SOLUTION: I need to solve and graph these equations. I have not done this since high school and cannot remember how to solve these. (x+2)^2 + (y-4)^2 = 36 y=3x^2 y=x^2+x (x-1)^2

Question 171147: I need to solve and graph these equations. I have not done this since high school and cannot remember how to solve these.
(x+2)^2 + (y-4)^2 = 36
y=3x^2
y=x^2+x
(x-1)^2 + (y-8)^2 = 16
Answer by KnightOwlTutor(293) (Show Source): You can put this solution on YOUR website!
The first equation is a circle with radius 6 and the center is at (-2,4)
(x+2)^2 + (y-4)^2 = 36
The general equation is x^2+y^2= (radius)^2
This is a parabola
y=3x^2

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=0 is zero! That means that there is only one solution: .
Expression can be factored:

Again, the answer is: 0, 0. Here's your graph:

This is also a parabola
y=x^2+x

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0, -1. Here's your graph:

This is also an equation for a circle
(x-1)^2 + (y-8)^2 = 16
The center is at (1,8) and the radius is 4.
I hope that this helps!

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