Exercises: Drawing graphs of linear functions

Draw the five given points in a coordinate system and draw a straight line through the points.

Draw the four given points in a coordinate system and draw a straight line through the points.

Drawing the linear function

First read off the $yy$-intercept and the slope from the function term of the linear function.

In this case:

$f(x)=-2x+4f(x)=-2x+4$

You obtain a $yy$-intercept of $t=4t=4$ and a slope of $m=-2m=-2$.

Drawing the linear function

First read off the $yy$-intercept and the slope from the function term of the linear function.

In this case:

$f(x)={\displaystyle \frac{1}{2}}x-2f(x)=\backslash dfrac\{1\}\{2\}x-2$

You obtain a $yy$-intercept of $t=-2t=-2$ and a slope of $m={\displaystyle \frac{1}{2}}m=\backslash dfrac\{1\}\{2\}$.

Drawing the linear function

The function $h(x)=5h(x)=5$ represents a special case of a linear function. The slope of $h(x)h(x)$ is $00$.

This means that the function value does not change regardless of the variable $xx$.

So if you draw the function value $h(x)=5h(x)=5$ for each $xx$ in a coordinate system, you get a straight line that runs parallel to the $xx$ axis at the height $y=5y=5$.

Determine one point

$y=3x-2y=3x-2$

$-2-2$ is the $tt$ within the general line equation, or in other words, the $yy$-axis intercept.

$\Rightarrow P(0\mathrm{\mid }-2)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;P\backslash left(0\backslash vert-2\backslash right)$

Find the slope

Determine the slope $mm$ of the function

$y=3x-2y=3x-2$

$33$ is the $mm$ within the general line equation, or in other words, the slope.

$m=3m=3$

Draw the line

Go from the previously determined point one unit to the right and $33$ upwards, since $mm$ is equal to $33$. Here is a second point of the function.

Then connect the two points to form a straight line.

Transform the equation

$y=2-xy=2-x$

First, we do a little transformation, such that we get the form of a general line equation.

$y=-x+2y=-x+2$

Determine one point

$y=-x+2y=-x+2$

$+2+2$ is the $tt$ within the general line equation, or in other words, the $yy$-axis intercept.

$\Rightarrow P\left(0\mathrm{\mid }2\right)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;P\backslash left(0\backslash vert2\backslash right)$

Find the slope

Determine the slope $mm$ of the function

$y=-x+2y=-x+2$

$-1-1$ is the $mm$ within the general line equation, or in other words, the slope.

$m=-1m=-1$

Draw the line

From the previously determined point, go one unit to the right and $11$ downwards, since $mm$ is negative. Here is a second point of the function.

Then connect the two points to form a straight line.

Determine one point

$y=-\frac{3}{4}x-1y=-\backslash frac\{3\}\{4\}x-1$

$-1-1$ is the $tt$ within the general line equation, or in other words, the $yy$-axis intercept.

$\Rightarrow P(0\mathrm{\mid }-1)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;P\backslash left(0\backslash vert-1\backslash right)$

Find the slope

Determine the slope $mm$ of the function

$y=-\frac{3}{4}x-1y=-\backslash frac\{3\}\{4\}x-1$

$-\frac{3}{4}-\backslash frac\{3\}\{4\}$ is the $mm$ within the general line equation, or in other words, the slope.

$m=-\frac{3}{4}m=-\backslash frac\{3\}\{4\}$

Draw the line

From the previously determined point, go one unit to the right and $\frac{3}{4}\backslash frac34$ downwards, since $mm$ is negative. Here is a second point of the function.

Then connect the two points to form a straight line.

Determine one point

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

$+2+2$ is the $tt$ within the general line equation, or in other words, the $yy$-axis intercept.

$\Rightarrow P\left(0\mathrm{\mid }2\right)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;P\backslash left(0\backslash vert2\backslash right)$

Find the slope

Determine the slope $mm$ of the function

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

$-\frac{1}{2}-\backslash frac\{1\}\{2\}\backslash$ is the $mm$ within the general line equation, or in other words, the slope.

