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Math Help - Sketch the graph of the following:

  1. #1
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    Sketch the graph of the following:

    2x^2 + 2y^2 + x + y = 0



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  2. #2
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    Your question is not at all clear. You could "sketch" the graph by calculating several points: if x= 0, 2y^2+ y= y(2y+1)= 0 so y= 0 or -1/2. (0, 0) and (0, -\frac{1}{2}) are points on the graph. Taking y= 0, we get, in exactly the same way, (0,0) which we had before, (0,\frac{1}{2}). Get enough points that way and you can sketch the graph.

    But I suspect that you are intended to complete the square to see what kind of figure this is:
    If you divide through by 2 you get x^2+ \frac{1}{2}x+ y^2+ \frac{1}{2}y= 0. Presumably you know that (x+ a)^2= x^2+ 2ax+ a^2. Comparing that to x^2+ \frac{1}{2}x, they match if a= 1/4 and we can make this a "perfect square by adding \left(1/4\right)^2= 1/16. Doing that for both x and y, x^2+ \frac{1}{2}x+ \frac{1}{4}+ y^2+ \frac{1}{2}y+ \frac{1}{4}= \frac{1}{4}+ \frac{1}{4} or (x+\frac{1}{2})^2+ (y+\frac{1}{2})^2= \frac{1}{2}. What kind of figure is that?
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  3. #3
    Junior Member Rachel.F's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Your question is not at all clear. You could "sketch" the graph by calculating several points: if x= 0, 2y^2+ y= y(2y+1)= 0 so y= 0 or -1/2. (0, 0) and (0, -\frac{1}{2}) are points on the graph. Taking y= 0, we get, in exactly the same way, (0,0) which we had before, (0,\frac{1}{2}). Get enough points that way and you can sketch the graph.

    But I suspect that you are intended to complete the square to see what kind of figure this is:
    If you divide through by 2 you get x^2+ \frac{1}{2}x+ y^2+ \frac{1}{2}y= 0. Presumably you know that (x+ a)^2= x^2+ 2ax+ a^2. Comparing that to x^2+ \frac{1}{2}x, they match if a= 1/4 and we can make this a "perfect square by adding \left(1/4\right)^2= 1/16. Doing that for both x and y, x^2+ \frac{1}{2}x+ \frac{1}{4}+ y^2+ \frac{1}{2}y+ \frac{1}{4}= \frac{1}{4}+ \frac{1}{4} or (x+\frac{1}{2})^2+ (y+\frac{1}{2})^2= \frac{1}{2}. What kind of figure is that?

    Why didn't you put the squares on 1/4?
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  4. #4
    Senior Member Stroodle's Avatar
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    2x^2+2y^2+x+y=0

    2(x^2+\frac{1}{2}x)+2(y^2+\frac{1}{2}y)=0

    2(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{1}{16})+2(y^2+\frac{1}{2}y+\frac{1}{16}-\frac{1}{16})=0

    2[(x+\frac{1}{4})^2-\frac{1}{16}]+2[(y+\frac{1}{4})^2-\frac{1}{16}]=0

    2(x+\frac{1}{4})^2-\frac{1}{8}+2(y+\frac{1}{4})^2-\frac{1}{8}=0

    2(x+\frac{1}{4})^2+2(y+\frac{1}{4})^2=\frac{1}{4}

    (x+\frac{1}{4})^2+(y+\frac{1}{4})^2=\frac{1}{8}

    (x+\frac{1}{4})^2+(y+\frac{1}{4})^2=\left (\frac{1}{2\sqrt 2}\right )^2

    Which is a circle with it's centre at \left ( -\frac {1}{4}, -\frac {1}{4}\right ) with a radius of \frac{1}{2\sqrt 2}
    Last edited by Stroodle; June 17th 2009 at 07:29 PM.
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