Exercises: Linear functions and line equations

1
Read off the slope from the graph.
1. Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You start from the $y$ -axis intercept (here $y = 0$ ) and go $1$ to the right and $3$ upwards.
  The slope is therefore: $m = \dfrac{3}{1}=3$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
2. Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You start from the $y$ -axis intercept (here $y = 3$ ) and go $1$ to the right and $3$ downwards.
  The slope is therefore: $m=\dfrac{-3}{1}=-3$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
3. Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You start from the $y$ -axis intercept (here $y = - 3$ ) and go $2$ to the right and $3$ upwards.
  The slope is therefore: $m=\dfrac{3}{2}=1{,}5$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
4. Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You start from the $y$ -axis intercept (here $y = 4$ ) and go $1$ to the right and $1$ downwards.
  The slope is therefore: $m=\dfrac{-1}{1}=-1$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
5. Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You start from the $y$ -axis intercept (here $y = - 3$ ) and go $1$ to the right and $2$ upwards.
  The slope is therefore: $m=\dfrac{2}{1}=2$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
2
Consider the following graphs of linear functions:
1. Which of the four graphs belongs to the equation ${y}=\frac{5}{4}{x}-1$ ?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Funktionsgraphen linearer Funktionen zuordnen
  Finde den Funktionsgraphen, der zur Funktionsgleichung $y=\frac{5}{4}x−1$ gehört.
  Bestimme daher zunächst den y-Achsenabschnitt.
  $y(0)=\frac{5}{4}\cdot0-1$
  Multipliziere und subtrahiere.
  $\frac{5}{4}\cdot0-1=-1$
  Gebe den y-Achsenabschnitt an.
  Der y-Achsenabschnitt ist -1, der Graph schneidet also im Punkt (0,-1) die y-Achse. Somit können nur Graph 1 (rot) und Graph 2 (grün) zu der Funktion gehören.
  Betrachte nun die Steigung der Graphen. Gib zuerst die Funktionsgleichung der Funktion an.
  $y=\frac{5}{4}x−1$
  Gib die Steigung der Funktion an, indem du sie von der Geradengleichung abliest.
  $m=\frac{5}{4}$
  Graph 1 (rot)
  Bestimme die Steigung von Graph 1 (rot), indem du ein Steigungsdreieck konstruierst. Lies dazu den Wert an der Stelle $x_{2} = 1$ ab.
  $x_{2} = 1$
  $y_{2} = y (x_{2}) = - 0.25$
  Für die Konstruktion eines Steigungsdreieck benötigst du einen zweiten Punkt. Gib dazu den schon berechneten Punkt des y-Achsenabschnitts an.
  $x_{1} = 0$
  $y_{1} = - 1$
  Berechne nun die Steigung gemäß $m = \frac{y_2 - y_1}{x_2-x_1}$ .
  $m=\frac{-0.25-(-1)}{1-0}=\frac{0.75}{1}=0.75$
  Die Steigung des Graphen ist offenbar zu klein, als dass Graph 1 (rot) der zugehörige Funktionsgraph sein könnte.
  Graph 2 (grün)
  Bestimme die Steigung von Graph 2 (grün), indem du ein Steigungsdreieck konstruierst. Lies dazu den Wert an der Stelle $x_{2} = 1$ ab.
  $x_{2} = 1$
  $y_{2} = y (x_{2}) = 0.25$
  Gib den Punkt des y-Achsenabschnitts an.
  $x_{1} = 0$
  $y_{1} = - 1$
  Für die Konstruktion eines Steigungsdreieck benötigst du einen zweiten Punkt. Gib dazu den schon berechneten Punkt des y-Achsenabschnitts an.
  $m=\frac{0.25-(-1)}{1-0}=\frac{1.25}{1}=1.25=\frac{5}{4}$
  Ergebnis
  Der zugehörige Graph ist Graph 2 (grün), da seine Steigung und sein y-Achsenabschnitt mit der Geradengleichung übereinstimmen.
  Do you have a question?
  Write it here 👉
2. Determine (approximately) the function term to the graph 3.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  You have to find the general form of a line equation:
  The general equation of a straight line is $y = m x + t$ .
  $y$ -axis intercept
  Determine the y-axis intercept of the function graph 3 (blue).
  The $y$ -axis intercept is the function value at the intersection of the function graph with the $y$ -axis, i.e. the function value $f (x)$ at $x = 0$ .
  Read off the point where the function graph intersects the $y$ -axis.
  $(x, ∣ f (x)) = (0 ∣ 1.25)$
  Wrote down the y-axis intercept t.
  $t = 1.25$
  Slope
  Determine the slope/gradient of the function graph 3 (blue) using a gradient triangle and give the expression for the slope.
  $\displaystyle m = \frac{y_2 - y_1}{x_2-x_1}$
  Read off two points on graph 3 (blue) and then calculate the slope.
  $x_{1} = 0$
  $y_{1} = y (x_{1}) = 1.25$
  $x_{2} = - 1$
  $y_{2} = y (x_{2}) = 0.25$
  Now calculate the slope $m$ .
  $\displaystyle m = \frac{0.25 - 1.25}{-1 -0} = \frac{-1}{-1} = 1$
  Result
  The line equation of the function graph 3 (blue) is $\displaystyle y (x) = x + 1.25$ .
  Do you have a question?
  Write it here 👉
3
Determine the slope of the following straight lines.
1. $- 1$
  $2.5$
  $1$
  $- 2.5$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  The graph is rising. So only options $2.5$ and $1$ can be correct. If you go from the $y$ -axis intercept (here $y = - 2.5$ ) to the right by $1$ , you still have to go up by about $1$ to reach the straight line again.
  The slope is therefore: $m=\dfrac{1}{1}=1$ .
  Do you have a question?
  Write it here 👉
  Überlege dir, wie du ein Steigungsdreieck einzeichnen könntest.
2. $-\frac{2}{3}$
  $- 1.5$
  $3$
  $4.5$
  $\frac{3}{2}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You look for two points in the coordinate system whose coordinates you can read off easily. Here, for example, $(1 ∣ 3)$ and $(3 ∣ 0)$ . To get from $(1 ∣ 3)$ to $(3 ∣ 0)$ , you have to go $1$ to the right and $3$ down.
  The slope is therefore: $m=\frac{-3}{2}=-1.5$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
3. $- 1.2$
  $0.8$
  3
  $- 0.8$
  $2.4$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  The line is falling. Therefore, the slope can only be negative. The only possible correct solutions are $0.8$ or $- 1.2$ .
