Lösung 2.3:9b

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The points of intersection are those points on the curve which also lie on the
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<center> [[Image:2_3_9b-1(2).gif]] </center>
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<math>~x</math>
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-axis, i.e. they are those points which satisfy both the equation of the curve
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<math>y=x^{\text{2}}-\text{5}x+\text{6}</math>
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and the equation of the
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<math>~x</math>
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-axis
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<math>y=0</math>,
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<math>\left\{ \begin{matrix}
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y=x^{\text{2}}-\text{5}x+\text{6} \\
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y=0\quad \quad \quad \quad \\
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\end{matrix} \right.</math>
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This system of equations gives directly that
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<math>y=0</math>
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and that
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<math>~x</math>
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must satisfy the second-order equation
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<math>x^{\text{2}}-\text{5}x+\text{6}=0</math>
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. By completing the square, we obtain that the left-hand side is
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<math>\begin{align}
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& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
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& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
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\end{align}</math>
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and this gives that the equation has solutions
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<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.
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<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>
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and
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<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.
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The intersection points are therefore
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<math>\left( 2 \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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Version vom 11:57, 21. Sep. 2008

The points of intersection are those points on the curve which also lie on the \displaystyle ~x -axis, i.e. they are those points which satisfy both the equation of the curve \displaystyle y=x^{\text{2}}-\text{5}x+\text{6} and the equation of the \displaystyle ~x -axis \displaystyle y=0,


\displaystyle \left\{ \begin{matrix} y=x^{\text{2}}-\text{5}x+\text{6} \\ y=0\quad \quad \quad \quad \\ \end{matrix} \right.


This system of equations gives directly that \displaystyle y=0 and that \displaystyle ~x must satisfy the second-order equation \displaystyle x^{\text{2}}-\text{5}x+\text{6}=0 . By completing the square, we obtain that the left-hand side is


\displaystyle \begin{align} & x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\ & =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\ \end{align}


and this gives that the equation has solutions

\displaystyle x=\frac{5}{2}\pm \frac{1}{2}, i.e. \displaystyle x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2 and \displaystyle x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3.

The intersection points are therefore \displaystyle \left( 2 \right.,\left. 0 \right) and \displaystyle \left( 3 \right.,\left. 0 \right).


Image:2_3_9b-2(2).gif
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:9b“