Exercises: Linear functions and line equations

Determining of the function equation

Substitute $P_1$ and $P_2$ into the general line equation . $\begin{array}{l}\Rightarrow y=mx+t\\ \end{array}$

Use the addition method.

$\begin{array}{l}1)1=m\cdot 2+t\\ 2)4=m\cdot 5+t\end{array}$

Plug $m$ into $1)$.

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.

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 $t$ $\Rightarrow S_2(0|-1)$

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.

Determining of the function equation

Substitute $P_1$ and $P_2$ into the general line equation . $\begin{array}{l}\Rightarrow y=mx+t\\ \end{array}$

Use the addition method.

$\begin{array}{l}\begin{array}{l}1)-2=m\cdot (-3)+t\\ 2)3=m\cdot 2+t\end{array}\\ \end{array}$

Determining the axis intersection points

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

The intersection with the $x$-axis is $S_1(-1|0)$.

The point of intersection with the $y$-axis corresponds to the $y$-axis intercept $t$ $\Rightarrow S_2(0|-1)$ $\Rightarrow S\left(0|1\right)$

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.

Determining of the function equation

Substitute $P_1$ and $P_2$ into the general line equation . $\begin{array}{l}\Rightarrow y=mx+t\\ \end{array}$

Use the addition method.

$\begin{array}{l}\begin{array}{l}1)3=-2m+t\\ 2)-1=4m+t\end{array}\\ \end{array}$

$1)-2)4=-6m$ $|:(-6)$

Shorten the fractions. $\Rightarrow$$\begin{array}{l}m=\frac{4}{-6}\\ \end{array}$

Plug $m$ into $1)$. $\Rightarrow$ $\begin{array}{l}m=-\frac{2}{3}\\ \end{array}$

Multiply. $\Rightarrow$ $\begin{array}{l}3=-2(-\frac{2}{3})+t\\ \end{array}$

$3=\frac{4}{3}+\mathrm{t}$ $|-\frac{4}{3}$

Subtract. $\Rightarrow$ $\begin{array}{l}t=3-\frac{4}{3}\\ \end{array}$

Setze m und t in die allgemeine Geradengleichung ein. $\Rightarrow$ $\begin{array}{l}t=1\frac{2}{3}\\ \end{array}$

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

Determining the axis intersection points

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

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

$-\frac{2}{3}x=-1\frac{2}{3}=-\frac{5}{3}$ $|:(-\frac{2}{3})$

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

$\begin{array}{l}{x}_N=-\frac{5}{3}\cdot (-\frac{3}{2})=\frac{5}{2}=2.5\\ \end{array}$

The intersection with the $x$-axis is ${\mathrm{S}}_1\left(\frac{5}{2}|0\right)$

The point of intersection with the $y$-axis corresponds to the $y$-axis intercept $t$ $\Rightarrow S_2(0|-1)$  $\Rightarrow S\left(0|\frac{5}{3}\right)$

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.

Determining of the function equation

Substitute $P_1$ and $P_2$ into the general line equation . $\begin{array}{l}\Rightarrow y=mx+t\\ \end{array}$

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

$1)-2)-2=-7m$ $|:(-7)$

Plug $m$ into $2)$. $\Rightarrow$ $\begin{array}{l}m=\frac{2}{7}\\ \end{array}$

$1=\frac{2}{7}\cdot 3+t$ $|-\frac{6}{7}$

Subtract. $\Rightarrow$ $\begin{array}{l}t=1-\frac{6}{7}\\ \end{array}$

Plug $m$ and $t$ into the general line equation. $\begin{array}{l}\Rightarrow t=\frac{1}{7}\\ \end{array}$

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

Determining the axis intersection points

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

$\frac{2}{7}x+\frac{1}{7}=0$ $|-\frac{1}{7}$

$\frac{2}{7}x=-\frac{1}{7}$ $|:\left(\frac{2}{7}\right)$

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

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

$x_N=-\frac{1}{2}$

The intersection with the $x$-axis is $S_1(-\frac{1}{2}|0)$.

The point of intersection with the $y$-axis corresponds to the $y$-axis intercept $t$ $\Rightarrow S_2(0|-1)$  $\Rightarrow S_2\left(0|\frac{1}{7}\right)$

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.

Determining of the function equation

Substitute $P_1$ and $P_2$ into the general line equation . $\begin{array}{l}\Rightarrow y=mx+t\\ \end{array}$

Use the addition method. $\begin{array}{l}\Rightarrow \begin{array}{l}1)\frac{9}{2}=-3m+t\\ 2)-1=4m+t\end{array}\\ \end{array}$

$1)-2)\frac{11}{2}=-7m$ $|:(-7)$

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

Multiply. $\begin{array}{l}\Rightarrow -1=-\frac{11}{14}\cdot 4+t\\ \end{array}$

$-1=-\frac{44}{14}+t$ $|+\frac{44}{14}$

Plug $m$ and $t$ into the general equation of a straight line. $\begin{array}{l}\Rightarrow t=\frac{30}{14}\\ \end{array}$

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

Determining the axis intersection points

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

$-\frac{11}{14}x+\frac{30}{14}=0$ $|-\frac{30}{14}$

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

Multiply. $\begin{array}{l}\Rightarrow x=-\frac{30}{14}\cdot (-\frac{14}{11})\\ \end{array}$

$x_N=\frac{30}{11}$

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 $t$ $\Rightarrow S_2(0|-1)$  $\Rightarrow S\left(0|\frac{30}{14}\right)$

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.

Determining of the function equation

Substitute $P_1$ and $P_2$ into the general line equation . $\begin{array}{l}\Rightarrow y=mx+t\\ \end{array}$

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

$-6=-7.5m$ $|:(-7.5)$

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

Plug $m$ into $1)$. $\Rightarrow m=0.8$

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

$-2=-3.2+t$ $|+3.2$

Plug $m$ and $t$ into the general equation of a straight line. $\Rightarrow t=1.2$

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

Determining the axis intersection points

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

$0.8x+1.2=0$ $|-1.2$

$0.8x=-1.2$ $|:0.8$

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

$x_N=-1.5$

The intersection with the $x$-axis is $S_1(-1.5|0)$

The point of intersection with the $y$-axis corresponds to the $y$-axis intercept $t$ $\Rightarrow S_2(0|-1)$  $\Rightarrow \mathrm{S}\left(0|1.2\right)$

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.