If quadratic function satisfies f (x + 1) - f (x) = 2x, f (0) = 1, find f (x)=

If quadratic function satisfies f (x + 1) - f (x) = 2x, f (0) = 1, find f (x)=

Let f (x) = ax ^ 2 + BX + C
f(0) = 1
Then C = 1
f(x) = ax^2 + bx + 1
f(x+1) = a(x+1)^2 + b(x+1) + 1 = ax^2 + (b+2a)x + a + b + 1
f(x+1) - f(x)
= [ax^2 + (b+2a)x + a + b + 1] - [ ax^2 + bx + 1]
= 2ax + a + b
If f (x + 1) - f (x) = 2x holds for any x, then
2a = 2
a + b = 0
a = 1
b = -1
f(x) = x^2 - x + 1
Given that FX is a quadratic function and f (0) = 3, f (x + 2) - f (x) = 4x + 2, the analytic expression of F (x) is
F (x) = x ^ 2-x + 3 Let f (x) = ax ^ 2 + BX + C, then f (0) = C = 3f (x + 2) = a (x + 2) ^ 2 + B (x + 2) + 3 = ax ^ 2 + (4a + B) x + 4A + 2B + 3, then f (x + 2) - f (x) = ax ^ 2 + (4a + b) x + 4A + 2B + 3-ax ^ 2-bx-3 = 4ax + 4A + 2B, because f (x + 2) - f (x) = 4x + 2, so 4A = 4, that is, a = 14a + 2B = 2 = 4 + 2B, that is, B = - 1, that is, f (x) = x ^
Let f (x) = ax & # 178; + BX + 3
f(x+2)-f(x)=a(x+2)²-ax²+b(x+2)-bx=4ax+4a+2b=4x+2
That is, (2a-2) x = 1-b-2a
As the above formula is established
So 2a-2 = 0, and 1-b-2a = 0
The solution is a = 1, B = - 1
∴f(x)=x²-x+3
solution
Let f (x) = ax & # 178; + BX + C
∵f(0)=3
∴c=3
∵f(x+2)-f(x)=4x+2
∴a(x+2)²+b(x+2)+3-ax²-bx-3=4x+2
That is, 4ax + 4A + 2B = 4x + 2
∴4a=4,4a+2b=2
∴a=1，b=-1
∴f(x)=x²-x+3
Function y = maximum value of (12-2x) + root (x-1)
Seems to use the mean theorem, please master solution
Y = the maximum value of (12-2x) + the maximum value of (x-1) under the root sign
Domain: 1
How much is the maximum value of function y = root (5 + 3x-2x ^ 2)
When x = 3 / 4, the maximum value of 5 + 3x-2x ^ 2 is 49 / 8, so the maximum value of (5 + 3x-2x ^ 2) under y = root is √ (49 / 8) = 7 √ 2 / 4
X2-y2-2x + 2Y factorization
Hello
x²-y²-2x+2y
=（x+y）（x-y）-2（x-y）
=（x+y-2）（x-y）
If you don't understand, please ask. If you understand, please select it as a satisfactory answer in time! (*^__ ^*Thank you!
x2-y2-2x+2y
=(x-y)(x+y)-2(x-y)
=(x-y)(x+y-2)
The quadratic function y = 3x & # 178; - 6x-24 is known
(1) Through the formula, write out the opening direction, symmetry axis and vertex coordinates of parabola
(2) The intersection coordinates of parabola and x-axis and y-axis are obtained respectively
Y = 3x & # 178; - 6x-24 = 3 (x * x-2x-8) = 3 (x-1) ^ 2-27, so the opening direction of the parabola is upward, the axis of symmetry is x = 1, the vertex coordinates are (1, - 27) (2) let y = 0 get 3 (x-1) ^ 2-27, so (x-1) ^ 2 = 9 get x = 4 or - 2 let x = 0 get y = - 24, so the intersection of the parabola and X-axis coordinates is (4
Given the vector a = 3 B = 4, the angle between a and B is 135 degrees, find the value of (1) (3a-2b) · (a-2b) (2) a + B
1.=91+48*2^0.5
2.. =（25-12*2^0.5)^0.5
Solve the equation: 4x of 0.5 minus 1.5 minus 5x of 0.2 minus 0.8 equals minus 4.1
Wrong upstairs
4X of 0.5 minus 1.5 minus 5x of 0.2 minus 0.8 equals minus 4.1 times 4
8（4x-1.5）-20(5x-0.8)=-16.4
32x-12-100x+16=-16.4
100x-32x=16-12+16.4
68x=20.4
x=0.3
If you don't understand this question, you can ask,
Same as * 10,
80x-15-250x-8=-41,
-170x=-18
X = 18 out of 170?
Factorization factor x2 + 2XY + y2-2x-2y-1
This is a more important topic in factoring, and we must remember it
x2+2xy +y2-2x-2y-1
=（X+y）^2 -2(x+y)-1
=[(X+y)-1]^2
= (X+y-1) ^2
Let f (x-1) = 2x & # 178; + 3x-5, then f (x) =?
Let X-1 = t
x=t+1
f(t)=2(t+1)²+3(t+1)-5
f(t)=2t²+7t
f(x)=2x²+7x
Let t = X-1, then x = t + 1
∴f(t)＝2(t＋1)²＋3(t＋1)－5＝2t²＋4t＋2＋3t＋3－5＝2t²＋7t
∴f(x)＝2x²＋7x