Lösung 4.1:7b

Aus Online Mathematik Brückenkurs 1

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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 ''y''²- and ''y''-terms into a quadratic term by completing the square
The equation is almost in the standard form for a circle; all that is needed is for us to collect together the ''y''²- and ''y''-terms into a quadratic term by completing the square
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{{Displayed math||<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}}
After rewriting, the equation is
After rewriting, the equation is
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{{Displayed math||<math>x^2 + (y+2)^2 = 4</math>}}
+
{{Abgesetzte Formel||<math>x^2 + (y+2)^2 = 4</math>}}
and we see that the equation describes a circle having its centre at (0,-2) and radius <math>\sqrt{4}=2\,</math>.
and we see that the equation describes a circle having its centre at (0,-2) and radius <math>\sqrt{4}=2\,</math>.

Version vom 08:49, 22. Okt. 2008

The equation is almost in the standard form for a circle; all that is needed is for us to collect together the y²- and y-terms into a quadratic term by completing the square

\displaystyle y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}

After rewriting, the equation is

\displaystyle x^2 + (y+2)^2 = 4

and we see that the equation describes a circle having its centre at (0,-2) and radius \displaystyle \sqrt{4}=2\,.


Image:4_1_7_b.gif
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:7b“