  If you move $1$ to the right in the coordinate system from the $y$ -axis intercept, you have to move less than $1$ down to meet the straight line again. So the answer $- 1.2$ cannot be correct.
  If you look at the graph very very carefully, you get the result:
  $m=\frac{-0.8}{1}=-0.8$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
4. $2$
  $1.8$
  $- 3.6$
  $- 1.8$
  $2.2$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  The graph is rising, so the slope can only be positive. The answer options $2$ or $1.8$ or $2.2$ are therefore sensible at first glance.
  Find a point on the straight line whose coordinates you can read off easily. For example, the point $(3 ∣ 0)$ is suitable here. From here you have to go $1$ to the right and less than $2$ upwards to reach the straight line again. Therefore, only the answer option $m = 1.8$ makes sense.
  The slope is therefore: $m=\frac{1.8}{1}=1.8$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
5. $3$
  $-\frac{3}{7}$
  $7$
  $\frac{3}{7}$
  $-\frac{7}{3}$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  You start from the $y$ -axis intercept (here $y = 3$ ) and go $7$ to the right and $3$ downwards.
  The slope is therefore: $m=\dfrac{-3}{7}=-\dfrac{3}{7}$
  Do you have a question?
  Write it here 👉
  Think about how you could draw a gradient triangle.
4
Consider the following graphs for linear functions:
Which of the four graphs belongs to the equation $y=\frac{5}{4}x-1$
What is the equation of graph III?
Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
Part 1:
You can read off the slope and the $y$ -axis intercept of this graph from the equation.
Given line equation: $y=\frac54x-1$
First check for which functions the $y$ -intercept is $t = - 1$ by reading off the $y$ -value of each graph at the intersection with the $y$ -axis.
$m=\frac{5}{4}$
$t = - 1$
Now check which of the two graphs has the slope $m=\frac54$ by moving one to the right from the point $x = 0$ and checking which of the two $y$ -values increases by $\frac54$ .
Only graph I and II have the $y$ -axis intercept $- 1$ so you can exclude any other graph.
Both graphs start at the point $P\left(0;-1\right)$ . Since the line we are looking for has the gradient $\frac54$ , it also passes through the point $(0 + 4 ∣ - 1 + 5) = (4 ∣ 4)$ .
Only line II runs through this point.
$\Rightarrow$ Graph II is the one belonging to the given equation.
Part 2:
First look where the graph intersects the $y$ -axis to find the $y$ -axis intercept.
The line to be checked is III
Now read off how much the $y$ -value changes if you move one to the right starting from $x = 0$ . This will give you the slope.
The $y$ -value of the point the line intersects the $y$ -axis is $y = 1.25$ .
Therefore, $t = 1.25$ .
Set up the line equation.
The $y$ -value increases from $y = 1.25$ to $y = 2.25$ .
Thus the slope is $m=\frac{2.25-1.25}{1}=\frac{1}{1}=1$ .
$\Rightarrow$ Graph III has the line equation $y = x + 1.25$ .
5
Determine the equation of the line $g$ that is parallel to the line $h$ and passes through the point $P$ .
1. h: $y = 3 x - 2$ ; $P(1|0) \;$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = 3 x - 2$ ; $P (1 ∣ 0)$
  For the line to be parallel to $h$ , it must have the same gradient/slope.
  The gradient of a line is called $m$ in the general line equation.
  $m = 3$
  Setting up the equation
  Plug $m = 3$ and the point $P (1 ∣ 0) = (x ∣ y)$ into the general line equation.
  $\displaystyle 0$ $=$ $\displaystyle 3\cdot1+t$ $\displaystyle -3$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle -3$
  Substitute $m$ and $t$ into the general line equation.
  $\Rightarrow$ $y = 3 x - 3$
  Do you have a question?
  Write it here 👉
2. $h$ : $y = x - 4$ ; $P(1|2) \;$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = x - 4$ ; $P (1 ∣ 2)$
  For the line to be parallel to $h$ , it must have the same gradient/slope.
  The gradient of a line is called $m$ in the general line equation.
  $m = 1$
  Setting up the equation
  Plug $m = 1$ and the point $P (1 ∣ 2) = (x ∣ y)$ into the general line equation.
  $\displaystyle 2$ $=$ $\displaystyle 1+t$ $\displaystyle -1$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle 1$
  Substitute $m$ and $t$ into the general line equation.
  $\Rightarrow$ $y = x + 1$
  Do you have a question?
  Write it here 👉
3. $h$ : $y = 4 x$ ; $P(5|18) \;$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = 4 x$ ; $P (5 ∣ 18)$
  For the line to be parallel to $h$ , it must have the same gradient/slope.
  The gradient of a line is called $m$ in the general line equation.
  $m = 4$
  Setting up the equation
  Plug $m = 4$ and the point $P (5 ∣ 18) = (x ∣ y)$ into the general line equation.
  $\displaystyle 18$ $=$ $\displaystyle 4\cdot5+t$ $\displaystyle -20$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle -2$
  Substitute $m$ and $t$ into the general line equation.
  $\Rightarrow$ $y = 4 x - 2$
  Do you have a question?
  Write it here 👉
4. $h$ : $y = - 2 x + 1$ ; $P (- 1 ∣ 4)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Slope/Gradient of a line
  $y = - 2 x + 1$ ; $P (- 1 ∣ 4)$
  For the line to be parallel to $h$ , it must have the same gradient/slope.
  The gradient of a line is called $m$ in the general line equation.
  $m = - 2$
  Setting up the equation
  Plug $m = - 2$ and the point $P (- 1 ∣ 4) = (x ∣ y)$ into the general line equation.
  $\displaystyle 4$ $=$ $\displaystyle −2⋅(−1)+t$ $\displaystyle -2$
  ↓
  solve for $t$
  $\displaystyle t$ $=$ $\displaystyle 2$
  Substitute $m$ and $t$ into the general line equation.
  $\Rightarrow$ $y = - 2 x + 2$
  Do you have a question?
  Write it here 👉

Determining function equations.

