Exercises: Linear functions, zeros, axis intercepts

1
Read off the $y$ -axis intercept from the graph.
1. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ 0)$ , so $y = 0$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
2. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ 3)$ , so $y = 3$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
3. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ - 4)$ , so $y = - 4$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
4. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ - 3)$ , so $y = - 3$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
5. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ 4)$ , so $y = 4$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
6. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ 5)$ , so $y = 5$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
7. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ 2)$ , so $y = 2$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
8. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ 3)$ , so $y = 3$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
9. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ - 1)$ , so $y = - 1$ .
  Do you have a question?
  Write it here 👉
  You get the $y$ -axis intercept by looking at the intersection of the line with the y-axis.
10. Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  The $y$ -axis intercept is the $y$ -value of the intersection with the $y$ -axis $S (0 ∣ - 4)$ , so $y = - 4$ .
  Do you have a question?
  Write it here 👉
  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.
1. Für diese Aufgabe benötigst Du folgendes Grundwissen: Calculating zeros
  The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (0 ∣ 0)$ , so $x = 0$ .
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  Write it here 👉
  You determine the zero by looking at the intersection of the line with the $x$ -axis.
2. Für diese Aufgabe benötigst Du folgendes Grundwissen: Calculating zeros
  The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (1 ∣ 0)$ , so $x = 1$ .
  Do you have a question?
  Write it here 👉
  You determine the zero by looking at the intersection of the line with the $x$ -axis.
3. Für diese Aufgabe benötigst Du folgendes Grundwissen: Calculating zeros
  The zero is the $x$ -value of the intersection of the line with the $x$ -axis $(2\vert 0)$ , so $x = 2$ .
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  Write it here 👉
  You determine the zero by looking at the intersection of the line with the $x$ -axis.
4. Für diese Aufgabe benötigst Du folgendes Grundwissen: Calculating zeros
  The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (3 ∣ 0)$ , so $x = 3$ .
  Do you have a question?
  Write it here 👉
  You determine the zero by looking at the intersection of the line with the $x$ -axis.
5. Für diese Aufgabe benötigst Du folgendes Grundwissen: Calculating zeros
  The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (3 ∣ 0)$ , so $x = 3$ .
  Do you have a question?
  Write it here 👉
  You determine the zero by looking at the intersection of the line with the $x$ -axis.
6. The zero is the $x$ -value of the intersection of the line with the $x$ -axis $N (7 ∣ 0)$ , so $x = 7$ .
  Do you have a question?
  Write it here 👉
  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?
1. What is the $y$ -axis intercept of $a (x)$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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. $4$ .
  Do you have a question?
  Write it here 👉
  Notice where the straight line $G_{a}$ intersects the $y$ -axis.
  Determine the intersection.
2. What is the slope of $a (x)$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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 $m = - 1$
  Do you have a question?
  Write it here 👉
  Pick a gradient triangle (see below).
  Then determine the slope.
3. What is the function term of $a (x)$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Setting up the function term
  The function term of a linear function takes the form:
  $\displaystyle y=mx+t$
  Here $m$ is the slope and $t$ the $y$ -axis intercept.
  If you insert the values from the previous sub-tasks, you get:
  $\displaystyle y=-1 \cdot x +4$
  Simplified, that is:
  $\displaystyle y=-x+4$
  The function equation of $G_{a}$ is thus:
  $\displaystyle a(x)=-x+4$
  .
  Do you have a question?
  Write it here 👉
  Recap the basic knowledge about linear functions.
  Put your previous values into the function (see below for a more detailed explanation).
4. What is the y-axis intercept of c(x)?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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 $- 3$ .
  Do you have a question?
  Write it here 👉
  Notice where the straight line $G_{c}$ intersects the $y$ -axis.
  Determine the intersection.
5. What is the slope of $c (x)$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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: $m = 2$
  Do you have a question?
  Write it here 👉
  Pick a gradient triangle (see below).
  Then determine the slope.
6. What is the function term of $c (x)$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Setting up the function term
  The function term of a linear function takes the form:
  $\displaystyle y=mx+t$
  Here $m$ is the slope and $t$ the $y$ -axis intercept.
  If you insert the values from the previous sub-tasks, you get:
  $\displaystyle y=2 \cdot x -3$
  The functional equation of $G_{c}$ is thus:
  $\displaystyle c(x)=2x-3$
  Do you have a question?
  Write it here 👉
  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 $g:\;y=2x-3$ and $h:\;y=-0.5x+3$ .
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?
Für diese Aufgabe benötigst Du folgendes Grundwissen: 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 $x = 1$ .
$-1=2\cdot1-3$
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 $g$ : $\;87=2\cdot45-3\;$ This statement is true, so $E$ lies on $g$ .
Part 2
Completing coordinates
$h:\;y=-0.5x+3$ ; $P (5 ∣ ?)$
The given coordinate of the point (the $x$ -coordinate) is inserted into the function equation and the missing $y$ -coordinate is calculated from it.
$y=-0.5\cdot5+3=-2.5+3=0.5$
$\;\;\Rightarrow\;\;$ $P (5 ∣ 0,5)$
Q: $y=-0.5\cdot(-3.5)+3=1.75+3=4.75$ $\;\;\Rightarrow\;\;$ $Q (- 3.5 ∣ 4.75)$
R: $12=-0.5x+3\;\;\Rightarrow\;\;0.5x=-9\;\;\Rightarrow\;\;x=-18$ $\;\;\Rightarrow\;\;$ $R (- 18 ∣ 12)$
S: $-7.5=-0.5x+3\;\;\Rightarrow\;\;0.5x=10.5\;\;\Rightarrow\;\;x=21$ $\;\;\Rightarrow\;\;$ $S (21 ∣ - 7.5)$
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 $g$ : $1.8=2\cdot2.4-3$
$1.8 = 4.8 - 3$
$1.8 = 1.8$
on $h$ : $1.8=-0.5\cdot2.4+3$
$1.8 = - 1.2 + 3$
$1.8 = 1.8$
$\;\;\Rightarrow\;\;$ Both equations give correct statements, so the point $T$ lies on both straight lines.
$\;\;\Rightarrow\;\;$ $T$ Is the intersection point of the lines.

