SOLUTION: Solve for roots, axis of symmetry, and vertex -x^2+4x-3

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Question 731556: Solve for roots, axis of symmetry, and vertex
-x^2+4x-3
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
-x%5E2%2B4x-3
Solve for roots:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-4+%2B-+sqrt%28+4%5E2-4%2A%28-1%29%2A%28-3%29+%29%29%2F%282%2A%28-1%29%29+

x+=+%28-4+%2B-+sqrt%28+16-12+%29%29%2F%28-2%29+

x+=+%28-4+%2B-+sqrt%28+4+%29%29%2F%28-2%29+

x+=+%28-4+%2B-+2%29%2F%28-2%29+

roots:
x+=+%28-4+%2B+2%29%2F-2+
x+=+-2%2F-2+
x+=1+
and
x+=+%28-4+-+2%29%2F-2+
x+=+-6%2F-2+
x+=3+
vertex:
Convert to Vertex Form:
-x%5E2%2B4x-3...group the first 2 terms together, separating them from the constant term
%28-x%5E2%2B4x%29-3..factor out leading coefficient, for completing the square to work, the coefficient of x%5E2 must be 1
-1%28x%5E2-4x%29-3...Complete the square, Take half of x coefficient and square it. Notice to keep equation balanced you must add this number and subtract it making the net effect zero.
-1%28x%5E2-4x%2B4-4%29-3
-1%28%28x%5E2-4x%2B4%29%29%2B4-3
-1%28x-2%29%5E2%2B1..... vertex form; so, h=2 and k=1
axis of symmetry is x=h=2