Determining of the function equation

Substitute $P_1\{P\}\_1$ and $P_2\{P\}\_2$ into the general line equation . $\Rightarrow y=mx+t\backslash Rightarrow\; y=\{mx\}+t\backslash \backslash$

Use the addition method.

$\begin{array}{c}1)1=m\cdot 2+t\\ 2)4=m\cdot 5+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}1)\backslash ;1=m\backslash cdot2+t\backslash \backslash 2)\backslash ;4=m\backslash cdot5+t\backslash end\{array\}$

Plug $mm$ into $1)1)$.

Setze t und m in die allgemeine Geradengleichung ein.

$\mathrm{y}=\mathrm{x}-1\Rightarrow \mathrm{f}(x)=x-1\backslash mathrm\; y=\backslash mathrm\; x-1\backslash ;\backslash Rightarrow\backslash ;\backslash mathrm\; f(x)=x-1$

Determining the axis intersection points

Set $y=0y=0$ to determine the intersection with the $xx$-axis.

The intersection with the $xx$-axis is $S_1\left(1\mathrm{\mid }0\right)S\_1\backslash left(1|0\backslash right)$.

The point of intersection with the $yy$-axis corresponds to the $yy$-axis intercept $tt$ $\Rightarrow S_2(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0|-1\backslash right)$

Drawing the graph

• Connect either the two intersection points $S_{1}S\_1$ and $S_{2}S\_2$ or the two given points $P_{1}P\_1$ and $P_{2}P\_2$.

• Or choose one of these points and go one to the right and one up according to the slope and connect these two points.

Determining of the function equation

Substitute $P_1\{P\}\_1$ and $P_2\{P\}\_2$ into the general line equation . $\Rightarrow y=mx+t\backslash Rightarrow\; y=\{mx\}+t\backslash \backslash$

Use the addition method.

$\begin{array}{c}1)-2=m\cdot (-3)+t\\ 2)3=m\cdot 2+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}1)\backslash ;-2=m\backslash cdot(-3)+t\backslash \backslash 2)\backslash ;3=m\backslash cdot2+t\backslash end\{array\}\backslash \backslash$

Determining the axis intersection points

Set $y=0y=0$ to determine the intersection with the $xx$-axis.

The intersection with the $xx$-axis is $S_1(-1\mathrm{\mid }0)S\_1(-1\backslash ,|\backslash ,0)$.

The point of intersection with the $yy$-axis corresponds to the $yy$-axis intercept $tt$ $\Rightarrow S_2(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0|-1\backslash right)$ $\Rightarrow S\left(0\mathrm{\mid }1\right)\backslash Rightarrow\backslash ;S\backslash left(0\backslash ,|\backslash ,1\backslash right)$

Drawing the graph

• Connect either the two intersection points $S_{1}S\_1$ and $S_{2}S\_2$ or the two given points $P_{1}P\_1$ and $P_{2}P\_2$.

• Or choose one of these points and go one to the right and one up according to the slope and connect these two points.

Determining of the function equation

Substitute $P_1\{P\}\_1$ and $P_2\{P\}\_2$ into the general line equation . $\Rightarrow y=mx+t\backslash Rightarrow\; y=\{mx\}+t\backslash \backslash$

Use the addition method.

$\begin{array}{c}1)3=-2m+t\\ 2)-1=4m+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}1)\backslash ;3=-2m+t\backslash \backslash 2)\backslash ;-1=4m+t\backslash end\{array\}\backslash \backslash$

$1)-2)4=-6m1)-\backslash ;2)\backslash ;4=-6m$ $\mid :(-6)\backslash left|:(-6)\backslash right.$

Shorten the fractions. $\Rightarrow \backslash Rightarrow\backslash$$m=\frac{4}{-6}m=\backslash frac4\{-6\}\backslash \backslash$

Plug $mm$ into $1)1)$. $\Rightarrow \backslash Rightarrow$ $m=-\frac{2}{3}m=-\backslash frac23\backslash \backslash$

Multiply. $\Rightarrow \backslash Rightarrow$ $3=-2(-\frac{2}{3})+t3=-2\backslash left(-\backslash frac23\backslash right)+t\backslash \backslash$

$3=\frac{4}{3}+\mathrm{t}3=\backslash frac43+\backslash mathrm\; t$ $\mid -\frac{4}{3}\backslash left|-\backslash frac43\backslash right.$

Subtract. $\Rightarrow \backslash Rightarrow$ $t=3-\frac{4}{3}t=3-\backslash frac43\backslash \backslash$

Setze m und t in die allgemeine Geradengleichung ein. $\Rightarrow \backslash Rightarrow$ $t=1\frac{2}{3}t=1\backslash frac23\backslash \backslash$

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

Determining the axis intersection points

Set $y=0y=0$ to determine the intersection with the $xx$-axis.

