Setting up the function term

$m=\frac{3}{4};t=-2m=\backslash frac34;\backslash ;t=-2$

Plug $mm$ and $tt$ into the general line equation $y=mx+ty=mx+t$ .

$\Rightarrow y=\frac{3}{4}x-2\backslash Rightarrow\; y=\backslash frac\{3\}\{4\}x-2$

Drawing the line

Select any point on the line, e.g. the $yy$-axis interception point $(0\mathrm{\mid }-2)(0|-2)$. From there go 1 to the right and according to the slope $m=\frac{3}{4}m=\backslash frac34$ upwards (alternatively, also the quadruple length to avoid fractions: 4 to the right and 3 upwards). Connect the two points to form a straight line.

Setting up the function term

$m=0;t=3m=0;\backslash ;t=3$

Plug $mm$ and $tt$ into the general line equation $y=mx+ty=mx+t$ .

$\Rightarrow y=3\backslash Rightarrow\; y=3$

Drawing the line

The line runs through the $yy$-axis intercept $(0\mathrm{\mid }3)(0|3)$ parallel to the $xx$-axis because the line has a slope of 0.

Setting up the function term

Since the line is parallel to the $xx$-axis, so its slope is $0\Rightarrow m=00\; \backslash Rightarrow\; m=0$.

The line goes through the point $(-3\mathrm{\mid }-2)(-3\backslash vert-2)$. Since the slope is 0, the line has the $yy$-value $-2-2$ at $x=0x=0$.

Its $yy$-axis intercept is therefore $-2-2$

$m=0;t=-2m=0;\backslash ;t=-2$

Plug $mm$ and $tt$ into the general line equation $y=mx+ty=mx+t$ .

$y=-2y=-2$

Drawing the line

The line runs through the $yy$-axis interception point $(0\mathrm{\mid }-2)(0|-2)$ and is parallel to the x-axis.

Setting up the function term

This straight line cannot be described by a function equation. The gradient would be infinitely large.

However, we can still draw the straight line.

Drawing the line

The line runs through the point $(-4\mathrm{\mid }2)(-4\backslash vert2)$ and is parallel to the $yy$-axis.