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

Algebra ->  Quadratic Equations and Parabolas -> 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       Log On


   

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) About Me  (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 ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B0x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%280%29%5E2-4%2A3%2A0=0.

Discriminant d=0 is zero! That means that there is only one solution: x+=+%28-%280%29%29%2F2%5C3.
Expression can be factored: 3x%5E2%2B0x%2B0+=+3%28x-0%29%2A%28x-0%29

Again, the answer is: 0, 0. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B0%2Ax%2B0+%29


This is also a parabola
y=x^2+x
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A0=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+1+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%281%29-sqrt%28+1+%29%29%2F2%5C1+=+-1

Quadratic expression 1x%5E2%2B1x%2B0 can be factored:
1x%5E2%2B1x%2B0+=+1%28x-0%29%2A%28x--1%29
Again, the answer is: 0, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B0+%29




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!