Linear function $f(x)f(x)$

Set up the general form of a linear function.

$f(x)=m\cdot x+tf(x)=m\; \backslash cdot\; x+t$

Read off the y-intercept from the diagram.

$t=2t\; =\; 2$

Read off two points from the graph of the linear function to calculate the gradient:

You can use these points, for example:

$P_1(-3\mathrm{\mid }0)P\_1(-3|0)$ $x_1=-3x\_1=\; -3$ and $y_1=0y\_1\; =\; 0$

$P_2(0\mathrm{\mid }2)P\_2(0|2)$ $x_2=0x\_2=\; 0$ and $y_2=2y\_2\; =\; 2$

For the gradient, you then get by substituting:

${\displaystyle m=\frac{2-0}{0-(-3)}=\frac{2}{3}}\backslash displaystyle\; m\; =\; \backslash frac\{2\; -\; 0\}\{0\; -\; (-3)\}\; =\; \backslash frac\{2\}\{3\}$

Now plug the calculated values of $mm$ and $tt$ into the general form:

$f(x)={\displaystyle \frac{2}{3}}\cdot x+2f(x)\; =\; \backslash dfrac\{2\}\{3\}\backslash cdot\; x\; +\; 2$

Linear function $g(x)g(x)$

Set up the general form of a linear function.

$g(x)=m\cdot x+tg(x)=m\; \backslash cdot\; x+t$

Read off two points on the graph of the linear function to calculate the gradient.

${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}\backslash displaystyle\; m\; =\; \backslash frac\{y\_2\; -\; y\_1\}\{x\_2\; -\; x\_1\}$

For example, you can use these two points:

$P_1(2\mathrm{\mid }-4)P\_1(2|-4)$ $x_1=2x\_1=\; 2$ and $y_1=-4y\_1=-4$

$P_2(3.5\mathrm{\mid }0)P\_2(3.5|0)$ $x_2=3.5x\_2=3.5$ and $y_2=0y\_2\; =\; 0$

The gradient is then:

${\displaystyle m=\frac{0-(-4)}{3.5-2}=\frac{4}{1.5}=4\cdot \frac{2}{3}=\frac{8}{3}}\backslash displaystyle\; m\; =\; \backslash frac\{0\; -\; (-4)\}\{3.5\; -\; 2\}\; =\; \backslash frac\{4\}\{1.5\}\; =\; 4\; \backslash cdot\; \backslash frac\{2\}\{3\}\; =\; \backslash frac\{8\}\{3\}$

As the y-intercept is not visible, you have to calculate it. To do this, set up the equation of the line.

${\displaystyle g(x)=\frac{8}{3}\cdot x+t}\backslash displaystyle\; g(x)\; =\; \backslash frac\{8\}\{3\}\; \backslash cdot\; x\; +\; t$

Insert one of the points, for example $(2\mathrm{\mid }-4)(2|-4)$.

${\displaystyle -4=\frac{8}{3}\cdot 2+t}\backslash displaystyle\; -4\; =\; \backslash frac\{8\}\{3\}\; \backslash cdot\; 2\; +\; t$

Now solve for $tt$.

${\displaystyle t=-4-\frac{8}{3}\cdot 2=-\frac{12}{3}-\frac{16}{3}=-\frac{28}{3}}\backslash displaystyle\; t\; =\; -4-\; \backslash frac\{8\}\{3\}\; \backslash cdot\; 2\; =\; -\backslash frac\{12\}\{3\}-\; \backslash frac\{16\}\{3\}\; =\; -\backslash frac\{28\}\{3\}$

Substitute the values of $mm$ and $tt$ into the general form of the linear function, and you will get the linear equation:

${\displaystyle g(x)=\frac{8}{3}\cdot x-\frac{28}{3}}\backslash displaystyle\; g(x)=\backslash frac\{8\}\{3\}\; \backslash cdot\; x\; -\; \backslash frac\{28\}\{3\}$

Linear function $h(x)h(x)$

Set up the general form of a linear function.

$h(x)=m\cdot x+th(x)=m\; \backslash cdot\; x+t$

Read off two points from the graph of the linear function to calculate the gradient:

${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}\backslash displaystyle\; m\; =\; \backslash frac\{y\_2\; -\; y\_1\}\{x\_2\; -\; x\_1\}$

You can use these two points, for example:

$P_1(0\mathrm{\mid }-2)P\_1(0|-2)$ $x_1=0x\_1=\; 0$ and $y_1=-2y\_1\; =\; -2$

$P_2(4\mathrm{\mid }-3)P\_2(4|-3)$ $x_2=4x\_2=\; 4$ and $y_2=-3y\_2=-3$

Now use them to calculate the gradient:

${\displaystyle m=\frac{-3-(-2)}{4-0}=\frac{-3+2}{4}=-\frac{1}{4}}\backslash displaystyle\; m\; =\; \backslash frac\{-3\; -\; (-2)\}\{4\; -\; 0\}\; =\; \backslash frac\{-3\; +\; 2\}\{4\}\; =\; -\backslash frac\{1\}\{4\}$

Either read off $t=-2t=-2$ or calculate the value. To calculate it, set up the linear equation.

