Quadratic function f (x) = ax & # 178; + BX + C For any x ∈ R, is there a, B, C ∈ R, which satisfies that ① f (x-1) + F (- 1-x) = 0 and f (x) ≥ 0, and ② 0 ≤ f (x) - x It should be that there are 0 ≤ f (x) - x ≤ 1 / 2 (x-1) &;

Quadratic function f (x) = ax & # 178; + BX + C For any x ∈ R, is there a, B, C ∈ R, which satisfies that ① f (x-1) + F (- 1-x) = 0 and f (x) ≥ 0, and ② 0 ≤ f (x) - x It should be that there are 0 ≤ f (x) - x ≤ 1 / 2 (x-1) &;

If f (x) = 1 / 2x and 178; + 1 / 2, then f (x-1) + F (- 1-x) is equal to 1 / 2 (x-1) & 178; + 1 / 2 (- 1-x) & 178; so

The evolution of quadratic function y = ax & # 178; + BX + C is just like the change of function when it moves a few lattices to the left

Y = ax & # 178; -- shift D units to the left (d > 0) -- y = a (x + D) &# 178;, [left +]
——Up translation h units (H ＞ 0) - y = a (x + D) &# 178; + h. [up +]
Pithy formula: left "+" right "-", up "+" down "-"

By using the collocation method, the following quadratic functions are transformed into the vertex formula: y = ax & # 178; + BX + C

y=a(x²+b/a*x)+c
=a[x²+1/2*b/2a*x+(b/2a)²-(b/2a)²]+c
=a[(x+b/2a)²-b²/4a²]+c
=a(x+b/2a)²-b²/4a+c
=a(x+b/2a)²+(4ac-b²)/4a
So the vertex coordinates are (- B / 2a, (4ac-b & # 178;) / 4A)

In the vertex formula y = a (X-H) 2 + k of quadratic function y = ax & # 178; + BX + C, what does h stand for
ditto

-h=b/(2a)
h=-b/(2a)
The axis of symmetry