug $x_{S}$ into one of the two functions.
$y$ $=$ $3 \cdot 0.8 - 2$
$y$ $=$ $2.4 - 2$
$y_{S}$ $=$ $0.4$
$\Rightarrow S (0,8 | 0,4)$
The intersection point is at $S (0.8 | 0.4)$ .
7
Determine the intersection points with the coordinate axes of the following straight lines.
$y = - 2 x + 3.5$

Intersection with the $x$ -axis
Calculate the intersection of the line with the $x$ -axis.
Set the expression for the line equation equal to 0 and solve for $x$ .
$\begin{array}{rcll} 0 & = & - 2 x + 3.5 & | + 2 x \\ 2 x & = & 3.5 & | : 2 \\ x & = & 1.75 \end{array}$
$\Rightarrow$ The line intersects the $x$ -axis at $S_{x} (1.75 | 0)$ .
Intersection with the $y$ -axis
Calculate the intersection of the line with the $y$ -axis.
To do this, plug the value $x = 0$ into the expression for the straight line equation.
$y = - 2 \cdot 0 + 3.5$
$y = 3.5$
$\Rightarrow$ The line intersects the $y$ -axis at $S_{y} (0 | 3.5)$ .

$y = 5 x - 7$

Intersection with the $x$ -axis
Calculate the intersection of the line with the $x$ -axis.
Set the expression for the line equation equal to 0 and solve for $x$ .
$\begin{array}{rcll} 0 & = & 5 x - 7 & | + 7 \\ 5 x & = & 7 & | : 5 \\ x & = & \frac{7}{5} = 1.4 \end{array}$
$\Rightarrow$ The line intersects the $x$ -axis at $S_{x} (1.4 | 0)$ .
Intersection with the $y$ -axis
Calculate the intersection of the line with the $y$ -axis.
To do this, plug the value $x = 0$ into the expression for the straight line equation.
$y = 5 \cdot 0 - 7$
$y = - 7$
$\Rightarrow$ The line intersects the $y$ -axis at $S_{y} (0 | - 7)$ .

$y = \frac{3}{2} x + 2$

Intersection with the $x$ -axis
Calculate the intersection of the line with the $x$ -axis.
Set the expression for the line equation equal to 0 and solve for $x$ .
$\begin{array}{rcll} 0 & = & \frac{3}{2} x + 2 & | - 2 \\ \frac{3}{2} x & = & - 2 & | \cdot \frac{2}{3} \\ x & = & - \frac{4}{3} \end{array}$
$\Rightarrow$ The line intersects the $x$ -axis at $S_{x} ( - \frac{4}{3} - | 0)$ .
Intersection with the $y$ -axis
Calculate the intersection of the line with the $y$ -axis.
To do this, plug the value $x = 0$ into the expression for the straight line equation.
$y = \frac{3}{2} \cdot 0 + 2$
$y = 2$
$\Rightarrow$ The line intersects the $y$ -axis at $S_{y} (0 | 2)$ .

$y = - \frac{2}{5} x + \frac{5}{2}$

Intersection with the $x$ -axis
Calculate the intersection of the line with the $x$ -axis.
Set the expression for the line equation equal to 0 and solve for $x$ .
$\begin{array}{rcll} 0 & = & - \frac{2}{5} x + \frac{5}{2} & | + \frac{2}{5} x \\ \frac{2}{5} x & = & \frac{5}{2} & | \cdot \frac{5}{2} \\ x & = & \frac{25}{4} = 6.25 \end{array}$
$\Rightarrow$ The line intersects the $x$ -axis at $S_{x} (6.25 | 0)$ .
Intersection with the $y$ -axis
Calculate the intersection of the line with the $y$ -axis.
To do this, plug the value $x = 0$ into the expression for the straight line equation.
$y = - \frac{2}{5} \cdot 0 + \frac{5}{2}$
$y = \frac{5}{2} = 2.5$
$\Rightarrow$ The line intersects the $y$ -axis at $S_{y} (0 | 2.5)$ .

