SOLUTION: Solve and graph: y = x^2 + 4x

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Question 18825: Solve and graph:
y = x^2 + 4x
Answer by Alwayscheerful(414) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%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=%284%29%5E2-4%2A1%2A0=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%284%29-sqrt%28+16+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B4x%2B0 can be factored:
1x%5E2%2B4x%2B0+=+1%28x-0%29%2A%28x--4%29
Again, the answer is: 0, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B0+%29

Hope this helps!