A line has the gradient $a_{1}$ and passes through the point $P$ . Determine the equation of the function $f (x)$ , the points of intersection and draw the graph.

${a}_1=\frac{1}{2}$ ${P}\left(4|-2\right)\$

${a}_1=\frac34\;\;\;\;\;\;\;\;\;\;\;P\left( 1| -3\right)$

${a}_1=2\;\;\;\;\;\;\;\;\;\;P\left(3|-1\right)$

${a}_1=\frac45\;\;\;\;\;\;\;\;\;\;P\left(\frac32|4\right)$

Determining the equation of a function.

A line passes through the points $P_{1}$ and $P_{2}$ . Determine the equation of the function $f (x)$ , the points of intersection and draw the graph for the following values:

${P}_1\left(2|1\right)\;\;\;\;\;\;\;\;\;\;{P}_2\left(5|4\right)$
Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
Determining of the function equation
Substitute $P_{1}$ and $P_{2}$ into the general line equation . $\Rightarrow y={mx}+t\\$
Use the addition method.
$\def\arraystretch{1.25} \begin{array}{l}1)\;1=m\cdot2+t\\2)\;4=m\cdot5+t\end{array}$
$\displaystyle 1)-\;2)\;-3$ $=$ $\displaystyle -3m$ $\displaystyle :(-3)$
$\displaystyle m$ $=$ $\displaystyle 1$
Plug $m$ into $1)$ .
$\displaystyle 1$ $=$ $\displaystyle 1\cdot2+t$ $\displaystyle -2$
$\displaystyle t$ $=$ $\displaystyle -1$
Setze t und m in die allgemeine Geradengleichung ein.
$\mathrm y=\mathrm x-1\;\Rightarrow\;\mathrm f(x)=x-1$
Determining the axis intersection points
Set $y = 0$ to determine the intersection with the $x$ -axis.
$\displaystyle x-1$ $=$ $\displaystyle 1$ $\displaystyle +1$
$\displaystyle {x}_{{N}}$ $=$ $\displaystyle 0$
The intersection with the $x$ -axis is $S_1\left(1|0\right)$ .
The point of intersection with the $y$ -axis corresponds to the $y$ -axis intercept $t$ $\;\;\Rightarrow\;\;S_2\left(0|-1\right)$
Drawing the graph
Connect either the two intersection points $S_{1}$ and $S_{2}$ or the two given points $P_{1}$ and $P_{2}$ .
Or choose one of these points and go one to the right and one up according to the slope and connect these two points.
Do you have a question?
Write it here 👉

