Lösung 4.1:7b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K (Lösning 4.1:7b moved to Solution 4.1:7b: Robot: moved page) |
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| + | The equation is almost in the standard form for a circle; all that is needed is for us to collect together the | ||
| + | <math>y^{\text{2}}</math> | ||
| + | - and | ||
| + | <math>y</math> | ||
| + | -terms into a quadratic term by completing the square | ||
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| + | <math>y^{2}+4y=\left( y+2 \right)^{2}-2^{2}</math> | ||
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| + | After rewriting, the equation is | ||
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| + | <math>x^{2}+\left( y+2 \right)^{2}=4</math> | ||
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| + | and we see that the equation describes a circle having its centre at | ||
| + | <math>\left( 0 \right.,\left. -2 \right)</math> | ||
| + | and radius | ||
| + | <math>\sqrt{4}=2</math>. | ||
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<center> [[Image:4_1_7_b.gif]] </center> | <center> [[Image:4_1_7_b.gif]] </center> | ||
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Version vom 12:00, 27. Sep. 2008
The equation is almost in the standard form for a circle; all that is needed is for us to collect together the \displaystyle y^{\text{2}} - and \displaystyle y -terms into a quadratic term by completing the square
\displaystyle y^{2}+4y=\left( y+2 \right)^{2}-2^{2}
After rewriting, the equation is
\displaystyle x^{2}+\left( y+2 \right)^{2}=4
and we see that the equation describes a circle having its centre at
\displaystyle \left( 0 \right.,\left. -2 \right)
and radius
\displaystyle \sqrt{4}=2.
