SOLUTION: graph the quadritic equation y= x^2 +2x

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Question 156636: graph the quadritic equation
y= x^2 +2x
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
y= x^2 +2x
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%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=%282%29%5E2-4%2A1%2A0=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%282%29-sqrt%28+4+%29%29%2F2%5C1+=+-2

Quadratic expression 1x%5E2%2B2x%2B0 can be factored:
1x%5E2%2B2x%2B0+=+%28x-0%29%2A%28x--2%29
Again, the answer is: 0, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B0+%29

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The y-intercept is at x=0.
At x=0, y=0 so the y-intercept is (0,0)
The vertex is at the minimum (in this example).
The 1st derivative is 2x+2
2x+2=0
x = -1
at x = -1, y = -1, so the vertex is (-1,-1)