${P}_1\left(-3|-2\right)\;\;\;\;\;\;\;\;\;\;{P}_2\left(2\,|\,3\right)$

${P}_1\left(-2\,|\,3\right)\;\;\;\;\;\;\;\;\;\;{P}_2\left(4\,|-1\right)$
Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
Determining of the function equation
Substitute $P_{1}$ and $P_{2}$ into the general line equation . $\Rightarrow y={mx}+t\\$
Use the addition method.
$\def\arraystretch{1.25} \begin{array}{l}1)\;3=-2m+t\\2)\;-1=4m+t\end{array}\\$
$1)-\;2)\;4=-6m$ $\left|:(-6)\right.$
Shorten the fractions. $\Rightarrow\$ $m=\frac4{-6}\\$
Plug $m$ into $1)$ . $\Rightarrow$ $m=-\frac23\\$
Multiply. $\Rightarrow$ $3=-2\left(-\frac23\right)+t\\$
$3=\frac43+\mathrm t$ $\left|-\frac43\right.$
Subtract. $\Rightarrow$ $t=3-\frac43\\$
Setze m und t in die allgemeine Geradengleichung ein. $\Rightarrow$ $t=1\frac23\\$
${y}=-\frac{2}{3}{x}+1\frac{2}{3}$
Determining the axis intersection points
Set $y = 0$ to determine the intersection with the $x$ -axis.
$-\frac23x+1\frac23=0$ $\left|-1\frac23\right.$
$-\frac23x=-1\frac23=-\frac53$ $\left|:\left(-\frac23\right)\right.$
Divide the fractions $\Rightarrow$ $x=\displaystyle\frac{-\frac53}{-\frac23}\\$
${x}_N=-\frac53\cdot(-\frac32)=\frac52=2.5\\$
The intersection with the $x$ -axis is $\mathrm S_1\left(\frac52\,|\,0\right)$
The point of intersection with the $y$ -axis corresponds to the $y$ -axis intercept $t$ $\;\;\Rightarrow\;\;S_2\left(0|-1\right)$ $\Rightarrow\;\;S\left(0\,|\,\frac53\right)$
Drawing the graph
Connect either the two intersection points $S_{1}$ and $S_{2}$ or the two given points $P_{1}$ and $P_{2}$ .
Or choose one of these points and go 3 to the right and 2 down according to the slope and connect these two points.
Do you have a question?
Write it here 👉
${P}_1\left(-4\,|-1\right)\;\;\;\;\;\;\;\;\;\;{P}_2\left(3\,|\,1\right)$
Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
Determining of the function equation
Substitute $P_{1}$ and $P_{2}$ into the general line equation . $\Rightarrow y={mx}+t\\$
Use the addition method. $\Rightarrow$ $\def\arraystretch{1.25} \begin{array}{l}1)\;-1=-4m+t\\2)\;1=3m+t\end{array}\\$
$1)-\;2)\;-2=-7m$ $\left|:\left(-7\right)\right.$
Plug $m$ into $2)$ . $\Rightarrow$ $m=\frac27\\$
$1=\frac27\cdot3+t$ $\left|-\frac67\right.$
Subtract. $\Rightarrow$ $t=1-\frac67\\$
Plug $m$ and $t$ into the general line equation. $\Rightarrow t=\frac17\\$
${y}=\frac{2}{7}{x}+\frac{1}{7}\Rightarrow{f}(x)=\frac{2}{7}x+\frac{1}{7}$
Determining the axis intersection points
Set $y = 0$ to determine the intersection with the $x$ -axis.
$\frac27x+\frac17=0$ $\left|-\frac17\right.$
$\frac27x=-\frac17$ $\left|:\left(\frac27\right)\right.$
Divide the fractions. $\rightarrow$ Multiply by the inverse. $\Rightarrow x=\displaystyle\frac{-\frac17}{\frac27}\\$
Multiply. $\Rightarrow$ ${x}=-\frac{1}{7}\cdot\frac{7}{2}$
${x}_{{N}}=-\frac{1}{2}$
The intersection with the $x$ -axis is $S_1\left(-\frac12\,|\,0\right)$ .
The point of intersection with the $y$ -axis corresponds to the $y$ -axis intercept $t$ $\;\;\Rightarrow\;\;S_2\left(0|-1\right)$ $\Rightarrow\;\;S_2\left(0\,|\,\frac17\right)$
Drawing the graph
Connect either the two intersection points $S_{1}$ and $S_{2}$ or the two given points $P_{1}$ and $P_{2}$ .
Or choose one of these points and go 7 to the right and 2 up according to the slope and connect these two points.
Do you have a question?
Write it here 👉
${P}_1\left(-3\,|\,\frac92\right)\;\;\;\;\;\;\;\;\;\;{P}_2\left(4\,|-1\right)$
Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
Determining of the function equation
Substitute $P_{1}$ and $P_{2}$ into the general line equation . $\Rightarrow y={mx}+t\\$
Use the addition method. $\def\arraystretch{1.25} \Rightarrow\begin{array}{l}1)\;\frac92=-3m+t\\2)\;-1=4m+t\end{array}\\$
$1)-\;2)\;\frac{11}2=-7m$ $\left|:\left(-7\right)\right.$
Plug $m$ into $2)$ . $\Rightarrow m=\displaystyle\frac{\frac{11}2}{-7}=-\frac{11}{14}\\$
Multiply. $\Rightarrow-1=-\frac{11}{14}\cdot4+t\\$
$-1=-\frac{44}{14}+t$ $\left|+\frac{44}{14}\right.$
Plug $m$ and $t$ into the general equation of a straight line. $\Rightarrow t=\frac{30}{14}\\$
${y}=-\frac{11}{14}{x}+\frac{30}{14}$
Determining the axis intersection points
Set $y = 0$ to determine the intersection with the $x$ -axis.
$-\frac{11}{14}x+\frac{30}{14}=0$ $\left|-\frac{30}{14}\right.$
$-\frac{11}{14}x=-\frac{30}{14}$ $\left|:\left(-\frac{11}{14}\right)\right.$ Divide the fractions. That means, multiply by the inverse.
Multiply. $\Rightarrow x=-\frac{30}{14}\cdot\left(-\frac{14}{11}\right)\\$
${x}_{{N}}=\frac{30}{11}$
The intersection with the $x$ -axis is $S_1\left(\frac{30}{11}\,|\,0\right)$
The point of intersection with the $y$ -axis corresponds to the $y$ -axis intercept $t$ $\;\;\Rightarrow\;\;S_2\left(0|-1\right)$ $\Rightarrow\;\;S\left(0\,|\,\frac{30}{14}\right)$
Drawing the graph
Connect either the two intersection points $S_{1}$ and $S_{2}$ or the two given points $P_{1}$ and $P_{2}$ .
Or choose one of these points and go 14 to the right and 11 down according to the slope and connect these two points.
Do you have a question?
Write it here 👉
${P}_1\left(-4\,|-2\right)\;\;\;\;\;\;\;\;\;\;{P}_2\left(\frac72\,|\,4\right)$
Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
Determining of the function equation
Substitute $P_{1}$ and $P_{2}$ into the general line equation . $\Rightarrow y={mx}+t\\$
Use the addition method. $\def\arraystretch{1.25} \Rightarrow\begin{array}{l}1)\;-2=-4m+t\\2)\;4=3.5m+t\end{array}\\$
$- 6 = - 7.5 m$ $\left|:\left(-7.5\right)\right.$
Divide. $\Rightarrow m=\frac{-6}{-7.5}$
Plug $m$ into $1)$ . $\Rightarrow m=0.8$
Multiply. $\Rightarrow-2=0.8\cdot\left(-4\right)+t$
$- 2 = - 3.2 + t$ $\left|+3.2\right.$
Plug $m$ and $t$ into the general equation of a straight line. $\Rightarrow{t}=1.2$
${y}=0.8{x}+1.2\Rightarrow{f}(x)=0.8x+1.2$
Determining the axis intersection points
Set $y = 0$ to determine the intersection with the $x$ -axis.
$0.8 x + 1.2 = 0$ $\left|-1.2\right.$
$0.8 x = - 1.2$ $\left|:0.8\right.$
Dividiere. $\Rightarrow\mathrm x=\frac{-1.2}{0.8}$
${x}_{{N}}=-1.5$
The intersection with the $x$ -axis is $S_1\left(-1.5\,|\,0\right)$
The point of intersection with the $y$ -axis corresponds to the $y$ -axis intercept $t$ $\;\;\Rightarrow\;\;S_2\left(0|-1\right)$ $\Rightarrow\;\;\mathrm S\left(0\,|\,1.2\right)$
Drawing the graph
Connect either the two intersection points $S_{1}$ and $S_{2}$ or the two given points $P_{1}$ and $P_{2}$ .
Or choose one of these points and go 1 to the right and 0.8 up according to the slope and connect these two points.
Do you have a question?
Write it here 👉

