When x = 4, y has a maximum value of 2, and the image passes through the point (2,0), then the analytic expression of the function is ()

When x = 4, y has a maximum value of 2, and the image passes through the point (2,0), then the analytic expression of the function is ()

X = 4, ymax = 2
y=a(x-4)²+2
Over (2,0)
0=a(2-4)²+2
a=-1/2
y=(-1/2)(x-4)²+2
That is y = - X & sup2 / 2-4x-6

: given that the maximum value of quadratic function y = f (x) is 13, and f (3) = f (- 1) = 5, find the value of F (2)
pose a question

1) F (3) = f (- 1), axis of symmetry x = (3-1) / 2 = 2 / 2 = 1.2). The known maximum value is 13, which is substituted by the vertex formula y = a (x-1) ^ 2 + 13.3). F (3) = 5, a (3-1) ^ 2 + 13 = 5, 4A = 5-13 = - 8, a = - 2, y = - 2 (x-1) ^ 2 + 13.4. F (2) = - 2 (2-1) ^ 2 + 13 = - 2x1 + 23 = - 2 + 13 = 11

If the function f (x) = LG (x (X-2 / 3) + 1) x belongs to the closed interval of 1,3 / 2, then the maximum value of F (x) is

x(x-2/3)+1
=x^2-2x/3+1
=(x-1/3)^2+1-1/9
=(x-1/3)^2+8/9
When x = 3 / 2
Maximum = LG (9 / 4-1 + 1)
=lg(9/4)

The domain of the function f (x) = 1 / (lgx-1)

According to the denominator is not 0
If lgx-1 ≠ 0, then x ≠ 2
X-1 > 0 is x > 1
So the domain asks
(1,2）∪（2,+∞）