Lösung 2.3:9a

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{{NAVCONTENT_START}}
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A point lies on the
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<center> [[Image:2_3_9a.gif]] </center>
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<math>x</math>
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{{NAVCONTENT_STOP}}
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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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<math>0=x^{\text{2}}-\text{1}</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 11:43, 21. Sep. 2008

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


\displaystyle 0=x^{\text{2}}-\text{1}


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


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