SOLUTION: I need help on this program complete the square, then graph conic section
x^2+y^2+4x+6y-3=0
So far I have x^2+y^2+4x+6y=-3
(x^2+4x)+y^2+6y=-3
I'm confused at this point
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Algebra.Com
Question 250320: I need help on this program complete the square, then graph conic section
x^2+y^2+4x+6y-3=0
So far I have x^2+y^2+4x+6y=-3
(x^2+4x)+y^2+6y=-3
I'm confused at this point
When you do get it in this form (x-h)^2+(y-k) how do you go about graphing it?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
first you bring all the x's and then the y's together and put them in order of exponents.
x^2+4x+y^2+6y = 3
now complete the square for x
(4/2)^2=2^2=4
add 4 to both sides
(6/2)^2=3^2=9
add 9 to both sides
btw this is a circle
set x=0 and get y
set y=0 and get x
plot those and make a circle
center is at-2,-3
radius 4
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