Quadratic function y = f (x) satisfies: (1) f (0) = 1; (2) f (x + 1) - f (x) = 2x, find the analytic expression of F (x), and find the minimum value of F (x) in the interval [- 1,1]

Quadratic function y = f (x) satisfies: (1) f (0) = 1; (2) f (x + 1) - f (x) = 2x, find the analytic expression of F (x), and find the minimum value of F (x) in the interval [- 1,1]

Let y = ax & # 178; + BX + C
Because f (0) = 1
So C = 1
Because f (x + 1) - f (x) = 2x,
So ax & # 178; + 2aX + A + BX + B + 1 - ax & # 178; - BX - 1 = 2x
It is reduced to: 2aX + A + B = 2x
So a = 1, B = - 1
So f (x) = x & # 178; - x + 1
f(x) = x² - x + 1
=(x - 1/2)² + 3/4
When x = 1 / 2
fmin = f(1/2) = 3/4
A = 1b = - 1C = 1, f (x) = x ^ 2-x + 1, when x is 1 / 2, the minimum is 3 / 4
It is known that y = f (x) is a quadratic function, the equation f (x) = 0 has two equal real roots, and f '(x) = 2x + 2 (1) find the analytic expression of F (x); (2) find the area of the figure enclosed by the curve y = f (x) and the straight line x + Y-1 = 0
(1) Let f (x) = AX2 + BX + C (a ≠ 0) From B2 − 4ac = 02ax + B = 2x + 2, a = 1, B = 2, C = 1 (5) f (x) = x2 + 2x + 1 (6 points) (2) x = - 3 or x = 0 (8) s = ∫ 0 − 3 (− x + 1) DX − ∫ 0 − 3 (x2 + 2x + 1) DX (10 points) = (− 12x2 + x). 0 − 3 - (13x3 + x2 + x). 0 − 3 (12 points) = 92 (13 points)
Given that f (x) is a quadratic function and satisfies f (0) = 1, f (x + 1) - f (x) = 2x, the expression of F (x) is obtained
Let f (x) = AX2 + BX + C. from F (0) = 1, we can get C = 1 ∥ f (x) = AX2 + BX + 1 ∥ f (x + 1) = a (x + 1) 2 + B (x + 1) + 1 = AX2 + (2A + b) x + A + B + 1 ∥ f (x + 1) - f (x) = AX2 + (2a + b) x + A + B + 1-ax2-bx-1 = 2aX + A + B ∥ f (x + 1) - f (x) = 2x ∥ 2aX + A + B = 2x ∥ 2A = 2 and a
Factorization: - x to the second power - 4Y to the second power + 4xy
-The second power of X - the second power of 4Y + 4xy
=-(x²-4xy+4y²)
=-(x-2y)²
Given the quadratic function y = ax & # 178; + K, when x = 0, y = - 3, when x = 1, y = - 1, find the value of y when x = - 2
When x = 0, y = - 3, when x = 1, y = - 1
So - 3 = K
a+k=-1
So k = - 3, a = 2
So the quadratic function y = 2x & # 178; - 3
When x = - 2, y = 2 * 4-3 = 5
Because when x = 0, y = - 3, there is - 3 = a * 0 ^ 2 + K,
And because when x = 1, y = - 1, there is - 1 = a * 1 ^ 2 + K,
A = 2, k = - 3
So when x = - 2, y = 5
∵ when x = 0, y = - 3; when x = 1, y = - 1
∴ -3=k
a+k=-1
∴k=-3 a=2
The quadratic function y = 2x & # 178; - 3
When x = - 2
y=2*4-3=5
The product of three rational numbers a, B and C is negative and the sum is positive. When x = | a | a + | B | B + | C | C, we can find the algebraic formula
The value of X to the power of 2001-2 (x to the power of 2000) + 3
Its product is negative and its sum is positive
So one is less than 0 and two are greater than 0
So x = - 1 + 1 + 1 = 1
So the original formula = 1-2 + 3 = 2
The product of three rational numbers a, B and C is negative and the sum is positive,
Then one of a, B and C is negative
And x = | a | a + | B | B + | C | C
Then x = - 1 + 1 + 1 = 1
2001 power of X - 2 (2000 power of x) + 3
=1-2+3
=2
a*b*c＜0 a+b+c＞0
The three numbers are one negative and two positive
∴x=a/|a|+b/|b|+c/|c|
=2-1=1
∴x^2001 -2（x^2000）+3
=1-2+3
=2
The plot is a negative positive number
Therefore, a less than 0.2 is greater than 0
So x = - 1 + 1 + 1 = 1
So the original formula = 1 - 2 + 3 = 2
It is known that the solutions of the equations 2x + 3A = 4 and 3x + 5 = 2 are the same=
3x+5=2
3x=-3
x=-1
So the solution of 3x + 5 = 2 is x = - 1
Substituting it into the equation 2x + 3A = 4, we get
2×(-1)+3a=4
-2+3a=4
3a=4+2=6
A=2
3x+5=2
3x=-3
x=-1
Substituting x = - 1 into 2x + 3A = 4
-2+3a=4
3a=6
A=2
From 3x + 5 = 2 we get 3x = - 3, x = - 1. If x = - 1 is taken into the form, 3A = 6 and a = 2 are obtained
3X=2-5
3X=-3
X=-1
Substituting x = - 1 into 2x + 3A = 4
2*（-1）+3A=4
-2+3A=4
3A=6
A=2
2x+3a=4
2x＝4－3a
x＝2－﹙3／2﹚a
3x+5=2
3x＝﹣3
x＝﹣1
The solution of equation 2x + 3A = 4 is the same as that of equation 3x + 5 = 2
∴2－﹙3／2﹚a＝﹣1
－﹙3／2﹚a＝﹣3
a＝2
∴a＝2
How to calculate factorization-x ^ 2-4y ^ 2 + 4xy
-x^2-4y^2+4xy
=-( x^2+4y^2-4xy)
=-(x-2y)^2
Given that the quadratic function y = ax & # 178; + X + C (a, C are constants) passes through the point (0, - 2), (1,0) (1) to find the analytic expression of the quadratic function
The following equations can be obtained by taking two points into the function expression respectively;
C=-2
a+1+C=0
C = - 2, a = 1
The analytic formula is y = x ^ + X-2
According to the meaning of the title
-2=c
0=a+1+c
∴a=1
c=-2
∴y=-x²+x-2
-2=a0²+0+c
1=a0²-0+c
C=1
a=-2
The analytic expression of function y = - 2x & # 178; + X + 1
four thousand five hundred and fifty-five trillion and five hundred and fifty-five billion five hundred and fifty-five million five hundred and fifty-five thousand five hundred and fifty-five
It is known that the product of three rational numbers a, B and C is a positive number, and their sum is a positive number. When x = | a | / A + | B | / B + | C | / C, we can find the algebraic formula
The 19th power of 2013x - the value of 2011x + 2012
It is known that the product of three rational numbers a, B and C is a positive number, and their sum is a positive number,
∵abc＞0;
a+b+c＞0;
All positive numbers, or one positive number;
When x = | a | / A + | B | / B + | C | / C, find the algebraic formula
X = 3 or - 1;
x=3；
The 19th power of 2013x - the value of 2011x + 2012
=The 19th power of 2013 × 3 - 6033 + 2012
=2013 × 3 to the 19th power - 4021;
x=-1；
The 19th power of 2013x - the value of 2011x + 2012
=-2013+2011+2012
=2010；
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