${\displaystyle h(x)=-\frac{1}{4}\cdot x+t}\backslash displaystyle\; h(x)\; =\; -\backslash frac\{1\}\{4\}\; \backslash cdot\; x\; +\; t$

Insert a point that lies on the straight line, for example $(4\mathrm{\mid }-3)(4|-3)$.

${\displaystyle -3=-\frac{1}{4}\cdot 4+t}\backslash displaystyle\; -3\; =\; -\backslash frac\{1\}\{4\}\; \backslash cdot\; 4\; +\; t$

Now, solve for $tt$.

$t=-3+1=-2t\; =\; -3\; +\; 1\; =\; -2$

Setze $m=-{\displaystyle \frac{1}{4}}m\; =\; -\backslash dfrac14$ und $t=-2t\; =\; -2$ in die allgemeine Form ein und du erhältst die Geradengleichung:

${\displaystyle h(x)=-\frac{1}{4}\cdot x-2}\backslash displaystyle\; h(x)\; =\; -\backslash frac\{1\}\{4\}\; \backslash cdot\; x\; -2$

Linear function $f(x)f(x)$

Set up the general form of a linear function.

$f(x)=m\cdot x+tf(x)=m\; \backslash cdot\; x+t$

Read off two points from the graph of the linear function to calculate the gradient:

${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}\backslash displaystyle\; m\; =\; \backslash frac\{y\_2\; -\; y\_1\}\{x\_2\; -\; x\_1\}$

You can use these two points, for example:

$P_1(2\mathrm{\mid }4)P\_1(2|4)$ $x_1=2x\_1=\; 2$ and $y_1=4y\_1=4$

$P_2(\mathrm{2,5}\mathrm{\mid }0)P\_2(2\{,\}5|0)$ $x_2=\mathrm{2,5}x\_2=\; 2\{,\}5$ and $y_2=0y\_2=0$

The result is a gradient:

${\displaystyle m=\frac{0-4}{\mathrm{2,5}-2}=\frac{-4}{\mathrm{0,5}}=-4\cdot \frac{2}{1}=-8}\backslash displaystyle\; m\; =\; \backslash frac\{0\; -\; 4\}\{2\{,\}5\; -\; 2\}\; =\; \backslash frac\{-4\}\{0\{,\}5\}\; =\; -4\; \backslash cdot\; \backslash frac\{2\}\{1\}\; =\; -8$

As the y-intercept is not visible, you have to calculate it. Therefore, set up the equation of the line:

${\displaystyle f(x)=-8\cdot x+t}\backslash displaystyle\; f(x)\; =\; -8\; \backslash cdot\; x\; +\; t$

Insert one of the points, for example $(2\mathrm{\mid }4)(2|4)$ :

${\displaystyle 4=-8\cdot 2+t}\backslash displaystyle\; 4\; =\; -8\; \backslash cdot\; 2\; +\; t$

Solve for $tt$.

${\displaystyle t=4+8\cdot 2=4+16=20}\backslash displaystyle\; t\; =\; 4\; +\; 8\; \backslash cdot\; 2\; =\; 4\; +\; 16\; =\; 20$

Plug $m=-8m=-8$ and $t=20t=20$ into the general form, and you will get the linear equation as a result:

$f(x)=-8\cdot x+20f(x)\; =\; -8\backslash cdot\; x\; +\; 20$

Linear function $g(x)g(x)$

Set up the general form of a linear function.

$g(x)=m\cdot x+tg(x)=m\; \backslash cdot\; x+t$

Read off two points from the graph of the linear function to calculate the gradient:

${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}\backslash displaystyle\; m\; =\; \backslash frac\{y\_2\; -\; y\_1\}\{x\_2\; -\; x\_1\}$

You can use these two points, for example:

$P_1(0\mathrm{\mid }-2)P\_1(0|-2)$ $x_1=0x\_1=\; 0$ and $y_1=-2y\_1=-2$

$P_2(-4\mathrm{\mid }-3)P\_2(-4|-3)$ $x_2=-4x\_2=\; -4$ and $y_2=-3y\_2=-3$

Now use them to calculate the gradient:

${\displaystyle m=\frac{-3-(-2)}{-4-0}=\frac{-3+2}{-4}=\frac{1}{4}}\backslash displaystyle\; m\; =\; \backslash frac\{-3\; -\; (-2)\}\{-4\; -\; 0\}\; =\; \backslash frac\{-3\; +2\}\{-4\}\; =\; \backslash frac\{1\}\{4\}$

Read off the y-intercept from the diagram.

$t=-2t\; =\; -2$

Plug $m={\displaystyle \frac{1}{4}}m\; =\; \backslash dfrac14$ and $t=-2t\; =\; -2$ into the general form of the linear function, and you get the line equation of $gg$:

${\displaystyle g(x)=\frac{1}{4}\cdot x-2}\backslash displaystyle\; g(x)=\backslash frac\{1\}\{4\}\; \backslash cdot\; x\; -\; 2$