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What is the vertex for the function? (You will want to find the axis of symmetry first.) Write as an ordered pair, leaving no spaces, such as (1,-4). y = -3x2 + 6x
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axis of symmetry is x=-b/(2a)=-6/(2*-3)=1 plug in 1 to y=-3x^2+6x to get your y-coordinate for the vertex
Can you further explain?
if you have a parabola in the form ax^2+bx+c the axis of symmetry is: x=-b/(2a)
the vertex is at: \[(\frac{-b}{2a}, f(\frac{-b}{2a}))\]
you can also rewrite this to a form you might more easily recognize: \[a(x-h)^2+k\] where (h,k) is the vertex y=-3x^2+6x y=-3(x^2-2x) y=-3(x^2-2x+1-1) y=-3(x-1)^2-3
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sorry, last line should be: y=-3(x-1)^2+3
thats called completing the square
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