$m=-\frac{1}{2}m=-\backslash frac\{1\}\{2\}$

Draw the line

From the previously determined point, go one unit to the right and $\frac{1}{2}\backslash frac12$ downwards, since $mm$ is negative. Here is a second point of the function.

Then connect the two points to form a straight line.

Determine one point

$\mathrm{y}=\frac{3}{4}\mathrm{x}+1\backslash mathrm\{y\}=\backslash frac\{3\}\{4\}\backslash mathrm\{x\}+1$

$11$ is the $tt$ within the general line equation, or in other words, the $yy$-axis intercept.

$\Rightarrow P\left(0\mathrm{\mid }1\right)\backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;P\backslash left(0\backslash vert1\backslash right)$

Find the slope

Determine the slope $mm$ of the function

$\mathrm{y}=\frac{3}{4}\mathrm{x}+1\backslash mathrm\{y\}=\backslash frac\{3\}\{4\}\backslash mathrm\{x\}+1$

$\frac{3}{4}\backslash frac\{3\}\{4\}$ is the $mm$ within the general line equation, or in other words, the slope.

$m=\frac{3}{4}m=\backslash frac\{3\}\{4\}$

Draw the line

From the previously determined point go one unit to the right and $\frac{3}{4}\backslash frac34$ upwards, since $mm$ is positive. Here is a second point of the function.

Then connect the two points to form a straight line.

$\mathrm{f}(\mathrm{x})=2\mathrm{x}-5\backslash mathrm\{f\}(\backslash mathrm\{x\})=2\backslash mathrm\{x\}-5$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid -5\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;-5\backslash right.\backslash right)$

$\Rightarrow m_f=2\backslash Rightarrow\backslash ;m\_f=2$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}\left(2.5\mid 0\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(2\{.\}5\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=-\mathrm{x}-3\backslash mathrm\{f\}(\backslash mathrm\{x\})=-\backslash mathrm\{x\}-3$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid -3\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;-3\backslash right.\backslash right)$

$\Rightarrow m_f=-1\backslash Rightarrow\backslash ;m\_f=-1$

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}(-3\mid 0)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(-3\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=\frac{1}{2}\mathrm{x}+1\backslash mathrm\{f\}(\backslash mathrm\{x\})=\backslash frac\{1\}\{2\}\backslash mathrm\{x\}+1$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid 1\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;1\backslash right.\backslash right)$

$\Rightarrow m_f=\frac{1}{2}\backslash Rightarrow\backslash ;m\_f=\backslash frac12$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}(-2\mid 0)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(-2\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=-\frac{1}{2}\mathrm{x}-2\backslash mathrm\{f\}(\backslash mathrm\{x\})=-\backslash frac\{1\}\{2\}\backslash mathrm\{x\}-2$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid -2\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;-2\backslash right.\backslash right)$

$\Rightarrow m_f=-\frac{1}{2}\backslash Rightarrow\backslash ;m\_f=-\backslash frac12$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0..

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}(-4\mid 0)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(-4\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=\frac{1}{3}\mathrm{x}-\frac{1}{2}\backslash mathrm\{f\}(\backslash mathrm\{x\})=\backslash frac\{1\}\{3\}\backslash mathrm\{x\}-\backslash frac\{1\}\{2\}$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid -\frac{1}{2}\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;-\backslash frac12\backslash right.\backslash right)$

$\Rightarrow m_f=\frac{1}{3}\backslash Rightarrow\backslash ;m\_f=\backslash frac13$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}\left(\frac{3}{2}\mid 0\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(\backslash frac32\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=-\frac{1}{4}\mathrm{x}+\frac{3}{2}\backslash mathrm\{f\}(\backslash mathrm\{x\})=-\backslash frac\{1\}\{4\}\backslash mathrm\{x\}+\backslash frac\{3\}\{2\}$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid \frac{3}{2}\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;\backslash frac32\backslash right.\backslash right)$