$-\frac{2}{3}x+1\frac{2}{3}=0-\backslash frac23x+1\backslash frac23=0$ $\mid -1\frac{2}{3}\backslash left|-1\backslash frac23\backslash right.$

$-\frac{2}{3}x=-1\frac{2}{3}=-\frac{5}{3}-\backslash frac23x=-1\backslash frac23=-\backslash frac53$ $\mid :(-\frac{2}{3})\backslash left|:\backslash left(-\backslash frac23\backslash right)\backslash right.$

Divide the fractions $\Rightarrow \backslash Rightarrow$ $x={\displaystyle \frac{-\frac{5}{3}}{-\frac{2}{3}}}x=\backslash displaystyle\backslash frac\{-\backslash frac53\}\{-\backslash frac23\}\backslash \backslash$

$x_N=-\frac{5}{3}\cdot (-\frac{3}{2})=\frac{5}{2}=2.5\{x\}\_N=-\backslash frac53\backslash cdot(-\backslash frac32)=\backslash frac52=2.5\backslash \backslash$

The intersection with the $xx$-axis is ${\mathrm{S}}_1\left(\frac{5}{2}\mathrm{\mid }0\right)\backslash mathrm\; S\_1\backslash left(\backslash frac52\backslash ,|\backslash ,0\backslash right)$

The point of intersection with the $yy$-axis corresponds to the $yy$-axis intercept $tt$ $\Rightarrow S_2(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0|-1\backslash right)$  $\Rightarrow S\left(0\mathrm{\mid }\frac{5}{3}\right)\backslash Rightarrow\backslash ;\backslash ;S\backslash left(0\backslash ,|\backslash ,\backslash frac53\backslash right)$

Drawing the graph

• Connect either the two intersection points $S_{1}S\_1$ and $S_{2}S\_2$ or the two given points $P_{1}P\_1$ and $P_{2}P\_2$.

• Or choose one of these points and go 3 to the right and 2 down according to the slope and connect these two points.

Determining of the function equation

Substitute $P_1\{P\}\_1$ and $P_2\{P\}\_2$ into the general line equation . $\Rightarrow y=mx+t\backslash Rightarrow\; y=\{mx\}+t\backslash \backslash$

Use the addition method. $\Rightarrow \backslash Rightarrow$ $\begin{array}{c}1)-1=-4m+t\\ 2)1=3m+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}1)\backslash ;-1=-4m+t\backslash \backslash 2)\backslash ;1=3m+t\backslash end\{array\}\backslash \backslash$

$1)-2)-2=-7m1)-\backslash ;2)\backslash ;-2=-7m$ $\mid :(-7)\backslash left|:\backslash left(-7\backslash right)\backslash right.$

Plug $mm$ into $2)2)$. $\Rightarrow \backslash Rightarrow$ $m=\frac{2}{7}m=\backslash frac27\backslash \backslash$

$1=\frac{2}{7}\cdot 3+t1=\backslash frac27\backslash cdot3+t$ $\mid -\frac{6}{7}\backslash left|-\backslash frac67\backslash right.$

Subtract. $\Rightarrow \backslash Rightarrow$ $t=1-\frac{6}{7}t=1-\backslash frac67\backslash \backslash$

Plug $mm$ and $tt$ into the general line equation. $\Rightarrow t=\frac{1}{7}\backslash Rightarrow\; t=\backslash frac17\backslash \backslash$

$y=\frac{2}{7}x+\frac{1}{7}\Rightarrow f(x)=\frac{2}{7}x+\frac{1}{7}\{y\}=\backslash frac\{2\}\{7\}\{x\}+\backslash frac\{1\}\{7\}\backslash Rightarrow\{f\}(x)=\backslash frac\{2\}\{7\}x+\backslash frac\{1\}\{7\}$

Determining the axis intersection points

Set $y=0y=0$ to determine the intersection with the $xx$-axis.

