Exercises: Linear functions, zeros, axis intercepts

Draw the graphs of the following lines including the point of intersection with the $y$ -axis and a gradient triangle. Calculate the point of intersection with the $x$ -axis and check the result using the graph.

$f (x) = 2 x - 5$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = 2 x - 5$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 5)$
$\Rightarrow m_{f} = 2$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$2 x - 5$ $=$ $0$ $+ 5$
$2 x$ $=$ $5$ $: 2$
$x_{0}$ $=$ $2.5$
$\Rightarrow P_{x} (2.5 | 0)$
$f (x) = - x - 3$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - x - 3$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 3)$
$\Rightarrow m_{f} = - 1$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$- x - 3$ $=$ $0$ $+ x$
$- 3$ $=$ $x_{0}$
$\Rightarrow P_{x} (- 3 | 0)$
$f (x) = \frac{1}{2} x + 1$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{1}{2} x + 1$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | 1)$
$\Rightarrow m_{f} = \frac{1}{2}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\frac{1}{2} x + 1$ $=$ $0$ $- 1$
$\frac{1}{2} x$ $=$ $- 1$ $\cdot 2$
$x_{0}$ $=$ $- 2$
$\Rightarrow P_{x} (- 2 | 0)$
$f (x) = - \frac{1}{2} x - 2$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - \frac{1}{2} x - 2$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 2)$
$\Rightarrow m_{f} = - \frac{1}{2}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$- \frac{1}{2} x - 2$ $=$ $0$ $+ 2$
$- \frac{1}{2} x$ $=$ $2$ $\cdot (- 2)$
$x_{0}$ $=$ $- 4$
$\Rightarrow P_{x} (- 4 | 0)$
$f (x) = \frac{1}{3} x - \frac{1}{2}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{1}{3} x - \frac{1}{2}$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - \frac{1}{2})$
$\Rightarrow m_{f} = \frac{1}{3}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\frac{1}{3} x - \frac{1}{2}$ $=$ $0$ $+ \frac{1}{2}$
$\frac{1}{3} x$ $=$ $\frac{1}{2}$ $\cdot 3$
$x_{0}$ $=$ $\frac{3}{2}$
$\Rightarrow P_{x} (\frac{3}{2} | 0)$
$f (x) = - \frac{1}{4} x + \frac{3}{2}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - \frac{1}{4} x + \frac{3}{2}$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | \frac{3}{2})$
$\Rightarrow m_{f} = - \frac{1}{4}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$- \frac{1}{4} x + \frac{3}{2}$ $=$ $0$ $- \frac{3}{2}$
$- \frac{1}{4} x$ $=$ $- \frac{3}{2}$ $\cdot (- 4)$
$x_{0}$ $=$ $6$
$\Rightarrow P_{x} (6 | 0)$
$f (x) = \frac{2}{3} x + 2$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{2}{3} x + 2$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | 2)$
$\Rightarrow m_{f} = \frac{2}{3}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\frac{2}{3} x + 2$ $=$ $0$ $- 2$
$\frac{2}{3} x$ $=$ $- 2$ $: \frac{2}{3}$
$x_{0}$ $=$ $- 3$
$\Rightarrow P_{x} (- 3 | 0)$
$f (x) = - \frac{3}{4} x - 1$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - \frac{3}{4} x - 1$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 1)$
$\Rightarrow m_{f} = - \frac{3}{4}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$- \frac{3}{4} x - 1$ $=$ $0$ $+ 1$
$- \frac{3}{4} x$ $=$ $1$ $: (- \frac{3}{4})$
$x_{0}$ $=$ $- \frac{4}{3}$
$\Rightarrow P_{x} (- \frac{4}{3} | 0)$
$f (x) = - 3 x + \frac{5}{10}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = - 3 x + \frac{5}{10}$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | \frac{5}{10})$
$\Rightarrow m_{f} = - 3$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$- 3 x + \frac{5}{10}$ $=$ $0$ $- \frac{5}{10}$
$- 3 x$ $=$ $- \frac{1}{2}$ $: (- 3)$
$x_{0}$ $=$ $- \frac{1}{6}$
$\Rightarrow P_{x} (\frac{1}{6} | 0)$
$f (x) = \frac{5}{7} x - \frac{12}{4}$

For this task you need the following basic knowledge: Lines in coordinate systems
$f (x) = \frac{5}{7} x - \frac{12}{4} = \frac{5}{7} x - 3$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow P_{y} (0 | - 3)$
$\Rightarrow m_{f} = \frac{5}{7}$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\frac{5}{7} x - 3$ $=$ $0$ $+ 3$
$\frac{5}{7} x$ $=$ $3$ $: \frac{5}{7}$
$x_{0}$ $=$ $\frac{21}{5}$
$\Rightarrow P_{x} (\frac{21}{5} | 0)$

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$\frac{1}{2} x + 1$	$=$	$0$	$- 1$
$\frac{1}{2} x$	$=$	$- 1$	$\cdot 2$
$x_{0}$	$=$	$- 2$

$- \frac{1}{2} x - 2$	$=$	$0$	$+ 2$
$- \frac{1}{2} x$	$=$	$2$	$\cdot (- 2)$
$x_{0}$	$=$	$- 4$

$\frac{1}{3} x - \frac{1}{2}$	$=$	$0$	$+ \frac{1}{2}$
$\frac{1}{3} x$	$=$	$\frac{1}{2}$	$\cdot 3$
$x_{0}$	$=$	$\frac{3}{2}$

$- \frac{1}{4} x + \frac{3}{2}$	$=$	$0$	$- \frac{3}{2}$
$- \frac{1}{4} x$	$=$	$- \frac{3}{2}$	$\cdot (- 4)$
$x_{0}$	$=$	$6$

$\frac{2}{3} x + 2$	$=$	$0$	$- 2$
$\frac{2}{3} x$	$=$	$- 2$	$: \frac{2}{3}$
$x_{0}$	$=$	$- 3$

$- \frac{3}{4} x - 1$	$=$	$0$	$+ 1$
$- \frac{3}{4} x$	$=$	$1$	$: (- \frac{3}{4})$
$x_{0}$	$=$	$- \frac{4}{3}$

$- 3 x + \frac{5}{10}$	$=$	$0$	$- \frac{5}{10}$
$- 3 x$	$=$	$- \frac{1}{2}$	$: (- 3)$
$x_{0}$	$=$	$- \frac{1}{6}$

$\frac{5}{7} x - 3$	$=$	$0$	$+ 3$
$\frac{5}{7} x$	$=$	$3$	$: \frac{5}{7}$
$x_{0}$	$=$	$\frac{21}{5}$