Question

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

Answer 1

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:

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

i did not really understand the perfect square formula however

Answer 6

\(\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?

Answer 7

i have never seen that equation before

Answer 8

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

Answer 9

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}\)

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Thank you for leading me one step closer to the solution

Answer 14

\[x^2 -6x + y^2 +2y +6 = 0\]

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