Exercises: Finding line equations

Linear function $f(x)$

Set up the general form of a linear function.

$f(x)=m\cdot x+t$

Read off the y-intercept from the diagram.

$t=2$

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

You can use these points, for example:

$P_1(-3|0)$ $x_1=-3$ and $y_1=0$

$P_2(0|2)$ $x_2=0$ and $y_2=2$

For the gradient, you then get by substituting:

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

Now plug the calculated values of $m$ and $t$ into the general form:

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

Linear function $g(x)$

Set up the general form of a linear function.

$g(x)=m\cdot x+t$

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

$${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}$$

For example, you can use these two points:

$P_1(2|-4)$ $x_1=2$ and $y_1=-4$

$P_2(3.5|0)$ $x_2=3.5$ and $y_2=0$

The gradient is then:

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

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

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

Insert one of the points, for example $(2|-4)$.

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

Now solve for $t$.

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

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

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

Linear function $h(x)$

Set up the general form of a linear function.

$h(x)=m\cdot x+t$

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

$${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}$$

You can use these two points, for example:

$P_1(0|-2)$ $x_1=0$ and $y_1=-2$

$P_2(4|-3)$ $x_2=4$ and $y_2=-3$

Now use them to calculate the gradient:

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

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

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

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

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

Now, solve for $t$.

$t=-3+1=-2$

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

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

Linear function $f(x)$

Set up the general form of a linear function.

$f(x)=m\cdot x+t$

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

$${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}$$

You can use these two points, for example:

$P_1(2|4)$ $x_1=2$ and $y_1=4$

$P_2(\mathrm{2,5}|0)$ $x_2=\mathrm{2,5}$ and $y_2=0$

The result is a gradient:

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

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

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

Insert one of the points, for example $(2|4)$ :

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

Solve for $t$.

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

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

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

Linear function $g(x)$

Set up the general form of a linear function.

$g(x)=m\cdot x+t$

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

$${\displaystyle m=\frac{y_2-y_1}{x_2-x_1}}$$

You can use these two points, for example:

$P_1(0|-2)$ $x_1=0$ and $y_1=-2$

$P_2(-4|-3)$ $x_2=-4$ and $y_2=-3$

Now use them to calculate the gradient:

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

Read off the y-intercept from the diagram.

$t=-2$

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

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