8
Draw the following straight lines and give the function term.
1. $G_{f}$ has the slope $\frac34$ and intersects the $y$ -axis at $- 2$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Setting up the function term
  $m=\frac34;\;t=-2$
  Plug $m$ and $t$ into the general line equation $y = m x + t$ .
  $\Rightarrow y=\frac{3}{4}x-2$
  Drawing the line
  Select any point on the line, e.g. the $y$ -axis interception point $(0 ∣ - 2)$ . From there go 1 to the right and according to the slope $m=\frac34$ upwards (alternatively, also the quadruple length to avoid fractions: 4 to the right and 3 upwards). Connect the two points to form a straight line.
  Do you have a question?
  Write it here 👉
2. $G_{f}$ has the slope $0$ and intersects the $y$ -axis at $3$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Setting up the function term
  $m=0;\;t=3$
  Plug $m$ and $t$ into the general line equation $y = m x + t$ .
  $\Rightarrow y=3$
  Drawing the line
  The line runs through the $y$ -axis intercept $(0 ∣ 3)$ parallel to the $x$ -axis because the line has a slope of 0.
  Do you have a question?
  Write it here 👉
3. $G_{f}$ passes through the point $P (-4\vert2)$ and is parallel to the $ x$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Setting up the function term
  Since the line is parallel to the $x$ -axis, so its slope is $0 \Rightarrow m=0$ .
  The line goes through the point $(-3\vert-2)$ . Since the slope is 0, the line has the $y$ -value $- 2$ at $x = 0$ .
  Its $y$ -axis intercept is therefore $- 2$
  $m=0;\;t=-2$
  Plug $m$ and $t$ into the general line equation $y = m x + t$ .
  $y = - 2$
  Drawing the line
  The line runs through the $y$ -axis interception point $(0 ∣ - 2)$ and is parallel to the x-axis.
  Do you have a question?
  Write it here 👉
4. $G_{f}$ passes through the point $P (-4\vert2)$ and is parallel to the $y$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Setting up the function term
  This straight line cannot be described by a function equation. The gradient would be infinitely large.
  
  However, we can still draw the straight line.
  Drawing the line
  The line runs through the point $(-4\vert2)$ and is parallel to the $y$ -axis.
  Do you have a question?
  Write it here 👉

Consider the four points $A (40 ∣ 220)$ , $B (100 ∣ 250)$ , $C (200 ∣ 300)$ , $D (80 ∣ 240)$ .

Draw the points $A$ - $D$ in a suitable coordinate system.
Determine the equation of the line passing through points $A$ - $D$ .
Give three more points that lie on the straight line.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation

Coordinate system

Geogebra File: https://assets.serlo.org/legacy/5373_YY7IuAMwyu.xml

Consider $A (40 ∣ 220)$ and $B (100 ∣ 250)$ .

Choose ... from A

$\displaystyle x_1$	$=$	$\displaystyle 40$
$\displaystyle y_2$	$=$	$\displaystyle 220$

and ... from B

$\displaystyle x_2$	$=$	$\displaystyle 100$
$\displaystyle y_2$	$=$	$\displaystyle 250$

$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{250-220}{100-40}=\frac{30}{60}=0.5$

Since the points $A$ - $D$ all lie on a common line, it is sufficient to select only two points (for example $A$ and $B$ ). With their help you determine the slope of the line. To do this, subtract the $y$ -coordinate of point $A$ from the $y$ -coordinate of point $B$ , and the $x$ -coordinate of point $A$ from the $x$ -coordinate of point $B$ .

Then determine the $y$ -axis intercept by inserting a point on the straight line (for example C) into the straight line equation $\ \ y = mx + t \$ and solving for $t$ .

Insert point $C$ into the line equation $y = m x + t$ together with the previously calculated $m = 0.5$ :

$\displaystyle y$	$=$	$\displaystyle 0.5\cdot x+t$
$\displaystyle 300$	$=$	$\displaystyle 0.5\cdot200+t$
$\displaystyle \Leftrightarrow\ t$	$=$	$\displaystyle 200$

With this we have determined both $m$ and $t$ , so that our line equation is:

$\ y=0.5\cdot x+200$

Insert any three $x$ -values into the line equation to get the respective $y$ -value, e.g. $x_{1} = 0$ , $x_{2} = - 100$ and $x_{3} = 300$ .

$\displaystyle y$	$=$	$\displaystyle 0.5\cdot x+200$
$\displaystyle y_1$	$=$	$\displaystyle 0.5\cdot x_1+200$
	$=$	$\displaystyle 0.5\cdot0+200\ =\ 200$
$\displaystyle y_2$	$=$	$\displaystyle 0.5\cdot x_2+200$
	$=$	$\displaystyle 0.5\cdot\left(-100\right)+200\ =150$
$\displaystyle y_3$	$=$	$\displaystyle 0.5\cdot x_3+200$
	$=$	$\displaystyle 0.5\cdot300+200\ =350$

This gives us the following three points $D$ , $E$ and $F$ :

$D (0 ∣ 200), E (- 100 ∣ 150)$ and $F (300 ∣ 350)$

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Check that the straight line through $P_{1}$ and $P_{2}$ goes through the origin.

