Lösung 2.3:8b

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As a starting point, we can take the curve
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As a starting point, we can take the curve <math>y=x^{2}+2</math> which is a parabola with a minimum at (0,2) and is sketched further down. Compared with that curve,
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<math>y=x^{2}+2</math>
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<math>y = (x-1)^{2}+2</math> is the same curve in which we must consistently choose ''x'' to be one unit greater in order to get the same ''y''-value. The curve <math>y = (x-1)^{2}+2</math> is thus shifted one unit to the right compared with <math>y=x^{2}+2</math>.
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which is a parabola with a minimum at
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<math>\left( 0 \right.,\left. 2 \right)</math>
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and is sketched further down. Compared with that curve,
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<math>y=\left( x-1 \right)^{2}+2</math>
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is the same curve in which we must consistently choose
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<math>x</math>
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to be one unit greater in order to get the same
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<math>y</math>
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-value. The curve
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<math>y=\left( x-1 \right)^{2}+2</math>
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is thus shifted one unit to the right compared with
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<math>y=x^{2}+2</math>.
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[[Image:2_3_8_b.gif|center]]
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{| align="center"
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||[[Image:2_3_8_b-1.gif|center]]
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|width="10px"|&nbsp;
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||[[Image:2_3_8_b-2.gif|center]]
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|-
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||<small>The graph of ''f''(''x'')&nbsp;=&nbsp;''x''²&nbsp;+&nbsp;2</small>
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||
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||<small>The graph of ''f''(''x'')&nbsp;=&nbsp;(''x''&nbsp;-&nbsp;1)²&nbsp;+&nbsp;2</small>
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|}

Version vom 12:56, 29. Sep. 2008

As a starting point, we can take the curve \displaystyle y=x^{2}+2 which is a parabola with a minimum at (0,2) and is sketched further down. Compared with that curve, \displaystyle y = (x-1)^{2}+2 is the same curve in which we must consistently choose x to be one unit greater in order to get the same y-value. The curve \displaystyle y = (x-1)^{2}+2 is thus shifted one unit to the right compared with \displaystyle y=x^{2}+2.


 
The graph of f(x) = x² + 2 The graph of f(x) = (x - 1)² + 2
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:8b“