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

To obtain a general line equation $y = m \cdot x + t$ , multiply out the bracket
$y = 2 (x - \frac{2}{3})$
$y = 2 x - \frac{4}{3}$
Intersection with the $x$ -axis
Calculate the intersection of the line with the $x$ -axis.
Set the expression for the line equation equal to 0 and solve for $x$ .
$\begin{array}{rcll} 0 & = & 2 x - \frac{4}{3} & | + \frac{4}{3} \\ 2 x & = & \frac{4}{3} & | : 2 \\ x & = & \frac{2}{3} \end{array}$
$\Rightarrow$ The line intersects the $x$ -axis at $S_{x} (\frac{2}{3} | 0)$ .
Intersection with the $y$ -axis
Calculate the intersection of the line with the $y$ -axis.
To do this, plug the value $x = 0$ into the expression for the straight line equation.
$y = 2 \cdot 0 - \frac{4}{3}$
$y = - \frac{4}{3}$
$\Rightarrow$ The line intersects the $y$ -axis at $S_{y} (0 | - \frac{4}{3})$ .

$y = - \frac{4}{3} - \frac{1}{2} x$

Transform the equation
To obtain a general line equation $y = m \cdot x + t$ , swap both elements on the right side.
$y = - \frac{4}{3} - \frac{1}{2} x$
$= - \frac{1}{2} x - \frac{4}{3}$
Intersection with the $x$ -axis
Calculate the intersection of the line with the $x$ -axis.
Set the expression for the line equation equal to 0 and solve for $x$ .
$\begin{array}{rcll} 0 & = & - \frac{4}{3} - \frac{1}{2} x & | + \frac{1}{2} x \\ \frac{1}{2} x & = & - \frac{4}{3} & | \cdot 2 \\ x & = & - \frac{8}{3} = - 2 \frac{2}{3} \end{array}$
$\Rightarrow$ The line intersects the $x$ -axis at $S_{x} (- \frac{8}{3} | 0)$ .
Intersection with the $y$ -axis
Calculate the intersection of the line with the $y$ -axis.
To do this, plug the value $x = 0$ into the expression for the straight line equation.
$y = - \frac{1}{2} \cdot 0 - \frac{4}{3}$
$y = - \frac{4}{3}$
$\Rightarrow$ The line intersects the $y$ -axis at $S_{y} (0 | - \frac{4}{3})$ .
8
Set up the function equation for the line through the points $P (- 25 | 30)$ and $Q (55 | - 30)$ and calculate the intersection of the line with the $x$ -axis.

For this task you need the following basic knowledge: Linear function
$P (- 25 | 30)$ ; $Q (55 | - 30)$
Determine the slope $m$ of the general line equation using the difference quotient .
$m = \frac{30 - (- 30)}{- 25 - 55} = \frac{60}{- 80} = - \frac{3}{4}$
Insert $m$ and the coordinates of a point, e.g. $P (- 25 | 30)$ , into the general line equation.
$30 = \frac{- 3}{4} (- 25) + t$
Simplify: $\frac{- 3}{4} (- 25) = \frac{- 3 (- 25)}{4} = \frac{3 \cdot 25}{4} = \frac{75}{4}$
$30$ $=$ $\frac{75}{4} + t$ $- \frac{75}{4}$
↓
solve for $t$
$t$ $=$ $30 - \frac{75}{4}$
$t$ $=$ $\frac{45}{4}$
$\Rightarrow y = - \frac{3}{4} x + \frac{45}{4}$
Intersection with the $x$ -axis
At the intersection with the $x$ -axis the $y$ -value is 0.
$0 = \frac{- 3}{4} x + \frac{45}{4}$
Solve for $x$ . To do this, we use $| \cdot \frac{- 4}{3}$
Note that both summands must be multiplied.
$0 = \frac{(- 3) (- 4)}{4 \cdot 3} x + \frac{45 (- 4)}{4 \cdot 3}$
$\frac{(- 3) (- 4)}{4 \cdot 3} = \frac{3 \cdot 4}{12} = \frac{12}{12} = 1$
$0 = x + \frac{- 180}{12}$
$0 = x - 15$
Add 15
$x = 15$
$\Rightarrow$ The line intersects the $x$ -axis at $S (15 | 0)$ .
9
Transform the equation into the form $y = a x + b$ .
$2 x - y = 6$