$\Rightarrow m_f=-\frac{1}{4}\backslash Rightarrow\backslash ;m\_f=-\backslash frac14$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}\left(6\mid 0\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(6\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=\frac{2}{3}\mathrm{x}+2\backslash mathrm\{f\}(\backslash mathrm\{x\})=\backslash frac\{2\}\{3\}\backslash mathrm\{x\}+2$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid 2\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;2\backslash right.\backslash right)$

$\Rightarrow m_f=\frac{2}{3}\backslash Rightarrow\backslash ;m\_f=\backslash frac23$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}(-3\mid 0)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(-3\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=-\frac{3}{4}\mathrm{x}-1\backslash mathrm\{f\}(\backslash mathrm\{x\})=-\backslash frac\{3\}\{4\}\backslash mathrm\{x\}-1$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid -1\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;-1\backslash right.\backslash right)$

$\Rightarrow m_f=-\frac{3}{4}\backslash Rightarrow\backslash ;m\_f=-\backslash frac34$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}(-\frac{4}{3}\mid 0)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(-\backslash frac43\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=-3\mathrm{x}+\frac{5}{10}\backslash mathrm\{f\}(\backslash mathrm\{x\})=-3\backslash mathrm\{x\}+\backslash frac\{5\}\{10\}$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid \frac{5}{10}\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;\backslash frac5\{10\}\backslash right.\backslash right)$

$\Rightarrow m_f=-3\backslash Rightarrow\backslash ;m\_f=-3$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}\left(\frac{1}{6}\mid 0\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(\backslash frac16\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

$\mathrm{f}(\mathrm{x})=\frac{5}{7}\mathrm{x}-\frac{12}{4}=\frac{5}{7}\mathrm{x}-3\backslash mathrm\{f\}(\backslash mathrm\{x\})=\backslash frac\{5\}\{7\}\backslash mathrm\{x\}-\backslash frac\{12\}\{4\}=\backslash frac\{5\}\{7\}\backslash mathrm\{x\}-3$

First read off the $yy$-axis intercept and the slope from the function equation.

$\Rightarrow {\mathrm{P}}_{\mathrm{y}}\left(0\mid -3\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash ;\backslash left(0\backslash ;\backslash left|\backslash ;-3\backslash right.\backslash right)$

$\Rightarrow m_f=\frac{5}{7}\backslash Rightarrow\backslash ;m\_f=\backslash frac57$

Calculate the point of intersection with the $xx$-axis. Set the function term equal to 0.

$\Rightarrow {\mathrm{P}}_{\mathrm{x}}\left(\frac{21}{5}\mid 0\right)\backslash Rightarrow\backslash ;\{\backslash mathrm\; P\}\_\backslash mathrm\; x\backslash ;\backslash left(\backslash frac\{21\}5\backslash ;\backslash left|\backslash ;0\backslash right.\backslash right)$

The $yy$-value of the straight line is always 2.5. Therefore, the straight line is parallel to the $xx$-axis.

The table of values for $y=0.3\cdot xy=0.3\backslash cdot\; x$ with step length $\mathrm{\Delta }x=1\backslash Delta\; x\; =1$:

Points of the function $ff$ in the coordinate system and graph of the function $ff$:

The graph of the function $ff$ is a line that passes through the origin $\left(0\mathrm{\mid }0\right)\backslash left(0\backslash vert0\backslash right)$ of the coordinate system.

Insert the coordinates of $PP$ and $QQ$ into the function equation:

Point $P\left(5\mathrm{\mid }1.4\right)P\backslash left(\backslash textcolor\{006400\}\{5\}|\backslash textcolor\{cc0000\}\{1.4\}\backslash right)$:

Answer: The point $PP$ does not lie on the graph of the function $ff$.

Point $Q\left(8\mathrm{\mid }2.4\right)Q\backslash left(\backslash textcolor\{006400\}\{8\}|\backslash textcolor\{cc0000\}\{2.4\}\backslash right)$ :

Answer: The point $QQ$ lies on the graph of the function $ff$.

Additional illustration of the graph with the two points $PP$ and $QQ$

(Ths is not required by the problem setting, but just provided for better visualization.)