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Question 285487: the problem says to write the equation in standard form and graph it
x^2+y^2+14y= -13
Found 2 solutions by Theo, richwmiller:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! This looks like the equation of a circle.
Standard form of the equation of a circle is:
(x-h)^2 + (y-k)^2 = r^2
Your equation is:
x^2+y^2+14y = -13
you isolate the x terms and y terms together to get:
(x^2) + (y^2 + 14y) = -13
You complete the squares on your x^2 term to get (x-0)^2 = x^2
You complete the squares on your (y^2 + 14y) term to get (y+7)^2 = 6^2 + 14y + 49
You add 49 to the right side of your equation, and your equation becomes:
(x-0)^2 + (y+7)^2 = 36
The center of your circle is (h,k) which equals (0,-7).
To graph this equation you have to solve for y.
Subtract (x-0)^2 from both sides of this equation to get:
(y+7)^2 = -(x-0)^2 + 36
Take the square root of both sides of this equation to get:
y+7 = +/- sqrt(-(x-0)^2 + 36)
Subtract 7 from both sides of this equation to get:
y = +/- (sqrt(-(x-0)^2 + 36)) - 7
A graph of this equation looks like this:
You can see that the center of this circle is at (0,-7) because when x = 0, the top of the circle = (0,-1) and the bottom of the circle = (0,-13). Each one of these points is 6 units away from (0,-7). When y = -7, the left side of the circle = (-6,-7) and the right side of the circle = (6,-7). Each one of these is 6 units away from the center of the circle at (0,-7).
Standard form of the equation is:
(x-0)^2 + (y+7)^2 = 36
Graph of the equation is as shown above, using the formula as shown below:
y = +/- (sqrt(-(x-0)^2 + 36)) - 7 which is really 2 equations as shown below:
y = + (sqrt(-(x-0)^2 + 36)) - 7
y = - (sqrt(-(x-0)^2 + 36)) - 7
Answer by richwmiller(17219)  (Show Source):
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