$2 x - y$ $=$ $6$ $- 2 x$
$- y$ $=$ $6 - 2 x$ $\cdot (- 1)$
$y$ $=$ $2 x - 6$

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

$x$ $=$ $\frac{1}{2} (y + 1)$ $\cdot 2$
↓
The fraction on the right side vanishes because $\frac{1}{2} \cdot 2 = 1$ .
$2 x$ $=$ $y + 1$ $- 1$
↓
Swap the left and right sides.
$y$ $=$ $2 x - 1$

$\frac{2}{5} y = 2 x - 1$

$\frac{2}{5} y$ $=$ $2 x - 1$ $\cdot \frac{5}{2}$
↓
Note: On the right-hand side, each element is multiplied by $\frac{5}{2}$ :
$\begin{array}{l} \frac{2}{5} \cdot y \cdot \frac{5}{2} = \frac{10}{10} \cdot y = 1 \cdot y; \\ 2 x \cdot \frac{5}{2} = \frac{10}{2} x = 5 x; \\ - 1 \cdot \frac{5}{2} = - \frac{5}{2} \end{array}$
$y$ $=$ $5 x - \frac{5}{2}$

$y = 3 (2 x - 1)$

$y$ $=$ $3 (2 x - 1)$
↓
Multiply out the bracket.
$y$ $=$ $6 x - 3$
10
Two lines $f (x)$ and $g (x)$ intersect on the $x$ -axis in $x = 4$ .
Determine possible function terms.

For this task you need the following basic knowledge: Linear function
There are several solutions to this problem. We are looking for two different linear functions that both pass through the point $(4 | 0)$ .
A very simple example would be $f (x) = 0$ , i.e. the $x$ -axis and $g (x) = x - 4$ .
$f (x)$ obviously runs through $(4 | 0)$ . For $g (x)$ this can also be checked very easily: $g (4) = 4 - 4 = 0$ .
Other possible functions are:
$y = - x + 4$ , $y = 2 x - 8$ (generally $y = a x - 4 a$ for any $a$ ).
Consider the point through which both lines must pass.
Determine two different functions that go through this point.
11
Consider the linear function $f (x) = 3 - \frac{12}{7} x$ .
Draw the graph and mark the function value $f (- 1)$ .

Draw the graph and mark the function value $f (- 1)$ .

Is the point $P (\sqrt{7} | - 1,54)$ on the graph of $f (x)$ ?

Check whether the point $P (\sqrt{7} | - 1.54)$ lies on the graph of the function.
To do this, insert the $x$ -value $\sqrt{7}$ into the function equation $y (x) = 3 - \frac{12}{7} x$ .
$3 - \frac{12}{7} \cdot \sqrt{7}$ $\approx$ $3 - \frac{12}{7} \cdot 2.65$
↓
multiply
$\approx$ $3 - 4.543$
↓
subtract
$\approx$ $- 1.543$
Result
The point $P (\sqrt{7} | - 1.54)$ is not on the graph because the function value of $\sqrt{7}$ is a non-terminating decimal fraction and not the finite decimal fraction $- 1.54$ .
If you have calculated with rounded values, you might have mistakenly conclude that the point is on the graph. However, this is not the case.
12
Consider the lines $g : y = 2 x - 3$ and $h : y = - 0.5 x + 3$ .
Check whether the points $A (1 | - 1)$ , $B (0.5 | 1.5)$ , $C (- 6 | 5)$ , $D (- 102 | 55)$ and $E (45 | 87)$ lie on one of the straight lines.

