Solution 2.3:8b

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Current revision (12:57, 29 September 2008) (edit) (undo)
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{| align="center"
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|align="center"|[[Image:2_3_8_b-1.gif|center]]
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||[[Image:2_3_8_b-2.gif|center]]
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|align="center"|[[Image:2_3_8_b-2.gif|center]]
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||<small>The graph of ''f''(''x'')&nbsp;=&nbsp;''x''²&nbsp;+&nbsp;2</small>
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|align="center"|<small>The graph of ''f''(''x'')&nbsp;=&nbsp;''x''²&nbsp;+&nbsp;2</small>
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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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|align="center"|<small>The graph of ''f''(''x'')&nbsp;=&nbsp;(''x''&nbsp;-&nbsp;1)²&nbsp;+&nbsp;2</small>
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Current revision

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
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