${P}_1\left(2|4\right);\;{P}_2\left(-1.5|-3\right)$

${P}_1\left(-1|3.5\right);\;{P}_2\left(2|-2\right)$

11
Determine the function equation.
A straight line has the $y$ -axis intercept $t$ and passes through the point $P$ . Determine the function equation $f (x)$ and draw the graph.
1. $t = - 1$ $P = (2 ∣ 3)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Determining the function equation
  General line equation:
  $y=m\cdot x+t$
  Substitute $t$ into the line equation.
  $f(x)=m\cdot x-1$
  Now plug the point $P$ into $f (x)$ .
  $3=m\cdot2-1$
  Solve for $m$ .
  $m = (3 + 1) : 2 = 2$
  $f (x) = 2 x - 1$
  Plot:
  Do you have a question?
  Write it here 👉
2. $t = 3$ $P (- 4 ∣ - 3)$
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Determining the function equation
  General line equation:
  $y=m\cdot x+t$
  Substitute $t$ into the line equation.
  $f(x)=m\cdot x+3$
  Now plug the point $P$ into $f (x)$ .
  $-3=m\cdot(-4)+3$
  Solve for $m$ .
  $m=\frac{-6}{-4}=\frac{3}{2}$
  $f(x)=\frac{3}{2}x+3$
  Plot:
  Do you have a question?
  Write it here 👉
12
Consider the following function graphs:
1. Which of the four graphs belongs to the equation $y=\frac{5}{4}x-1$ `?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  The line equation is: $y=\frac54x-1$
  You can read off the slope and the $y$ -axis intercept of this graph from the equation.
  $m=\frac{5}{4}$
  $t = - 1$
  First check which functions have a $y$ -intercept of $t = - 1$ by reading off the $y -$ value of each graph at the interaction with the $y$ -axis.
  Only graphs I and II have the $y -$ intercept $- 1$ so you can exclude any other graph.
  Now check which of the two graphs has the slope $m=\frac54$ by moving one to the right from the point $x = 0$ and checking which of the two $y$ -values increases by $\frac54$ .
  Both graphs start at the point $P\left(0;-1\right)$ . Since the straight line you are looking for has a slope of $\frac54$ , it also passes through the point $(0 + 4 ∣ - 1 + 5) = (4 ∣ 4)$ .
  But only line II is running through this point.
  $\Rightarrow$ Graph II is the one belonging to the given equation.
  Do you have a question?
  Write it here 👉
2. What is the equation to Graph III?
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  Line to be checked: Graph III
  First read off where the graph intersects the $y$ -axis to find the $y$ -axis intercept.
  The $y$ -value of the point where the $y$ -axis is intersected amounts to $y = 1.25$ .
  Now read off by how much the $y$ -value changes when you go from $x = 0$ , one to the right. This will give you the slope.
  The $y$ -value increases from $y = 1.25$ to $y = 2.25$ .
  Thus the slope is $m=\frac{2.25-1.25}{1}=\frac11=1$ .
  Set up the line equation.
  ⇒ Graph III has the equation $y = x + 1.25$
  Do you have a question?
  Write it here 👉

Consider the points $A (40 ∣ 220)$ , $B (100 ∣ 250)$ , $C (200 ∣ 300)$ , $D (80 ∣ 240)$ .

Draw the points $A$ - $D$ in a suitable coordinate system.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
Do you have a question?
Write it here 👉

Determine the equation of the line passing through points $A$ - $D$ .

Give three more points that lie on the straight line.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Geradengleichung
Insert any three $x$ -values into the line equation to get the respective $y$ -value, e.g. $x_{1} = 0$ , $x_{2} = - 100$ and $x_{3} = 300$ .
$\displaystyle y$ $=$ $\displaystyle 0.5\cdot x+200$
$\displaystyle y_1$ $=$ $\displaystyle 0.5\cdot x_1+200$
$=$ $\displaystyle 0.5\cdot0+200\ =\ 200$
$\displaystyle y_2$ $=$ $\displaystyle 0.5\cdot x_2+200$
$=$ $\displaystyle 0.5\cdot\left(-100\right)+200\ =150$
$\displaystyle y_3$ $=$ $\displaystyle 0.5\cdot x_3+200$
$=$ $\displaystyle 0.5\cdot300+200\ =350$
This gives us the following three points $D$ , $E$ and $F$ :
$D (0 ∣ 200), E (- 100 ∣ 150)$ and $F (300 ∣ 350)$
Do you have a question?
Write it here 👉

14
Draw the following lines and determine the function term.
1. $G_{f}$ has the slope $\frac34$ and intersects the $y$ -axis at $- 2$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Seting up the function term
  $m=\frac34;\;t=-2$
  Insert $m$ and $t$ into the general line equation $y = m x + t$ .
  $\Rightarrow y=\frac34x-2$
  Draw the line
  Select any point on the line, e.g. the $y$ -axis intercept (0|-2). Go from there 1 to the right and according to the slope $m=\frac34$ upwards (alternatively, also the 4-fold length to avoid fractions: 4 to the right and 3 upwards). Connect the two points to get a straight line.
  Do you have a question?
  Write it here 👉
2. $G_{f}$ has a slope of 0 and intersects the $y$ -axis at 3.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Seting up the function term
  $m=0;\;t=3$
  Insert $m$ and $t$ into the general line equation $y = m x + t$ .
  $\Rightarrow y=3$
  Draw the line
  The line runs through the $y$ -axis intercept (0|3) parallel to the $x$ -axis because the straight line has a slope of 0.
  Do you have a question?
  Write it here 👉
3. $G_{f}$ passes through the point $P (-3\vert-2)$ and is parallel to the $x$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Seting up the function term
  Since the straight line is parallel to the $x$ -axis, its slope is 0 $\Rightarrow m=0$ .
  The line passes through the point $(-3\vert-2)$ . Since the slope is 0, the line has the $y$ -value -2 at $x = 0$ .
  Its $y$ -axis intercept is therefore at $-2 \Rightarrow t=-2$ .
  $m=0;\;t=-2$
  Insert $m$ and $t$ into the general line equation $y = m x + t$ .
  $y = - 2$
  Draw the line
  The line runs through the $y$ -intercept $(0 ∣ - 2)$ parallel to the $x$ -axis because the straight line has a slope of 0.
  Do you have a question?
  Write it here 👉
4. $G_{f}$ passes through the point $P (-4\vert2)$ and is parallel to the $y$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Linear function
  Seting up the function term
  This straight line cannot be described by a function equation. The gradient would be infinitely large.
  