For this task you need the following basic knowledge: Line equation
Checking with sketch
Select any point on the straight line, e.g. the $y$ -axis sections $(0 | - 3)$ and $(0 | 3)$ . Go from there 1 to the right and, corresponding to the slopes, $m_{g} = 2$ upwards or $m_{h} = - 0.5$ downwards. Connect the two points to form a straight line.
If you look at the course of the straight line and, for example, the position of point $A$ , you will see that it will hardly lie on the straight line $h$ , but probably on $g$ . Similarly, you can decide for other points whether a mathematical check is worthwhile: Point $D (- 102 / 55)$ can only lie on $h$ , for example.
Computational verification
Plug $A (1 | - 1)$ into $g$ :
Plug the coordinates y=-1 and x=1 into the equation.
$- 1 = 2 \cdot 1 - 3$
That is a true statement.
⇒ $A$ is on $g$ .
$B$ does not lie on any of the straight lines. This can be clearly seen in the sketch.
Plug $C$ into $h$ :
$5 = - 0.5 \cdot (- 6) + 3$
⇒ $5 = 3 + 3$
This statement is false, so $C$ is not on $h$ .
Plug $D$ into $h$ :
$55 = - 0.5 \cdot (- 102) + 3$
⇒ $55 = 51 + 3$
This statement is false, so $D$ is not on $h$ .
Plug $E$ into $g$ :
$87 = 2 \cdot 45 - 3$
This statement is correct, so $E$ lies on $g$ .

Complete the coordinates so that the points lie on $h$ : $P (5 | ?)$ , $Q (- 3,5 | ?)$ , $R (? | 12)$ , $S (? | - 7,5)$ .

For this task you need the following basic knowledge: Line equation
Complete coordinates
$h : y = - 0.5 x + 3$ ; $P (5 | ?)$
The given coordinate of the point (the $x$ -coordinate) is inserted into the function equation and the missing $y$ -coordinate is calculated from it.
$y = - 0.5 \cdot 5 + 3 = - 2.5 + 3 = 0.5$
⇒ $P (5 | 0,5)$
Q: $y = - 0.5 \cdot (- 3.5) + 3 = 1.75 + 3 = 4.75$ $\Rightarrow$ $Q (- 3.5 | 4.75)$
R: $12 = - 0.5 x + 3 \Rightarrow 0.5 x = - 9 \Rightarrow x = - 18$ $\Rightarrow$ $R (- 18 | 12)$
S: $- 7.5 = - 0.5 x + 3 \Rightarrow 0.5 x = 10.5 \Rightarrow x = 21$ $\Rightarrow$ $S (21 | - 7.5)$

Show that $T (2.4 | 1.8)$ lies on both straight lines. What does this mean?

For this task you need the following basic knowledge: Line equation
Proof for $T$
$T (2.4 | 1.8)$
Plug the coordinates of $T$ into both line equations. If the statements are true, $T$ lies on the lines.
into $g$ :
$1.8$ $=$ $2 \cdot 2.4 - 3$
$1.8$ $=$ $4.8 - 3$
$1.8$ $=$ $1.8$
into $h$ :
$1.8$ $=$ $- 0.5 \cdot 2.4 + 3$
$1.8$ $=$ $- 1.2 + 3$
$1.8$ $=$ $1.8$
$\Rightarrow$ Both equations give correct statements, so the point $T$ lies on both lines.
$\Rightarrow$ $T$ Is the intersection point of the lines.