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.

$f(x)\;=\;2x-5$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Lines in coordinate systems
$f (x) = 2 x - 5$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow\;{P}_y\;\left(0\;\left|\;-5\right.\right)$
$\Rightarrow\;m_f=2$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\displaystyle 2x-5$ $=$ $\displaystyle 0$ $\displaystyle +5$
$\displaystyle 2x$ $=$ $\displaystyle 5$ $\displaystyle \colon 2$
$\displaystyle x_0$ $=$ $\displaystyle 2.5$
$\Rightarrow\;{P}_x\;\left(2{.}5\;\left|\;0\right.\right)$
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$f (x) = - x - 3$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Lines in coordinate systems
$f (x) = - x - 3$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow\;{P}_y\;\left(0\;\left|\;-3\right.\right)$
$\Rightarrow\;m_f=-1$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\displaystyle -x-3$ $=$ $\displaystyle 0$ $\displaystyle +x$
$\displaystyle -3$ $=$ $\displaystyle x_0$
$\Rightarrow\;{P}_x\;\left(-3\;\left|\;0\right.\right)$
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Write it here 👉

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

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

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

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

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

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

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

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

Draw the lines $y = 3 x - 2$ and $y=-\frac34x+1$ into a coordinate system. Determine the zeros and the point of intersection.

Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.

$\displaystyle 3x-2$	$=$	$\displaystyle 0$	$\displaystyle +2$
$\displaystyle 3x$	$=$	$\displaystyle 2$	$\displaystyle :3$
$\displaystyle \mathrm{x}_{\mathrm{N1}}$	$=$	$\displaystyle \frac{2}{3}$

Proceed in the same way for the second line.

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

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.

$\displaystyle 3\mathrm{x}-2$	$=$	$\displaystyle -\frac{3}{4}\mathrm{x}+1$	$\displaystyle +\frac{3}{4}x\ +2$
$\displaystyle 3\mathrm{x}+\frac{3}{4}\mathrm{x}$	$=$	$\displaystyle 1+2$
$\displaystyle 3.75x$	$=$	$\displaystyle 3$	$\displaystyle :3.75$
$\displaystyle x$	$=$	$\displaystyle \frac{3}{3.75}$
$\displaystyle \mathrm{x}_{\mathrm{S}}$	$=$	$\displaystyle 0.8$
	↓	Plug $x_{S}$ into one of the two functions.
$\displaystyle y$	$=$	$\displaystyle 3\cdot0.8-2$
$\displaystyle y$	$=$	$\displaystyle 2.4-2$
$\displaystyle \mathrm{y}_{\mathrm{S}}$	$=$	$\displaystyle 0.4$

$\;\;\Rightarrow\;\;S\left(0{,}8\vert0{,}4\right)$

The intersection point is at $S\left(0.8|0.4\right)$ .