$\frac{2}{7}x+\frac{1}{7}=0\backslash frac27x+\backslash frac17=0$ $\mid -\frac{1}{7}\backslash left|-\backslash frac17\backslash right.$

$\frac{2}{7}x=-\frac{1}{7}\backslash frac27x=-\backslash frac17$ $\mid :\left(\frac{2}{7}\right)\backslash left|:\backslash left(\backslash frac27\backslash right)\backslash right.$

Divide the fractions. $\to \backslash rightarrow$ Multiply by the inverse. $\Rightarrow x={\displaystyle \frac{-\frac{1}{7}}{\frac{2}{7}}}\backslash Rightarrow\; x=\backslash displaystyle\backslash frac\{-\backslash frac17\}\{\backslash frac27\}\backslash \backslash$

Multiply. $\Rightarrow \backslash Rightarrow$ $x=-\frac{1}{7}\cdot \frac{7}{2}\{x\}=-\backslash frac\{1\}\{7\}\backslash cdot\backslash frac\{7\}\{2\}$

$x_N=-\frac{1}{2}\{x\}\_\{\{N\}\}=-\backslash frac\{1\}\{2\}$

The intersection with the $xx$-axis is $S_1(-\frac{1}{2}\mathrm{\mid }0)S\_1\backslash left(-\backslash frac12\backslash ,|\backslash ,0\backslash right)$.

The point of intersection with the $yy$-axis corresponds to the $yy$-axis intercept $tt$ $\Rightarrow S_2(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0|-1\backslash right)$  $\Rightarrow S_2\left(0\mathrm{\mid }\frac{1}{7}\right)\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0\backslash ,|\backslash ,\backslash frac17\backslash right)$

Drawing the graph

• Connect either the two intersection points $S_{1}S\_1$ and $S_{2}S\_2$ or the two given points $P_{1}P\_1$ and $P_{2}P\_2$.

• Or choose one of these points and go 7 to the right and 2 up according to the slope and connect these two points.

Determining of the function equation

Substitute $P_1\{P\}\_1$ and $P_2\{P\}\_2$ into the general line equation . $\Rightarrow y=mx+t\backslash Rightarrow\; y=\{mx\}+t\backslash \backslash$

Use the addition method. $\Rightarrow \begin{array}{c}1)\frac{9}{2}=-3m+t\\ 2)-1=4m+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash Rightarrow\backslash begin\{array\}\{l\}1)\backslash ;\backslash frac92=-3m+t\backslash \backslash 2)\backslash ;-1=4m+t\backslash end\{array\}\backslash \backslash$

$1)-2)\frac{11}{2}=-7m1)-\backslash ;2)\backslash ;\backslash frac\{11\}2=-7m$ $\mid :(-7)\backslash left|:\backslash left(-7\backslash right)\backslash right.$

Plug $mm$ into $2)2)$ . $\Rightarrow m={\displaystyle \frac{\frac{11}{2}}{-7}=-\frac{11}{14}}\backslash Rightarrow\; m=\backslash displaystyle\backslash frac\{\backslash frac\{11\}2\}\{-7\}=-\backslash frac\{11\}\{14\}\backslash \backslash$

Multiply. $\Rightarrow -1=-\frac{11}{14}\cdot 4+t\backslash Rightarrow-1=-\backslash frac\{11\}\{14\}\backslash cdot4+t\backslash \backslash$

$-1=-\frac{44}{14}+t-1=-\backslash frac\{44\}\{14\}+t$ $\mid +\frac{44}{14}\backslash left|+\backslash frac\{44\}\{14\}\backslash right.$

Plug $mm$ and $tt$ into the general equation of a straight line. $\Rightarrow t=\frac{30}{14}\backslash Rightarrow\; t=\backslash frac\{30\}\{14\}\backslash \backslash$

$y=-\frac{11}{14}x+\frac{30}{14}\{y\}=-\backslash frac\{11\}\{14\}\{x\}+\backslash frac\{30\}\{14\}$

Determining the axis intersection points

Set $y=0y=0$ to determine the intersection with the $xx$-axis.

$-\frac{11}{14}x+\frac{30}{14}=0-\backslash frac\{11\}\{14\}x+\backslash frac\{30\}\{14\}=0$ $\mid -\frac{30}{14}\backslash left|-\backslash frac\{30\}\{14\}\backslash right.$

$-\frac{11}{14}x=-\frac{30}{14}-\backslash frac\{11\}\{14\}x=-\backslash frac\{30\}\{14\}$ $\mid :(-\frac{11}{14})\backslash left|:\backslash left(-\backslash frac\{11\}\{14\}\backslash right)\backslash right.$ Divide the fractions. That means, multiply by the inverse.

