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Math Help - Graphing Conics

  1. #1
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    Graphing Conics

    4x^2 + 4y^2 +20x -16y + 37 =0
    I completed the square and got:
    (2x+10)^2 + (2y-8)^2 =127

    x int ?
    y int ?
    Vertices ?
    Center (-5, 4)
    Domain ?
    Range ?

    I tried making y=0 to find the x intercepts but didn't work out. Same thing for y intercepts
    Completely lost....
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  2. #2
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    Hello, Honorable24!

    4x^2 + 4y^2 +20x -16y + 37 \:=\:0

    I completed the square and got: . (2x+10)^2 + (2y-8)^2 \:=\:127 .??

    Find: . x\text{-intercepts,}\quad y\text{-intercepts,}\quad \text{Vertices,}\quad \text{Center,}\quad \text{Domain,}\quad \text{Range}

    We have: . 4x^2 + 20x + 4y^2 - 16y \;=\;-37

    . . . . 4(x^2 + 5x \quad) + 4(y^2 - 4y \quad) \;=\;-37

    4\left(x^2 + 5x + {\color{blue}\tfrac{25}{4}}\right)^2 + 4\left(y^2 - 4y + {\color{red}4}\right) \;=\;-37 + {\color{blue}25} + {\color{red}16}

    . . . . . . . . 4\left(x + \tfrac{5}{2}\right)^2 + 4(y-2)^2 \;=\;4

    . . . . . . . . . \left(x + \tfrac{5}{2}\right)^2 + (y-2)^2 \;=\;1


    This is a circle: . \text{Center: }\left(-\tfrac{5}{2},\:2\right),\;\text{ radius } 2


    There are no x-intercepts.

    There are no y-intercepts.

    A circle does not have vertices.

    \text{Center: }\:\left(-\tfrac{5}{2},\:2\right)

    \text{Domain: }\;x \in \left[-\tfrac{7}{2},\:-\tfrac{3}{2}\right]

    \text{Range: }\;y \in[1,\:3]

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  3. #3
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    x^2-6x + y^2 +4x=3

    How do you do completing the square for this
    I don't know how to do these types becuase there is no single y term
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  4. #4
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    Quote Originally Posted by Honorable24 View Post
    Thanks

    x^2-6x + y^2 +4x=3

    How do you do completing the square for this
    I don't know how to do these types becuase there is no single y term
    I recommend you go back and check the problem. I suspect it should be x^2- 6x+ y^2+ 4y= 3.

    If x^2- 6x+ y^2+ 4x= 3 really is the problem, -6x+ 4x= -2x so this is the same as x^2- 2x+ y^2= 3. The fact that there is no "y to the first power" term is not a problem. It only means that y is already a "perfect square"" y^2= (y- 0)^2. Complete the square in x.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Honorable24!


    We have: . 4x^2 + 20x + 4y^2 - 16y \;=\;-37

    . . . . 4(x^2 + 5x \quad) + 4(y^2 - 4y \quad) \;=\;-37

    4\left(x^2 + 5x + {\color{blue}\tfrac{25}{4}}\right)^2 + 4\left(y^2 - 4y + {\color{red}4}\right) \;=\;-37 + {\color{blue}25} + {\color{red}16}

    . . . . . . . . 4\left(x + \tfrac{5}{2}\right)^2 + 4(y-2)^2 \;=\;4

    . . . . . . . . . \left(x + \tfrac{5}{2}\right)^2 + (y-2)^2 \;=\;1


    This is a circle: . \text{Center: }\left(-\tfrac{5}{2},\:2\right),\;\text{ radius } 1 <<<<<<<< typo


    There are no x-intercepts.

    There are no y-intercepts.

    A circle does not have vertices.

    \text{Center: }\:\left(-\tfrac{5}{2},\:2\right)

    \text{Domain: }\;x \in \left[-\tfrac{7}{2},\:-\tfrac{3}{2}\right]

    \text{Range: }\;y \in[1,\:3]

    ... I don't want to pick at you but there is a tiny typo.
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