Lösung 2.3:9c

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{{NAVCONTENT_START}}
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To determine all the points on the curve
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<center> [[Image:2_3_9c.gif]] </center>
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<math>y=3x^{2}-12x+9</math>
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which also lie on the
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<math>x</math>
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-axis we substitute the equation of the
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<math>x</math>
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-axis i.e.
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<math>y=0</math>
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in the equation of the curve and obtain that
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<math>x</math>
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must satisfy
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<math>3x^{2}-12x+9=0</math>
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After dividing by
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<math>3</math>
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and completing the square the right-hand side is
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<math>x^{2}-4x+3=\left( x-2 \right)^{2}-2^{2}+3=\left( x-2 \right)^{2}-1</math>
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and thus the equation has solutions
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<math>x=2\pm 1,</math>
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i.e.
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<math>x=2-1=1</math>
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and
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<math>x=2+1=3.</math>
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The points where the curve cut the
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<math>x</math>
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-axis are
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<math>\left( 1 \right.,\left. 0 \right)</math>
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and
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<math>\left( 3 \right.,\left. 0 \right)</math>
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[[Image:2_3_9_c.gif|center]]
[[Image:2_3_9_c.gif|center]]

Version vom 12:11, 21. Sep. 2008

To determine all the points on the curve \displaystyle y=3x^{2}-12x+9 which also lie on the \displaystyle x -axis we substitute the equation of the \displaystyle x -axis i.e. \displaystyle y=0 in the equation of the curve and obtain that \displaystyle x must satisfy


\displaystyle 3x^{2}-12x+9=0


After dividing by \displaystyle 3 and completing the square the right-hand side is


\displaystyle x^{2}-4x+3=\left( x-2 \right)^{2}-2^{2}+3=\left( x-2 \right)^{2}-1


and thus the equation has solutions


\displaystyle x=2\pm 1, i.e. \displaystyle x=2-1=1 and \displaystyle x=2+1=3.


The points where the curve cut the \displaystyle x -axis are


\displaystyle \left( 1 \right.,\left. 0 \right) and \displaystyle \left( 3 \right.,\left. 0 \right)


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