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Solution 2.3:9a

From Förberedande kurs i matematik 1

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m (Lösning 2.3:9a moved to Solution 2.3:9a: Robot: moved page)
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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]]

Revision as of 11:43, 21 September 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 y=x2−1 where y=0, i.e. all points which satisfy the equation


0=x2−1


This equation has solutions x=1, which means that the points of intersection are −10  and 10 .


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