Let the minimum value of the quadratic function y = f (x) be 4, and f (0) = f (2) = 6, find the analytic expression of F (x)

Let the minimum value of the quadratic function y = f (x) be 4, and f (0) = f (2) = 6, find the analytic expression of F (x)

The first way of thinking: F (0) = f (2) = 6, which means that the axis of symmetry is x = (0 + 2) / 2 = 1, and the minimum value is 4. So let the analytic formula be y = a (x-1) ^ 2 + 4f (0) = 6, and substitute it into a * 1 + 4 = 6, a = 2. So the analytic formula is y = 2 (x-1) ^ 2 + 4. The second way of thinking: because f (0) = f (2) = 6, substitute f (0) = 6 to get C = 6, and the axis of symmetry is x = (0 + 2) / 2

The maximum value of quadratic function f (x) is 8. The analytic expression of quadratic function can be obtained through a (- 2,0) B (1,6)

Let f (x): y = ax & sup2; + BX + C
Then: 4a-2b + C = 0
a+b+c=6
The maximum value is 8
The vertex ordinate is 8
Vertex ordinate formula: △ / 4A = 8
The equations are solved
a=-2/9
b=16/9
c=40/9
∴y=-2/9x²+16/9x+40/9

Given that the maximum value of quadratic function y = f (x) is 13, and f (3) = f (- 1) = 5, find the expression of F (x)

The maximum value of quadratic function y = f (x) is 13
Let f (x) = a (x + b) ² + 13
f（3）=f（-1）=5
So f (3) = a (3 + b) &# 178; + 13 = 5
f(-1)=a(-1+b)²+13=5
So a (3 + b) &# 178; = a (- 1 + b) &# 178;
So (3 + b) &# 178; = (- 1 + b) &# 178;
So 9 + 6B + B & # 178; = 1-2b + B & # 178;
So 8b = - 8
b=-1
Substituting ①, the solution is a = - 2
So f (x) = - 2 (x-1) &# 178; + 13