Question 285487
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:


{{{graph(400,400,-10,10,-15,5,(sqrt(-(x-0)^2 + 36)) - 7,-(sqrt(-(x-0)^2 + 36)) - 7,-7)}}}


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