Lösung 2.3:7b

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We rewrite the expression by completing the square:
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We rewrite the expression by completing the square,
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{{Displayed math||<math>\begin{align}
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-x^{2}+3x-4
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&= -\bigl(x^{2}-3x+4\bigr)\\[5pt]
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&= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\Bigl(\frac{3}{2}\Bigr)^{2}+4\Bigr)\\[5pt]
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&= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{9}{4}+\frac{16}{4}\Bigr)\\[5pt]
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&= -\Bigl(\Bigl(x-\frac{3}{2}\Bigr)^{2}+\frac{7}{4}\Bigr)\\[5pt]
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&= -\Bigl(x-\frac{3}{2}\Bigr)^{2}-\frac{7}{4}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Now, we see that the first term <math>-(x-\tfrac{3}{2})^{2}</math> is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is <math>-7/4</math> and that occurs when <math>x-\tfrac{3}{2}=0\,</math>, i.e. <math>x=\tfrac{3}{2}\,</math>.
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& -x^{2}+3x-4=-\left( x^{2}-3x+4 \right)=-\left( \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \right) \\
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& =-\left( \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4} \right)=-\left( \left( x-\frac{3}{2} \right)^{2}+\frac{7}{4} \right)=-\left( x-\frac{3}{2} \right)^{2}-\frac{7}{4} \\
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\end{align}</math>
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Now, we see that the first term
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<math>-\left( x-\frac{3}{2} \right)^{2}</math>
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is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is
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<math>-{7}/{4}\;</math>
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and that occurs when
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<math>x-\frac{3}{2}=0</math>, i.e.
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<math>x=\frac{3}{2}</math>.
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Version vom 11:55, 29. Sep. 2008

We rewrite the expression by completing the square,

Vorlage:Displayed math

Now, we see that the first term \displaystyle -(x-\tfrac{3}{2})^{2} is a quadratic with a minus sign in front, so that term is always less than or equal to zero. This means that the polynomial's largest value is \displaystyle -7/4 and that occurs when \displaystyle x-\tfrac{3}{2}=0\,, i.e. \displaystyle x=\tfrac{3}{2}\,.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:7b“