Solution 2.3:8b
From Förberedande kurs i matematik 1
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{| align="center" | {| align="center" | ||
- | ||[[Image:2_3_8_b-1.gif|center]] | + | |align="center"|[[Image:2_3_8_b-1.gif|center]] |
|width="10px"| | |width="10px"| | ||
- | ||[[Image:2_3_8_b-2.gif|center]] | + | |align="center"|[[Image:2_3_8_b-2.gif|center]] |
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- | ||<small>The graph of ''f''(''x'') = ''x''² + 2</small> | + | |align="center"|<small>The graph of ''f''(''x'') = ''x''² + 2</small> |
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- | ||<small>The graph of ''f''(''x'') = (''x'' - 1)² + 2</small> | + | |align="center"|<small>The graph of ''f''(''x'') = (''x'' - 1)² + 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 |