anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
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anonymous
  • 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:
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anonymous
  • anonymous
how did you convert x^2 -6x + y^2 +2y +6 = 0 into (x−3)^2+(y+1)^2=2^2?
jdoe0001
  • jdoe0001
do you know what a "perfect trinomial square" is? sometimes called only a "perfect square"

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anonymous
  • anonymous
i have heard of it. I think the formula is (-b/2)^2
anonymous
  • anonymous
i did not really understand the perfect square formula however
jdoe0001
  • jdoe0001
\(\begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad \qquad\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow \end{array}\) does that ring a bell?
anonymous
  • anonymous
i have never seen that equation before
anonymous
  • anonymous
i dont understand how to to input into the equation: x^2 -6x + y^2 +2y +6 = 0
jdoe0001
  • jdoe0001
hmmm, well, I happen to be using a and b how about \(\begin{array}{cccccllllll} {\color{brown}{ x}}^2& + &2{\color{brown}{ x}}{\color{blue}{ y}}&+&{\color{blue}{ y}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ x}}&& &&{\color{blue}{ y}}\\ &\to &({\color{brown}{ x}} + {\color{blue}{ y}})^2&\leftarrow \end{array}\qquad \qquad \qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ x}}^2& - &2{\color{brown}{ x}}{\color{blue}{ y}}&+&{\color{blue}{ y}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ x}}&& &&{\color{blue}{ y}}\\ &\to &({\color{brown}{ x}} - {\color{blue}{ y}})^2&\leftarrow \end{array}\)
anonymous
  • 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
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
okay, i will review how to do a perfect trinomial and complete the square to get a circle equation.
anonymous
  • anonymous
Thank you for leading me one step closer to the solution
IrishBoy123
  • 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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