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7
Determine the intersection points with the coordinate axes of the following straight lines.
1. $y = - 2 x + 3.5$
  
  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 $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}0&=&-2x+3.5&|+2x\\ 2x&=&3.5&|:2\\x&=&1.75\end{array}$
  $\;\;\Rightarrow\;\;$ 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.
  $y=-2\cdot0+3.5$
  $y = 3.5$
  $\;\;\Rightarrow\;\;$ The line intersects the $y$ -axis at $S_{y} (0 ∣ 3.5)$ .
  Do you have a question?
  Write it here 👉
2. $y = 5 x - 7$
  
  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 $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}0&=&5x-7&|+7\\5x&=&7&|:5\\x&=&\frac 7 5 = 1.4 \end{array}$
  $\;\;\Rightarrow\;\;$ 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.
  $y=5\cdot0-7$
  $y = - 7$
  $\;\;\Rightarrow\;\;$ The line intersects the $y$ -axis at $S_{y} (0 ∣ - 7)$ .
  Do you have a question?
  Write it here 👉
3. $y=\frac32x+2$
  
  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 $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}0&=&\frac32x+2&|-2\\\frac 3 2x&=&-2&|\cdot \frac 2 3\\x&=&-\frac 43\end{array}$
  $\;\;\Rightarrow\;\;$ The line intersects the $x$ -axis at $S_x( -\frac43−|0)$ .
  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.
  $y=\frac32\cdot0+2$
  $y = 2$
  $\;\;\Rightarrow\;\;$ The line intersects the $y$ -axis at $S_{y} (0 ∣ 2)$ .
  Do you have a question?
  Write it here 👉
4. $y=-\frac{2}{5}x+\frac{5}{2}$
  
  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 $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}0&=&-\frac25x+\frac52&|+\frac 25 x\\\frac25x&=&\frac 52&|\cdot \frac 52\\x&=&\frac{25}{4}=6.25\end{array}$
  $\;\;\Rightarrow\;\;$ 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.
  $y=-\frac25\cdot0+\frac52$
  $y=\frac{5}{2}=2.5$
  $\;\;\Rightarrow\;\;$ The line intersects the $y$ -axis at $S_{y} (0 ∣ 2.5)$ .
  Do you have a question?
  Write it here 👉
5. $y=2(x-\frac23)$
  
  To obtain a general line equation $y=m \cdot x+t$ , multiply out the bracket
  $y=2(x-\frac23)$
  $y=2x-\frac43$
  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 $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}0&=&2x-\frac43&|+\frac 43\\2x&=&\frac 43&|:2\\x&=&\frac 23\end{array}$
  $\;\;\Rightarrow\;\;$ 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.
  $y=2\cdot0-\frac43$
  $y=-\frac 4 3$
  $\;\;\Rightarrow\;\;$ The line intersects the $y$ -axis at $S_y(0|-\frac{4}{3})$ .
  Do you have a question?
  Write it here 👉
6. $y=-\frac43-\frac12x$
  
  Transform the equation
  To obtain a general line equation $y=m\cdot x+t$ , swap both elements on the right side.
  $y=-\frac43-\frac12x$
  $\phantom{y}=-\frac12x-\frac43$
  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 $x$ .
  $\def\arraystretch{1.25} \begin{array}{rcll}0&=&-\frac43-\frac12x&|+\frac 12 x\\\frac 12 x &=& -\frac 43 &|\cdot 2\\x&=&-\frac 83 = -2\frac 23\end{array}$
  $\;\;\Rightarrow\;\;$ 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.
  $y=-\frac12\cdot0-\frac43$
  $y=-\frac 4 3$
  $\;\;\Rightarrow\;\;$ The line intersects the $y$ -axis at $S_y(0|-\frac{4}{3})$ .
  Do you have a question?
  Write it here 👉

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.

Transform the equation into the form $y = a x + b$ .

