A problem of finding function expression in the first year of senior high school Given that the minimum value of the function y = 1-2a-2ax + 2x ^ 2 (- 1 ≤ x ≤ 1) is f (a), the expression of F (a) is obtained (what I don't know is that if you bring a into a function, it means that f (a) = 1-2a, and how to find an expression.)

A problem of finding function expression in the first year of senior high school Given that the minimum value of the function y = 1-2a-2ax + 2x ^ 2 (- 1 ≤ x ≤ 1) is f (a), the expression of F (a) is obtained (what I don't know is that if you bring a into a function, it means that f (a) = 1-2a, and how to find an expression.)

Here y ≠ f (x)
Let y = g (x) = 1-2a-2ax + 2x ^ 2 (- 1 ≤ x ≤ 1)
The axis of symmetry is a / 2
When - 1 ≤ A / 2 ≤ 1, the minimum value f (a) = g (A / 2) = - A ^ 2 / 2 - 2A + 1
When a / 21, the minimum f (a) = g (1) = 3-4A
In conclusion, when A2, f (a) = 3-4A

According to the following conditions, the expression of quadratic function is obtained: (1) the image passes through points (0.0), (1,2) and (2,6);
(2) The vertex coordinates of the image are (- 2,3) and pass through points (2, - 3)

(1) Let f (x) = ax ^ 2 + BX + C and substitute three points to get f (x) = x ^ 2 + X
(2) Let f (x) = a (x + 2) ^ 2 + 3, and substitute (2, - 3) into the solution to get f (x) = - 3 / 5 (x + 2) ^ 2 + 3

The expression of quadratic function is obtained according to the following conditions. (1) the image of quadratic function is known to pass through points (0,11 / 3), (1,3),
Find the expression of quadratic function according to the following conditions
(1) The graph of quadratic function is known to pass through points (0,11 / 3), (1,3), (2,0)
(2) It is known that a quadratic function has a maximum value of 7 when x = - 1, and y = 4 when x = - 2
(3) It is known that the image of quadratic function intersects with X axis at points (2,0), (- 3,0) and passes through (1,5)

1) From (0,11 / 3), let y = ax ^ 2 + BX + 11 / 3
Substituting (1,3); a + B + 11 / 3 = 3, a + B = - 2 / 3
Substituting (2,0): 4A + 2B + 11 / 3 = 0, we get: 2A + B = - 11 / 6
A = - 11 / 6 + 2 / 3 = - 7 / 6
So B = - 2 / 3-A = - 2 / 3 + 7 / 6 = 1 / 2
So y = - 6 / 7X ^ 2 + 1 / 2x + 11 / 3
2) Let y = a (x + 1) ^ 2 + 7 by vertex formula
Substituting (- 2,4): a + 6 = 4, a = - 2
So y = - 2 (x + 1) ^ 2 + 7
3) Let y = a (X-2) (x + 3) from the zero form
Substituting (1,5): a (- 1) * 4 = 5, we get a = - 5 / 4
y=(-5/4)(x-2)(x+3)

As shown in the figure, the image of a function is as shown in the figure

It is known from the function image that the straight line passes through: (- 3,0), (0,2)
Let the analytic formula be y = KX + B
Two points are substituted into the analytical formula
0=-3k+b
2=b
The solution is k = 2 / 3
b=2
So the analytic formula is y = 2x / 3 + 2

Write the function expression according to the given function image

The function expression can be obtained through the fitting function of Excel

Given that the image of function y = 2x + B passes through points, (a, 7) and (- 2, a), the expression of this function is obtained

Two points in
7=2a+b
a=-4+b
The solution is a = 1, B = 5
So the expression of this function is y = 2x + 5

Given that the abscissa of the intersection of line L and line y = 2x + 1 is 2, and the ordinate of the focus of line y = 2x + 2 is 2, the expression of line L is obtained
Given that the abscissa of the intersection of line L and line y = 2x + 1 is 2, and the ordinate of the intersection of line L and line y = 2x + 2 is 2, the expression of line L is obtained

When the abscissa of the line y = 2x + 1 is 2, the ordinate is 2 * 2 + 1 = 5
When the ordinate of the focus of the line y = 2x + 2 is 2, the abscissa is (2-2) / 2 = 0
So the line L passes the point (2,5) (0,2)
Let the equation of line l be
y=kx+b
Put in two points
2k+b=5
0+b=2
The solution is k = 3 / 2, B = 2
The equation of line L is y = 3 / 2x + 2

Given that the line L is parallel to the line y = 2x and passes through the point (0, - 4), the expression of L is obtained

Solution
Let the linear equation be y = KX + a
Because it is parallel to the straight line, y = 2x
Then k = 2
The straight line is y = 2x + a
If passing through point (0,4), then
-4＝0+A
A＝-4
The expression of L is y = 2X-4

Given that the line L is parallel to the line y = - 2x and intersects the Y axis at the point (0,2), the expression of the line L is obtained

Because l is parallel to y = - 2x
So the slope is the same
So k = - 2
Let y = - 2x + B
Substituting (0,2) into
b=2
So:
y=-2x+2
(hope to adopt)

University function expression
It is known that f (0) '= 1, f (x + y) = f (x) * (e ^ y) + F (y) * (e ^ x)
Find the expression of F (x)

Let x = y = 0; f (0) = 0;
Let y = det (small quantity)
f(x+det)=f(x)*(e^det)+f(det)*e^x;
f(x+det)-f(x)=f(x)*(e^det-1)+f(det)*e^x
Divide both sides of the equation by DET and take the limit Lim
lim[f(x+det)-f(x)]/det=f(x)*lim[(e^(0+det)-e^0)/det]+lim[f(0+det)-f(0)/det]*e^x；
So f '(x) = f (x) + e ^ X; let y = f (x)
y-y`=-e^x; (1)
y(0)=0; (2)
y`(0)=1; (3)
Next, solve the differential equation step by step. There's no more nonsense,