Select all the functions that have a graph with the vertex and the axis of symmetry to the left of the vertex and axis of symmetry of the graph of f(x) = (x – 1)2 + 1.
A. g(x) = 2(x − 1)2 − 1
B. g(x) = −(x + 1)2 − 2
C. g(x) = 2(x − 1)2 + 2
D. g(x) = −(x − 2)2 + 1
E. g(x) = −(x + 2)2 + 2
F. g(x) = 2(x + 1)2 + 2

step-by-step explanation:

1

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

(hint): the answer is in the question.