SOLUTION: Please help me with the following question: Determine the vertex of y=x^2-8x+22. Thank you!

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Question 43880: Please help me with the following question: Determine the vertex of y=x^2-8x+22. Thank you!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the vertex of y=x^2-8x+22. Thank you!
Put the equation in vertex form by completing the square, as follows:
y-22=x^2-8x
y-22+16=x^2-8x+4^2
y-6=(x-4)^2
Vertex is at (4,6).
This is what it looks like.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%2B22+=+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=%28-8%29%5E2-4%2A1%2A22=-24.

The discriminant -24 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -24 is + or - sqrt%28+24%29+=+4.89897948556636.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-8%2Ax%2B22+%29

Cheers,
Stan H.