## 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

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:

2. anonymous

how did you convert x^2 -6x + y^2 +2y +6 = 0 into (x−3)^2+(y+1)^2=2^2?

3. anonymous

do you know what a "perfect trinomial square" is? sometimes called only a "perfect square"

4. anonymous

i have heard of it. I think the formula is (-b/2)^2

5. anonymous

i did not really understand the perfect square formula however

6. anonymous

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

7. anonymous

i have never seen that equation before

8. anonymous

i dont understand how to to input into the equation: x^2 -6x + y^2 +2y +6 = 0

9. anonymous

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

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. anonymous

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

okay, i will review how to do a perfect trinomial and complete the square to get a circle equation.

13. anonymous

Thank you for leading me one step closer to the solution

14. IrishBoy123

$x^2 -6x + y^2 +2y +6 = 0$ 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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