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

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

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

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

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

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

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

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

$3 x - 2$	$=$	$0$	$+ 2$
$3 x$	$=$	$2$	$: 3$
$x_{N1}$	$=$	$\frac{2}{3}$

$y$	$=$	$- \frac{3}{4} x + 1$
	↓	Set y=0 to determine the zero.
$- \frac{3}{4} x + 1$	$=$	$0$	$- 1$
$- \frac{3}{4} x$	$=$	$- 1$	$: (- \frac{3}{4})$
$x$	$=$	$\frac{1}{\frac{3}{4}}$
	↓	You divide by a fraction -> multiply by the reciprocal value
$x_{N2}$	$=$	$\frac{4}{3}$

$30$	$=$	$\frac{75}{4} + t$	$- \frac{75}{4}$
	↓	solve for $t$
$t$	$=$	$30 - \frac{75}{4}$
$t$	$=$	$\frac{45}{4}$

$3 - \frac{12}{7} \cdot \sqrt{7}$	$\approx$	$3 - \frac{12}{7} \cdot 2.65$
	↓	multiply
	$\approx$	$3 - 4.543$
	↓	subtract
	$\approx$	$- 1.543$

$1.8$	$=$	$2 \cdot 2.4 - 3$
$1.8$	$=$	$4.8 - 3$
$1.8$	$=$	$1.8$

$1.8$	$=$	$- 0.5 \cdot 2.4 + 3$
$1.8$	$=$	$- 1.2 + 3$
$1.8$	$=$	$1.8$

$3 x - 2$	$=$	$- \frac{3}{4} x + 1$	$+ \frac{3}{4} x + 2$
$3 x + \frac{3}{4} x$	$=$	$1 + 2$
$3.75 x$	$=$	$3$	$: 3.75$
$x$	$=$	$\frac{3}{3.75}$
$x_{S}$	$=$	$0.8$
	↓	Plug $x_{S}$ into one of the two functions.
$y$	$=$	$3 \cdot 0.8 - 2$
$y$	$=$	$2.4 - 2$
$y_{S}$	$=$	$0.4$

$2 x - y$	$=$	$6$	$- 2 x$
$- y$	$=$	$6 - 2 x$	$\cdot (- 1)$
$y$	$=$	$2 x - 6$

$x$	$=$	$\frac{1}{2} (y + 1)$	$\cdot 2$
	↓	The fraction on the right side vanishes because $\frac{1}{2} \cdot 2 = 1$ .
$2 x$	$=$	$y + 1$	$- 1$
	↓	Swap the left and right sides.
$y$	$=$	$2 x - 1$

$y$	$=$	$3 (2 x - 1)$
	↓	Multiply out the bracket.
$y$	$=$	$6 x - 3$

Exercises: Linear functions, zeros, axis intercepts

﻿Determining the y-axis intercept

Determining the slope

Setting up the function term

﻿﻿Determining the y-axis intercept

Determining the slope

Setting up the function term

Part 1

Drawing a sketch

Check by computation

Part 2

Completing coordinates

Part 3

Proof for T

Draw the graphs

Determining the zeros

Determining of the intersection point

Intersection with the x-axis

Intersection with the y-axis

Intersection with the x-axis

Intersection with the y-axis

Intersection with the x-axis

Intersection with the y-axis

Intersection with the x-axis

Intersection with the y-axis

Intersection with the x-axis

Intersection with the y-axis

Transform the equation

Intersection with the x-axis

Intersection with the y-axis

Result

Checking with sketch

Computational verification

Complete coordinates

Proof for T

Determining the $y$ -axis intercept

Determining the $y$ -axis intercept

Proof for $T$

Intersection with the $x$ -axis

Intersection with the $y$ -axis

Intersection with the $x$ -axis

Intersection with the $y$ -axis

Intersection with the $x$ -axis

Intersection with the $y$ -axis

Intersection with the $x$ -axis

Intersection with the $y$ -axis

Intersection with the $x$ -axis

Intersection with the $y$ -axis

Intersection with the $x$ -axis

Intersection with the $y$ -axis

Proof for $T$