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.

$f(x)\;=\;2x-5$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Lines in coordinate systems
$f (x) = 2 x - 5$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow\;{P}_y\;\left(0\;\left|\;-5\right.\right)$
$\Rightarrow\;m_f=2$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\displaystyle 2x-5$ $=$ $\displaystyle 0$ $\displaystyle +5$
$\displaystyle 2x$ $=$ $\displaystyle 5$ $\displaystyle \colon 2$
$\displaystyle x_0$ $=$ $\displaystyle 2.5$
$\Rightarrow\;{P}_x\;\left(2{.}5\;\left|\;0\right.\right)$
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$f (x) = - x - 3$

Für diese Aufgabe benötigst Du folgendes Grundwissen: Lines in coordinate systems
$f (x) = - x - 3$
First read off the $y$ -axis intercept and the slope from the function equation.
$\Rightarrow\;{P}_y\;\left(0\;\left|\;-3\right.\right)$
$\Rightarrow\;m_f=-1$
Calculate the intersection point with the $x$ -axis. This is done by setting the function term equal to 0.
$\displaystyle -x-3$ $=$ $\displaystyle 0$ $\displaystyle +x$
$\displaystyle -3$ $=$ $\displaystyle x_0$
$\Rightarrow\;{P}_x\;\left(-3\;\left|\;0\right.\right)$
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$f\left(x\right)=\frac{1}{2}x+1$

$f\left(x\right)=-\frac{1}{2}x-2$

$f\left(x\right)=\frac{1}{3}x-\frac{1}{2}$

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

$f\left(x\right)=\frac{2}{3}x+2$

$f\left(x\right)=-\frac{3}{4}x-1$

$f\left(x\right)=-3x+\frac{5}{10}$

$f\left(x\right)=\frac{5}{7}x-\frac{12}{4}$

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$\displaystyle \frac{1}{2}x+1$	$=$	$\displaystyle 0$	$\displaystyle -1$
$\displaystyle \frac{1}{2}x$	$=$	$\displaystyle -1$	$\displaystyle \cdot2$
$\displaystyle x_0$	$=$	$\displaystyle -2$

$\displaystyle -\frac{1}{2}x-2$	$=$	$\displaystyle 0$	$\displaystyle +2$
$\displaystyle -\frac{1}{2}x$	$=$	$\displaystyle 2$	$\displaystyle \cdot(-2)$
$\displaystyle x_0$	$=$	$\displaystyle -4$

$\displaystyle \frac{1}{3}x-\frac{1}{2}$	$=$	$\displaystyle 0$	$\displaystyle +\frac{1}{2}$
$\displaystyle \frac{1}{3}x$	$=$	$\displaystyle \frac{1}{2}$	$\displaystyle \cdot3$
$\displaystyle x_0$	$=$	$\displaystyle \frac{3}{2}$

$\displaystyle -\frac{1}{4}x+\frac{3}{2}$	$=$	$\displaystyle 0$	$\displaystyle -\frac{3}{2}$
$\displaystyle -\frac{1}{4}x$	$=$	$\displaystyle -\frac{3}{2}$	$\displaystyle \cdot(-4)$
$\displaystyle x_0$	$=$	$\displaystyle 6$

$\displaystyle -\frac{3}{4}x-1$	$=$	$\displaystyle 0$	$\displaystyle +1$
$\displaystyle -\frac{3}{4}x$	$=$	$\displaystyle 1$	$\displaystyle \colon (-\frac34)$
$\displaystyle x_0$	$=$	$\displaystyle -\frac{4}{3}$

$\displaystyle 2x-5$	$=$	$\displaystyle 0$	$\displaystyle +5$
$\displaystyle 2x$	$=$	$\displaystyle 5$	$\displaystyle \colon 2$
$\displaystyle x_0$	$=$	$\displaystyle 2.5$

$\displaystyle -x-3$	$=$	$\displaystyle 0$	$\displaystyle +x$
$\displaystyle -3$	$=$	$\displaystyle x_0$

$\displaystyle \frac{2}{3}x+2$	$=$	$\displaystyle 0$	$\displaystyle -2$
$\displaystyle \frac{2}{3}x$	$=$	$\displaystyle -2$	$\displaystyle \colon\frac23$
$\displaystyle x_0$	$=$	$\displaystyle -3$

$\displaystyle -3x+\frac{5}{10}$	$=$	$\displaystyle 0$	$\displaystyle -\frac{5}{10}$
$\displaystyle -3x$	$=$	$\displaystyle -\frac{1}{2}$	$\displaystyle \colon (-3)$
$\displaystyle x_0$	$=$	$\displaystyle -\frac{1}{6}$

$\displaystyle \frac{5}{7}x-3$	$=$	$\displaystyle 0$	$\displaystyle +3$
$\displaystyle \frac{5}{7}x$	$=$	$\displaystyle 3$	$\displaystyle \colon \frac57$
$\displaystyle x_0$	$=$	$\displaystyle \frac{21}{5}$

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