$2 x - y = 6$

$x=\frac12(y+1)$

$\frac25y=2x-1$

$y = 3 (2 x - 1)$

$\displaystyle y$ $=$ $\displaystyle 3\left(2x-1\right)$
↓
Multiply out the bracket.
$\displaystyle y$ $=$ $\displaystyle 6x-3$
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Write it here 👉

10
Two lines $f\left(x\right)$ and $g\left(x\right)$ intersect on the $x$ -axis in $x = 4$ .
Determine possible function terms.

Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.

Consider the linear function ${f}\left({x}\right)=3-\frac{12}{7}{x}$ .

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

Draw the graph and mark the function value $f\left(-1\right)$ .
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Is the point $P\left(\sqrt7 \;| -1{,}54\right)$ on the graph of $f\left(x\right)$ ?

12
Consider the lines $g:\;y=2x-3$ and $h:\;y=-0.5x+3$ .
1. 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.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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.
  $-1=2\cdot1-3$
  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 $h$ :
  $55=-0.5\cdot\left(-102\right)+3$
  ⇒ $55 = 51 + 3$
  This statement is false, so $D$ is not on $h$ .
  Plug $E$ into $g$ :
  $87=2\cdot45-3$
  This statement is correct, so $E$ lies on $g$ .
  Do you have a question?
  Write it here 👉
2. Complete the coordinates so that the points lie on $h$ : $P (5 ∣ ?)$ , $Q (- 3,5 ∣ ?)$ , $R (? ∣ 12)$ , $S (? ∣ - 7,5)$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  Complete coordinates
  $h:\;y=-0.5x+3$ ; $P (5 ∣ ?)$
  The given coordinate of the point (the $x$ -coordinate) is inserted into the function equation and the missing $y$ -coordinate is calculated from it.
  $y=-0.5\cdot5+3=-2.5+3=0.5$
  ⇒ $P (5 ∣ 0,5)$
  Q: $y=-0.5\cdot(-3.5)+3=1.75+3=4.75$ $\;\;\Rightarrow\;\;$ $Q (- 3.5 ∣ 4.75)$
  R: $12=-0.5x+3\;\;\Rightarrow\;\;0.5x=-9\;\;\Rightarrow\;\;x=-18$ $\;\;\Rightarrow\;\;$ $R (- 18 ∣ 12)$
  S: $-7.5=-0.5x+3\;\;\Rightarrow\;\;0.5x=10.5\;\;\Rightarrow\;\;x=21$ $\;\;\Rightarrow\;\;$ $S (21 ∣ - 7.5)$
  Do you have a question?
  Write it here 👉
3. Show that $T (2.4 ∣ 1.8)$ lies on both straight lines. What does this mean?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: 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 $g$ :
  $\displaystyle 1.8$ $=$ $\displaystyle 2\cdot2.4-3$
  $\displaystyle 1.8$ $=$ $\displaystyle 4.8-3$
  $\displaystyle 1.8$ $=$ $\displaystyle 1.8$
  into $h$ :
  $\displaystyle 1.8$ $=$ $\displaystyle -0.5\cdot2.4+3$
  $\displaystyle 1.8$ $=$ $\displaystyle -1.2+3$
  $\displaystyle 1.8$ $=$ $\displaystyle 1.8$
  $\;\;\Rightarrow\;\;$ Both equations give correct statements, so the point $T$ lies on both lines.
  $\;\;\Rightarrow\;\;$ $T$ Is the intersection point of the lines.
  Do you have a question?
  Write it here 👉

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$\displaystyle \frac{1}{3}x-\frac{1}{2}$	$=$	$\displaystyle 0$	$\displaystyle +\frac{1}{2}$
$\displaystyle \frac{1}{3}x$	$=$	$\displaystyle \frac{1}{2}$	$\displaystyle \cdot3$
$\displaystyle x_0$	$=$	$\displaystyle \frac{3}{2}$

$\displaystyle -\frac{1}{4}x+\frac{3}{2}$	$=$	$\displaystyle 0$	$\displaystyle -\frac{3}{2}$
$\displaystyle -\frac{1}{4}x$	$=$	$\displaystyle -\frac{3}{2}$	$\displaystyle \cdot(-4)$
$\displaystyle x_0$	$=$	$\displaystyle 6$

