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anonymous

  • one year ago

can someone help me graph this please: which of the following could be the graph of the equation x^2 -6x + y^2 +2y +6 = 0

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  1. anonymous
    • one year ago
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    When you have a function like that then it should be some kind of a circle. Because the circles equations is: \[(x-a)^2+(y-b)^2=r^2\] In this equation a and b is the circles center coordinates and r is the length of the circle. And that function you have is the same as: \[(x-3)^2+(y+1)^2=2^2\] Therefore the graph should be a circle with center in (3,-1) with the length 2. That looks like this:

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  2. anonymous
    • one year ago
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    how did you convert x^2 -6x + y^2 +2y +6 = 0 into (x−3)^2+(y+1)^2=2^2?

  3. jdoe0001
    • one year ago
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    do you know what a "perfect trinomial square" is? sometimes called only a "perfect square"

  4. anonymous
    • one year ago
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    i have heard of it. I think the formula is (-b/2)^2

  5. anonymous
    • one year ago
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    i did not really understand the perfect square formula however

  6. jdoe0001
    • one year ago
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    \(\begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad \qquad\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow \end{array}\) does that ring a bell?

  7. anonymous
    • one year ago
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    i have never seen that equation before

  8. anonymous
    • one year ago
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    i dont understand how to to input into the equation: x^2 -6x + y^2 +2y +6 = 0

  9. jdoe0001
    • one year ago
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    hmmm, well, I happen to be using a and b how about \(\begin{array}{cccccllllll} {\color{brown}{ x}}^2& + &2{\color{brown}{ x}}{\color{blue}{ y}}&+&{\color{blue}{ y}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ x}}&& &&{\color{blue}{ y}}\\ &\to &({\color{brown}{ x}} + {\color{blue}{ y}})^2&\leftarrow \end{array}\qquad \qquad \qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ x}}^2& - &2{\color{brown}{ x}}{\color{blue}{ y}}&+&{\color{blue}{ y}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ x}}&& &&{\color{blue}{ y}}\\ &\to &({\color{brown}{ x}} - {\color{blue}{ y}})^2&\leftarrow \end{array}\)

  10. anonymous
    • one year ago
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    i am really confused. I dont understand how that would convert x^2 -6x + y^2 +2y +6 = 0 into (x−3)^2+(y+1)^2=2^2

  11. jdoe0001
    • one year ago
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    well... you may want to cover what a "perfect trinomial" is, before doing this exercise then the form is changed by using that, and "completing the square", to get a circle equation one could tell is a circle, because the tell-tale part is, the "x" and the "y" raised at the 2 exponent, both have the same coefficient

  12. anonymous
    • one year ago
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    okay, i will review how to do a perfect trinomial and complete the square to get a circle equation.

  13. anonymous
    • one year ago
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    Thank you for leading me one step closer to the solution

  14. IrishBoy123
    • one year ago
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    \[x^2 -6x + y^2 +2y +6 = 0\] just complete the square for x and y \[(x -3)^2 - 9 + (y +1)^2 - 1 + 6 = 0\] then you have to solve & draw :p

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