Exercises: Linear functions, zeros, axis intercepts

Intersection with the $x$-axis

Calculate the intersection of the line with the $x$-axis.

Set the expression for the line equation equal to 0 and solve for $x$.

$\begin{array}{rcll}0& =& -2x+3.5& |+2x\\ 2x& =& 3.5& |:2\\ x& =& 1.75& \end{array}$

$\Rightarrow$ The line intersects the $x$-axis at $S_x(1.75|0)$.

Intersection with the $y$-axis

Calculate the intersection of the line with the $y$-axis.

To do this, plug the value $x=0$ into the expression for the straight line equation.

$y=-2\cdot 0+3.5$

$y=3.5$

$\Rightarrow$ The line intersects the $y$-axis at $S_y(0|3.5)$.

Intersection with the $x$-axis

Calculate the intersection of the line with the $x$-axis.

Set the expression for the line equation equal to 0 and solve for $x$.

$\begin{array}{rcll}0& =& 5x-7& |+7\\ 5x& =& 7& |:5\\ x& =& \frac{7}{5}=1.4& \end{array}$

$\Rightarrow$ The line intersects the $x$-axis at $S_x(1.4|0)$.

Intersection with the $y$-axis

Calculate the intersection of the line with the $y$-axis.

To do this, plug the value $x=0$ into the expression for the straight line equation.

$y=5\cdot 0-7$

$y=-7$

$\Rightarrow$ The line intersects the $y$-axis at $S_y(0|-7)$.

Intersection with the $x$-axis

Calculate the intersection of the line with the $x$-axis.

Set the expression for the line equation equal to 0 and solve for $x$.

$\begin{array}{rcll}0& =& \frac{3}{2}x+2& |-2\\ \frac{3}{2}x& =& -2& |\cdot \frac{2}{3}\\ x& =& -\frac{4}{3}& \end{array}$

$\Rightarrow$ The line intersects the $x$-axis at $S_x(\text{}-\frac{4}{3}-|0)$.

Intersection with the $y$-axis

Calculate the intersection of the line with the $y$-axis.

To do this, plug the value $x=0$ into the expression for the straight line equation.

$y=\frac{3}{2}\cdot 0+2$

$y=2$

$\Rightarrow$ The line intersects the $y$-axis at $S_y(0|2)$.

Intersection with the $x$-axis

Calculate the intersection of the line with the $x$-axis.

Set the expression for the line equation equal to 0 and solve for $x$.

$\begin{array}{rcll}0& =& -\frac{2}{5}x+\frac{5}{2}& |+\frac{2}{5}x\\ \frac{2}{5}x& =& \frac{5}{2}& |\cdot \frac{5}{2}\\ x& =& \frac{25}{4}=6.25& \end{array}$

$\Rightarrow$ The line intersects the $x$-axis at $S_x(6.25|0)$.

Intersection with the $y$-axis

Calculate the intersection of the line with the $y$-axis.

To do this, plug the value $x=0$ into the expression for the straight line equation.

$y=-\frac{2}{5}\cdot 0+\frac{5}{2}$

$y=\frac{5}{2}=2.5$

$\Rightarrow$ The line intersects the $y$-axis at $S_y(0|2.5)$.

To obtain a general line equation $y=m\cdot x+t$, multiply out the bracket

$y=2(x-\frac{2}{3})$

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

Intersection with the $x$-axis

Calculate the intersection of the line with the $x$-axis.

Set the expression for the line equation equal to 0 and solve for $x$.

$\begin{array}{rcll}0& =& 2x-\frac{4}{3}& |+\frac{4}{3}\\ 2x& =& \frac{4}{3}& |:2\\ x& =& \frac{2}{3}& \end{array}$

$\Rightarrow$ The line intersects the $x$-axis at $S_x(\frac{2}{3}|0)$.

Intersection with the $y$-axis

Calculate the intersection of the line with the $y$-axis.

To do this, plug the value $x=0$ into the expression for the straight line equation.

$y=2\cdot 0-\frac{4}{3}$

$y=-\frac{4}{3}$

$\Rightarrow$ The line intersects the $y$-axis at $S_y(0|-\frac{4}{3})$.

Transform the equation

To obtain a general line equation $y=m\cdot x+t$, swap both elements on the right side.

$y=-\frac{4}{3}-\frac{1}{2}x$

$\phantom{y}=-\frac{1}{2}x-\frac{4}{3}$

Intersection with the $x$-axis

Calculate the intersection of the line with the $x$-axis.

Set the expression for the line equation equal to 0 and solve for $x$.

$\begin{array}{rcll}0& =& -\frac{4}{3}-\frac{1}{2}x& |+\frac{1}{2}x\\ \frac{1}{2}x& =& -\frac{4}{3}& |\cdot 2\\ x& =& -\frac{8}{3}=-2\frac{2}{3}& \end{array}$

$\Rightarrow$ The line intersects the $x$-axis at $S_x(-\frac{8}{3}|0)$.

Intersection with the $y$-axis

Calculate the intersection of the line with the $y$-axis.

To do this, plug the value $x=0$ into the expression for the straight line equation.

$y=-\frac{1}{2}\cdot 0-\frac{4}{3}$

$y=-\frac{4}{3}$

$\Rightarrow$ The line intersects the $y$-axis at $S_y(0|-\frac{4}{3})$.