$\displaystyle -3x+\frac{5}{10}$	$=$	$\displaystyle 0$	$\displaystyle -\frac{5}{10}$
$\displaystyle -3x$	$=$	$\displaystyle -\frac{1}{2}$	$\displaystyle \colon (-3)$
$\displaystyle x_0$	$=$	$\displaystyle -\frac{1}{6}$

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

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

$\displaystyle 2x-5$	$=$	$\displaystyle 0$	$\displaystyle +5$
$\displaystyle 2x$	$=$	$\displaystyle 5$	$\displaystyle \colon 2$
$\displaystyle x_0$	$=$	$\displaystyle 2.5$

$\displaystyle -x-3$	$=$	$\displaystyle 0$	$\displaystyle +x$
$\displaystyle -3$	$=$	$\displaystyle x_0$

$\displaystyle \frac{1}{2}x+1$	$=$	$\displaystyle 0$	$\displaystyle -1$
$\displaystyle \frac{1}{2}x$	$=$	$\displaystyle -1$	$\displaystyle \cdot2$
$\displaystyle x_0$	$=$	$\displaystyle -2$

$\displaystyle -\frac{1}{2}x-2$	$=$	$\displaystyle 0$	$\displaystyle +2$
$\displaystyle -\frac{1}{2}x$	$=$	$\displaystyle 2$	$\displaystyle \cdot(-2)$
$\displaystyle x_0$	$=$	$\displaystyle -4$

$\displaystyle \frac{2}{3}x+2$	$=$	$\displaystyle 0$	$\displaystyle -2$
$\displaystyle \frac{2}{3}x$	$=$	$\displaystyle -2$	$\displaystyle \colon\frac23$
$\displaystyle x_0$	$=$	$\displaystyle -3$

$\displaystyle -\frac{3}{4}x-1$	$=$	$\displaystyle 0$	$\displaystyle +1$
$\displaystyle -\frac{3}{4}x$	$=$	$\displaystyle 1$	$\displaystyle \colon (-\frac34)$
$\displaystyle x_0$	$=$	$\displaystyle -\frac{4}{3}$

$\displaystyle \frac{5}{7}x-3$	$=$	$\displaystyle 0$	$\displaystyle +3$
$\displaystyle \frac{5}{7}x$	$=$	$\displaystyle 3$	$\displaystyle \colon \frac57$
$\displaystyle x_0$	$=$	$\displaystyle \frac{21}{5}$

$\displaystyle x$	$=$	$\displaystyle \frac{1}{2}\left(y+1\right)$	$\displaystyle \cdot2$
	↓	The fraction on the right side vanishes because $\frac12\cdot2=1$ .
$\displaystyle 2x$	$=$	$\displaystyle y+1$	$\displaystyle -1$
	↓	Swap the left and right sides.
$\displaystyle y$	$=$	$\displaystyle 2x-1$

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

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

$\displaystyle 1.8$	$=$	$\displaystyle 2\cdot2.4-3$
$\displaystyle 1.8$	$=$	$\displaystyle 4.8-3$
$\displaystyle 1.8$	$=$	$\displaystyle 1.8$

$\displaystyle 1.8$	$=$	$\displaystyle -0.5\cdot2.4+3$
$\displaystyle 1.8$	$=$	$\displaystyle -1.2+3$
$\displaystyle 1.8$	$=$	$\displaystyle 1.8$

Exercises: Linear functions, zeros, axis intercepts

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﻿Determining the yyy-axis intercept

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Determining the slope

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Setting up the function term

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﻿﻿Determining the yyy-axis intercept

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Determining the slope

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Setting up the function term

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Part 1

Drawing a sketch

Check by computation

Part 2

Completing coordinates

Part 3

Proof for TTT

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Draw the graphs

Determining the zeros

Determining of the intersection point

Intersection with the xxx-axis

Intersection with the yyy-axis

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Intersection with the xxx-axis

Intersection with the yyy-axis

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Intersection with the xxx-axis

Intersection with the yyy-axis

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Intersection with the xxx-axis

Intersection with the yyy-axis

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Intersection with the xxx-axis

Intersection with the yyy-axis

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Transform the equation

Intersection with the xxx-axis

Intersection with the yyy-axis

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Result

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Checking with sketch

Computational verification

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Complete coordinates

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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 $T$