Multiply. $\Rightarrow x=-\frac{30}{14}\cdot (-\frac{14}{11})\backslash Rightarrow\; x=-\backslash frac\{30\}\{14\}\backslash cdot\backslash left(-\backslash frac\{14\}\{11\}\backslash right)\backslash \backslash$

$x_N=\frac{30}{11}\{x\}\_\{\{N\}\}=\backslash frac\{30\}\{11\}$

The intersection with the $xx$-axis is $S_1\left(\frac{30}{11}\mathrm{\mid }0\right)S\_1\backslash left(\backslash frac\{30\}\{11\}\backslash ,|\backslash ,0\backslash right)$

The point of intersection with the $yy$-axis corresponds to the $yy$-axis intercept $tt$ $\Rightarrow S_2(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0|-1\backslash right)$  $\Rightarrow S\left(0\mathrm{\mid }\frac{30}{14}\right)\backslash Rightarrow\backslash ;\backslash ;S\backslash left(0\backslash ,|\backslash ,\backslash frac\{30\}\{14\}\backslash right)$

Drawing the graph

• Connect either the two intersection points $S_{1}S\_1$ and $S_{2}S\_2$ or the two given points $P_{1}P\_1$ and $P_{2}P\_2$.

• Or choose one of these points and go 14 to the right and 11 down according to the slope and connect these two points.

Determining of the function equation

Substitute $P_1\{P\}\_1$ and $P_2\{P\}\_2$ into the general line equation . $\Rightarrow y=mx+t\backslash Rightarrow\; y=\{mx\}+t\backslash \backslash$

Use the addition method. $\Rightarrow \begin{array}{c}1)-2=-4m+t\\ 2)4=3.5m+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash Rightarrow\backslash begin\{array\}\{l\}1)\backslash ;-2=-4m+t\backslash \backslash 2)\backslash ;4=3.5m+t\backslash end\{array\}\backslash \backslash$

$-6=-7.5m-6=-7.5m$ $\mid :(-7.5)\backslash left|:\backslash left(-7.5\backslash right)\backslash right.$

Divide. $\Rightarrow m=\frac{-6}{-7.5}\backslash Rightarrow\; m=\backslash frac\{-6\}\{-7.5\}$

Plug $mm$ into $1)1)$. $\Rightarrow m=0.8\backslash Rightarrow\; m=0.8$

Multiply. $\Rightarrow -2=0.8\cdot (-4)+t\backslash Rightarrow-2=0.8\backslash cdot\backslash left(-4\backslash right)+t$

$-2=-3.2+t-2=-3.2+t$ $\mid +3.2\backslash left|+3.2\backslash right.$

Plug $mm$ and $tt$ into the general equation of a straight line. $\Rightarrow t=1.2\backslash Rightarrow\{t\}=1.2$

$y=0.8x+1.2\Rightarrow f(x)=0.8x+1.2\{y\}=0.8\{x\}+1.2\backslash Rightarrow\{f\}(x)=0.8x+1.2$

Determining the axis intersection points

Set $y=0y=0$ to determine the intersection with the $xx$-axis.

$0.8x+1.2=00.8x+1.2=0$ $\mid -1.2\backslash left|-1.2\backslash right.$

$0.8x=-1.20.8x=-1.2$ $\mid :0.8\backslash left|:0.8\backslash right.$

Dividiere. $\Rightarrow \mathrm{x}=\frac{-1.2}{0.8}\backslash Rightarrow\backslash mathrm\; x=\backslash frac\{-1.2\}\{0.8\}$

$x_N=-1.5\{x\}\_\{\{N\}\}=-1.5$

The intersection with the $xx$-axis is $S_1(-1.5\mathrm{\mid }0)S\_1\backslash left(-1.5\backslash ,|\backslash ,0\backslash right)$

The point of intersection with the $yy$-axis corresponds to the $yy$-axis intercept $tt$ $\Rightarrow S_2(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;S\_2\backslash left(0|-1\backslash right)$  $\Rightarrow \mathrm{S}\left(0\mathrm{\mid }1.2\right)\backslash Rightarrow\backslash ;\backslash ;\backslash mathrm\; S\backslash left(0\backslash ,|\backslash ,1.2\backslash right)$

Drawing the graph

• Connect either the two intersection points $S_{1}S\_1$ and $S_{2}S\_2$ or the two given points $P_{1}P\_1$ and $P_{2}P\_2$.

• 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.