Lösung 2.3:9a

Aus Online Mathematik Brückenkurs 1

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A point lies on the
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A point lies on the ''x''-axis if it has ''y''-coordinate 0 and we therefore look for all the points on the curve <math>y=x^{2}-1</math> where <math>y=0</math>, i.e. all points which satisfy the equation
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<math>x</math>
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-axis if it has
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<math>y</math>
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-coordinate
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<math>0</math>
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and we therefore look for all the points on the curve
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<math>y=x^{\text{2}}-\text{1}</math>
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where
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<math>y=0</math>, i.e. all points which satisfy the equation
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{{Displayed math||<math>0=x^{2}-1\,\textrm{.}</math>}}
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<math>0=x^{\text{2}}-\text{1}</math>
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This equation has solutions <math>x=\pm 1</math>, which means that the points of intersection are <math>(-1,0)</math> and <math>(1,0)</math>.
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This equation has solutions
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<math>x=\pm \text{1}</math>, which means that the points of intersection are
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<math>\left( -1 \right.,\left. 0 \right)</math>
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and
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<math>\left( 1 \right.,\left. 0 \right)</math>.
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[[Image:2_3_9_a.gif|center]]
[[Image:2_3_9_a.gif|center]]

Version vom 13:14, 29. Sep. 2008

A point lies on the x-axis if it has y-coordinate 0 and we therefore look for all the points on the curve \displaystyle y=x^{2}-1 where \displaystyle y=0, i.e. all points which satisfy the equation

Vorlage:Displayed math

This equation has solutions \displaystyle x=\pm 1, which means that the points of intersection are \displaystyle (-1,0) and \displaystyle (1,0).


Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:9a“