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Thread: 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!

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

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

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

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

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

    $\displaystyle 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} $

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

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


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


    There are no $\displaystyle x$-intercepts.

    There are no $\displaystyle y$-intercepts.

    A circle does not have vertices.

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

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

    $\displaystyle \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 $\displaystyle x^2- 6x+ y^2+ 4y= 3$.

    If $\displaystyle x^2- 6x+ y^2+ 4x= 3$ really is the problem, -6x+ 4x= -2x so this is the same as $\displaystyle x^2- 2x+ y^2= 3$. 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""$\displaystyle y^2= (y- 0)^2$. 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: .$\displaystyle 4x^2 + 20x + 4y^2 - 16y \;=\;-37$

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

    $\displaystyle 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} $

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

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


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


    There are no $\displaystyle x$-intercepts.

    There are no $\displaystyle y$-intercepts.

    A circle does not have vertices.

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

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

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

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