  However, we can still draw the straight line.
  Draw the line
  The line runs through the point $(-4\vert2)$ parallel to the $y$ -axis.
  Do you have a question?
  Write it here 👉
15
Determine the equation of the straight line through ...
1. the point $P (- 3 ∣ 4)$ and being parallel to the $x$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  Parallel to the $x$ -axis, means that the slope is $m = 0$ .
  Plug $m$ and $P$ into the general line equation.
  $\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} 4&=&0\cdot(-3)+t \\ 4&=& t \end{array}$
  Assemble to a line equation.
  $\displaystyle y=0\cdot x+4=4$
  Do you have a question?
  Write it here 👉
2. the point $Q (2 ∣ 5)$ and is parallel to the bisector of the 2nd quadrant (the diagonal pointing down).
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  Parallel to the bisector of the 2nd quadrant means the line has the same slope as the bisector.
  The slope of the bisector of the 2nd quadrant is -1
  $\displaystyle \Rightarrow m=-1$
  Plug $m$ into the straight line equation in order to calculate $t$ .
  $\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} 5&=&-2+t &|+2\\ t&=&7 \end{array}$
  Plug $m$ and $t$ into the general equation of a straight line.
  $\displaystyle y=-1\cdot x+7=-x+7$
  Do you have a question?
  Write it here 👉
3. the point $R (- 4 ∣ 2)$ and is parallel to the $y$ -axis.
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  Parallel to the $y$ -axis means that there is no function equation, since an $x$ -value is assigned an infinite number of $y$ -values.
  The straight line can therefore only be described as the $x$ -value of $R$ .
  $\displaystyle \;\;\Rightarrow\;\;x=-4$
  Do you have a question?
  Write it here 👉
4. the point $S (2 ∣ - 3)$ and is parallel to the bisector of the 1st quadrant (the diagonal pointing up).
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  The line being parallel to the bisector of the 1st quadrant means it has the same slope.
  The slope of the bisector of the 1st quadrant is $1$ .
  $\displaystyle \Rightarrow m=1$
  Substitute $m$ and $S$ into the general line equation and solve for $t$ .
  $\displaystyle \def\arraystretch{1.25} \begin{array}{rcl} -3&=&2+t &|-2\\ t&=&-5 \end{array}$
  Substitute $m$ and $t$ into the general line equation.
  $\displaystyle \;\Rightarrow\;\;y=x-5$
  Do you have a question?
  Write it here 👉
5. the origin and is parallel to the straight line $\overline{\mathrm{AB}}$ with $A (- 72 ∣ - 60)$ and $B (- 24 ∣ - 20)$ .
  
  Für diese Aufgabe benötigst Du folgendes Grundwissen: Line equation
  Passing through the origin means, the $y$ -axis intercept is $t = 0$ .
  Being parallel to the straight line $\overline{\mathrm{AB}}$ means that the line has the same slope as $\overline{\mathrm{AB}}$ .
  $\displaystyle A(-72|-60);B(-24|-20)$
  Calculate the slope $m$ using the difference quotient .
  $\displaystyle m=\frac{-20-(-60)}{-24-(-72)}=\frac{40}{48}=\frac{5}{6}$
  Substitute $m$ and $t$ into the general line equation.
  $\displaystyle \Rightarrow y=\frac{5}{6}x$
  Do you have a question?
  Write it here 👉

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$\displaystyle f\left(x\right)$	$=$	$\displaystyle 0$
	↓	We are looking for an $x$ with $f (x) = 0$ .
$\displaystyle \frac{1}{2}x-4$	$=$	$\displaystyle 0$	$\displaystyle +4$
$\displaystyle \frac{1}{2}x$	$=$	$\displaystyle 4$	$\displaystyle :\frac{1}{2}$
$\displaystyle x$	$=$	$\displaystyle \frac{4}{\frac{1}{2}}$
	↓	You divide by a fraction $\rightarrow$ multiply by the reciprocal.
$\displaystyle x$	$=$	$\displaystyle 4\cdot2$
$\displaystyle x$	$=$	$\displaystyle 8$

$\displaystyle \frac{3}{4}x-3.75$	$=$	$\displaystyle 0$	$\displaystyle +3.75$
$\displaystyle \frac{3}{4}x$	$=$	$\displaystyle 3.75$	$\displaystyle :\frac{3}{4}$
$\displaystyle x$	$=$	$\displaystyle 3.75:\frac{3}{4}$
	↓	You divide by a fraction $\rightarrow$ multiply by the reciprocal.
$\displaystyle x$	$=$	$\displaystyle 3.75\cdot\frac{4}{3}$
	↓	multiply
$\displaystyle x$	$=$	$\displaystyle 5$

$\displaystyle 2x-7$	$=$	$\displaystyle 0$	$\displaystyle +7$
$\displaystyle 2x$	$=$	$\displaystyle 7$	$\displaystyle :2$
$\displaystyle x$	$=$	$\displaystyle 7:2$
	↓	divide
$\displaystyle x$	$=$	$\displaystyle 3.5$

$\displaystyle \frac{4}{5}x+2.8$	$=$	$\displaystyle 0$	$\displaystyle -2.8$
$\displaystyle \frac{4}{5}x$	$=$	$\displaystyle -2.8$	$\displaystyle :\frac{4}{5}$
$\displaystyle x$	$=$	$\displaystyle -2.8:\frac{4}{5}$
	↓	divide
$\displaystyle x$	$=$	$\displaystyle -\frac{7}{2}$
$\displaystyle x$	$=$	$\displaystyle -3.5$

$\displaystyle -5.5$	$=$	$\displaystyle 3m$
$\displaystyle -\frac{11}{2}$	$=$	$\displaystyle 3m$	$\displaystyle :3$
$\displaystyle -\frac{11}{2\cdot3}$	$=$	$\displaystyle m$
$\displaystyle -\frac{11}{6}$	$=$	$\displaystyle m$

$\displaystyle 0$	$=$	$\displaystyle 3\cdot1+t$	$\displaystyle -3$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle -3$

$\displaystyle 2$	$=$	$\displaystyle 1+t$	$\displaystyle -1$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle 1$

$\displaystyle 18$	$=$	$\displaystyle 4\cdot5+t$	$\displaystyle -20$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle -2$

$\displaystyle 4$	$=$	$\displaystyle −2⋅(−1)+t$	$\displaystyle -2$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle 2$

$\displaystyle f\left(x\right)$	$=$	$\displaystyle {a}_1\cdot{x}+{t}$
	↓	plug $a_1 =\tfrac{1}{2}$ into the general line equation.
$\displaystyle f\left(x\right)$	$=$	$\displaystyle \frac{1}{2}x+t$
	↓	plug $P$ into $f (x)$
$\displaystyle -2$	$=$	$\displaystyle \frac{1}{2}\cdot4+t$	$\displaystyle -2$
	↓	solve for $t$
$\displaystyle t$	$=$	$\displaystyle -4$
	↓	Plug $t$ into $f (x)$ .
	$=$

