Solution 4.1:7b

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Current revision (11:23, 8 October 2008) (edit) (undo)
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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
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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
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<math>y^{\text{2}}</math>
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<math>y</math>
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-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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{{Displayed math||<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>}}
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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 (0,-2) and radius <math>\sqrt{4}=2\,</math>.
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and we see that the equation describes a circle having its centre at
 
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<math>\left( 0 \right.,\left. -2 \right)</math>
 
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and radius
 
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<math>\sqrt{4}=2</math>.
 
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{{NAVCONTENT_START}}
 
<center> [[Image:4_1_7_b.gif]] </center>
<center> [[Image:4_1_7_b.gif]] </center>
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{{NAVCONTENT_STOP}}
 

Current revision

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
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