Write the quadratic function in the form f (x)= a ( x - h) ^2 + k . then, give the vertex of its graph. f (x) = -3x ^2 + 18x - 25
writing in the form specified: f (x) =
vertex: _)

Y=-3x^2+18x-25 move constant to other side

y+25=-3x^2+18x  make leading coefficient 1 by dividing every thing by -3

(y+25)/-3=x^2-6x  halve the linear coefficient, square it, add it to both sides...ie (-6/2)^2=9, so add 9 to both sides

(y+25-27)/-3=x^2-6x+9  now the right side is a perfect square

(y-2)/-3=(x-3)^2  now multiply both sides by -3

y-2=-3(x-3)^2  add 2 to both sides

y=-3(x-3)^2+2

f(x)=-3(x-3)^2+2

So the vertex here is an absolute maximum for the parabola as anything squared and then multiplied by a negative will decrease the value of y.

So the absolute maximum for f(x) occurs at the vertex (3, 2)

I'm just guessing, but my answer would c. 0 to 300.