(x- h)^2+(y- k)^2= r^2
Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. In this case, we will need to complete the square twice — once for each variable.
The center of the circle is the point ( -3, -2), and its radius is 2.
b
First, notice that the only variable raised to the power of 2 is x. Therefore, we expect that the given equation represents a parabola and for this reason, we will try to rewrite it in the graphing form y=a(x-k)^2+h. To do so, we will first isolate the y-term.
Now that the y-term is isolated, we can simplify the right-hand side by completing the square. We have to add and subtract ( b2)^2. In this case, we have that the linearcoefficient b is -4.
b=-4 ⇒ (b/2)^2=(-4/2)^2=4
Let's do it!
The resulting equation matches the graphing form of a parabola. In this case, point ( -2, -5) is the vertex.
c Like in Part A, let's start by recalling the equation of a circle in a graphing form.
(x- h)^2+(y- k)^2= r^2
Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. In this case, we will need to complete the square twice — once for each variable.
