SOLUTION: how do I solve f(x)=-(x-3)^2 as a quadratic function?

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Question 1032404: how do I solve f(x)=-(x-3)^2 as a quadratic function?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Solving in this case means finding
the roots, or all +x+ when +f%28x%29+=+0+
+0+=+-%28x-3%29%5E2+
I can see that +x=3+ solves this
+0+=+-%28+3-3+%29%5E2+
+0+=+0+
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This is called a "double root" because a
quadratic should have 2 solutions
-----------------------------------
The fact that there is one solution is shown by
both the graph and the quadratic formula:
+x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
When there is just one solution, the discriminant is zero
+b%5E2+-+4a%2Ac+=+0+
---------------------
+f%28x%29+=+-%28x-3%29%5E2+
+f%28x%29+=+-%28+x%5E2+-+6x+%2B+9+%29+
+f%28x%29+=+-x%5E2+%2B+6x+-+9+
+a+=+1+
+b+=+6+
+c+=+-9+
----------------
The discriminant is:
+b%5E2+-+4a%2Ac+=+6%5E2+-+4%2A%28-1%29%2A%28-9%29+
+b%5E2+-+4a%2Ac+=+36+-+36+
+b%5E2+-+4a%2Ac+=+0+
------------------
So, referring to quadratic formula, the
single root is at:
+x+=+-b%2F%282a%29+
+x+=+-6%2F%282%2A%28-1%29%29+
+x+=+3+ ( what we decided already )
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Here's the plot:
+graph%28+400%2C+400%2C+-2%2C+10%2C+-10%2C+2%2C+-x%5E2+%2B+6x+-+9+%29+