$\displaystyle f\left(x\right)$	$=$	$\displaystyle a_1\cdot x+t$
	↓	$t$ : $y$ -axis intercept Plug $a_2=\frac34$ into the general line equation.
$\displaystyle f\left(x\right)$	$=$	$\displaystyle \frac{3}{4}\mathrm{x}+\mathrm{t}$
	↓	Plug $P (1 ∣ - 3)$ into $f (x)$ .
$\displaystyle -3$	$=$	$\displaystyle \frac{3}{4}\cdot1+t$
	↓	multiply
$\displaystyle -3$	$=$	$\displaystyle \frac{3}{4}+t$	$\displaystyle -\frac{3}{4}$
$\displaystyle t$	$=$	$\displaystyle -3-\frac{3}{4}$
	↓	add
$\displaystyle t$	$=$	$\displaystyle -3.75$
	↓	Plug $t$ into $f (x)$ .
	$=$

$\displaystyle f\left(x\right)$	$=$	$\displaystyle a_1\cdot x+t$
	↓	$t$ : $y$ -axis intercept Plug $a_{3} = 2$ into the general line equation.
$\displaystyle f\left(x\right)$	$=$	$\displaystyle 2x+t$
	↓	Plug $P (3 ∣ - 1)$ into $f (x)$ .
$\displaystyle -1$	$=$	$\displaystyle 2\cdot3+t$
	↓	multiply
$\displaystyle -1$	$=$	$\displaystyle 6+t$	$\displaystyle -6$
$\displaystyle t$	$=$	$\displaystyle -1-6$
	↓	subtract
$\displaystyle t$	$=$	$\displaystyle -7$
	↓	Plug $t$ into $f (x)$ .
	$=$

$\displaystyle f\left(x\right)$	$=$	$\displaystyle a_1\cdot x+t$
	↓	$t$ : $y$ -axis intercept Plug $a_4=\frac45$ into the general line equation
$\displaystyle f\left(x\right)$	$=$	$\displaystyle \frac{4}{5}x+t$
	↓	Plug $P\left(\frac32\|4\right)$ into $f (x)$ .
$\displaystyle 4$	$=$	$\displaystyle \frac{4}{5}\cdot\frac{3}{2}+t$
	↓	shorten by 2
$\displaystyle 4$	$=$	$\displaystyle \frac{2}{5}\cdot3+t$
	↓	multiply
$\displaystyle 4$	$=$	$\displaystyle \frac{6}{5}+t$	$\displaystyle -\frac{6}{5}$
$\displaystyle t$	$=$	$\displaystyle 4-\frac{6}{5}$
	↓	Write 4 as a fraction with a 4 in the denominator
$\displaystyle t$	$=$	$\displaystyle \frac{20}{5}-\frac{6}{5}$
	↓	subtract
$\displaystyle t$	$=$	$\displaystyle \frac{14}{5}$
$\displaystyle t$	$=$	$\displaystyle 2.8$
	↓	Setze t in f(x) ein.

$\displaystyle 1)-\;2)\;-3$	$=$	$\displaystyle -3m$	$\displaystyle :(-3)$
$\displaystyle m$	$=$	$\displaystyle 1$

$\displaystyle 1$	$=$	$\displaystyle 1\cdot2+t$	$\displaystyle -2$
$\displaystyle t$	$=$	$\displaystyle -1$

$\displaystyle x-1$	$=$	$\displaystyle 1$	$\displaystyle +1$
$\displaystyle {x}_{{N}}$	$=$	$\displaystyle 0$

$\displaystyle 1)-\;2)\;-5$	$=$	$\displaystyle -5m$	$\displaystyle :\left(-5\right)$
$\displaystyle m$	$=$	$\displaystyle 1$
	↓	Plug $m$ into $2)$ .
$\displaystyle 3$	$=$	$\displaystyle 2+t$	$\displaystyle -2$
$\displaystyle t$	$=$	$\displaystyle 1$
	↓	Plug $m$ and $t$ into the general line equation.
$\displaystyle y$	$=$	$\displaystyle {x}+1\Rightarrow{f}(x)=x+1$

$\displaystyle x+1$	$=$	$\displaystyle 0$	$\displaystyle -1$
$\displaystyle {x}_{{N}}$	$=$	$\displaystyle -1$

$\displaystyle 7$	$=$	$\displaystyle 3.5m$	$\displaystyle :3.5$
$\displaystyle m$	$=$	$\displaystyle \frac{7}{3.5}$
$\displaystyle m$	$=$	$\displaystyle 2$

$\displaystyle 4$	$=$	$\displaystyle 2\cdot2+t$
$\displaystyle 4$	$=$	$\displaystyle 4+t$	$\displaystyle -4$
$\displaystyle t$	$=$	$\displaystyle 4-4$
$\displaystyle t$	$=$	$\displaystyle 0$

$\displaystyle -2$	$=$	$\displaystyle 2\cdot\left(-\frac{11}{6}\right)+t$
$\displaystyle -2$	$=$	$\displaystyle -\frac{11}{3}+t$	$\displaystyle +\frac{11}{3}$
$\displaystyle -\frac{6}{3}+\frac{11}{3}$	$=$	$\displaystyle t$
$\displaystyle -\frac{5}{3}$	$=$	$\displaystyle t$

Exercises: Linear functions and line equations

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Graph 1 (rot)

Graph 2 (grün)

Ergebnis

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yyy-axis intercept

Slope

Result

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Part 1:

Part 2:

Setting up the equation

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Setting up the equation

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Setting up the equation

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Setting up the equation

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Determining the function equaition

Determining the yyy-axis intercept

Determining the xxx-axis intercept

Plot

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Determining the function equation

Determining the axis intercepts

Plot

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Determining the function equation

Determining the axis intercepts

Plot

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Determining the function equation

Determining the axis intercepts

Plot

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Determining of the function equation

Determining the axis intersection points

Drawing the graph

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Determining of the function equation

Determining the axis intersection points

Drawing the graph

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Determining of the function equation

Determining the axis intersection points

Drawing the graph

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Determining of the function equation

Determining the axis intersection points

Drawing the graph

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Determining of the function equation

Determining the axis intersection points

Drawing the graph

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Determining of the function equation

Determining the axis intersection points

Drawing the graph

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Setting up the function term

Drawing the line

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Setting up the function term

Drawing the line

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Setting up the function term

Drawing the line

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Setting up the function term

$y$ -axis intercept

Determining the $y$ -axis intercept

Determining the $x$ -axis intercept