SOLUTION: what is the vertex of this graph y=x^2-11x+28

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Question 639917: what is the vertex of this graph y=x^2-11x+28
Answer by reviewermath(1029) About Me  (Show Source):
You can put this solution on YOUR website!
a = 1, b = -11, c = 28
-b%2F2a+=+-%28-11%29%2F%282%281%29%29+=+11%2F2

The vertex is (11/2, -9/4).
Here's how to graph:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-11x%2B28+=+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=%28-11%29%5E2-4%2A1%2A28=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--11%2B-sqrt%28+9+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-11%29%2Bsqrt%28+9+%29%29%2F2%5C1+=+7
x%5B2%5D+=+%28-%28-11%29-sqrt%28+9+%29%29%2F2%5C1+=+4

Quadratic expression 1x%5E2%2B-11x%2B28 can be factored:
1x%5E2%2B-11x%2B28+=+1%28x-7%29%2A%28x-4%29
Again, the answer is: 7, 4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-11%2Ax%2B28+%29