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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When you have a function like that then it should be some kind of a circle. Because the circles equations is: \[(xa)^2+(yb)^2=r^2\] 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: \[(x3)^2+(y+1)^2=2^2\] Therefore the graph should be a circle with center in (3,1) with the length 2. That looks like this:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you convert x^2 6x + y^2 +2y +6 = 0 into (x−3)^2+(y+1)^2=2^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know what a "perfect trinomial square" is? sometimes called only a "perfect square"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have heard of it. I think the formula is (b/2)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i did not really understand the perfect square formula however

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad \qquad\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2&  &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}}  {\color{blue}{ b}})^2&\leftarrow \end{array}\) does that ring a bell?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have never seen that equation before

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand how to to input into the equation: x^2 6x + y^2 +2y +6 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm, well, I happen to be using a and b how about \(\begin{array}{cccccllllll} {\color{brown}{ x}}^2& + &2{\color{brown}{ x}}{\color{blue}{ y}}&+&{\color{blue}{ y}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ x}}&& &&{\color{blue}{ y}}\\ &\to &({\color{brown}{ x}} + {\color{blue}{ y}})^2&\leftarrow \end{array}\qquad \qquad \qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ x}}^2&  &2{\color{brown}{ x}}{\color{blue}{ y}}&+&{\color{blue}{ y}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ x}}&& &&{\color{blue}{ y}}\\ &\to &({\color{brown}{ x}}  {\color{blue}{ y}})^2&\leftarrow \end{array}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well... 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 telltale part is, the "x" and the "y" raised at the 2 exponent, both have the same coefficient

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, i will review how to do a perfect trinomial and complete the square to get a circle equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for leading me one step closer to the solution

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2 6x + y^2 +2y +6 = 0\] 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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