Let the quadratic function f (x) satisfy that f (x + 3) = f (x-1), and the sum of squares of the two real roots of F (x) = 0 is 10. If the image passes (0,3), find f (x)

Let the quadratic function f (x) satisfy that f (x + 3) = f (x-1), and the sum of squares of the two real roots of F (x) = 0 is 10. If the image passes (0,3), find f (x)

Because f (x + 3) = f (x-1), so the axis of symmetry x = - 1, Let f (x) = a (x + 1) ^ 2 + K, substitute (0,1), get a + k = 1,... (1) Let f (x) = a (x + 1) ^ 2 + k = 0, get ax ^ 2 + 2aX + A + k = 0, let two be x1, X2, then X1 + x2 = - 1, X1 * x2 = K / A, X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2 * x1 * x2 = (- 1) ^ 2-2k / a = 10,... (2)
Solve the equations 2x + 3Y = 15.5 5x + 6y = 35
2x＋3y=15.5 ①
5x＋6y=35 ②
①*2-②
（2x＋3y=15.5）*2
4X+6Y=31
-X=-4
X=4
Bring in (1)
Y=2.5
Binary linear equations, with the method of introduction, I want the process
It is known that the quadratic function f (x) satisfies f (0) = f (4), and the sum of two squares of F (x) = 0 is 10. The image passes through (0,3) points, and the analytic expression of F (x) is obtained
Such as the title
If f (0) = f (4), then the axis of symmetry is x = 2
y=a(x-2)^2+c=a(x^2-4x+4)+c=ax^2-4ax+4a+c
The sum of two squares is 10 = X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2 = 4 ^ 2-2 (4a + C) / A, a + C = 0
Over (0,3), then y (0) = 3 = 4A + C --- > 4a-a = 3 >, a = 1, C = - 1
So y = (X-2) ^ 2-1 = x ^ 2-4x + 3
Let f (x) = ax ^ 2 + BX + C, and let two of F (x) = 0 be x1, x2,
Then 16A + 4B + C = C (1)
x1^2+x2^2=(x1+x2)^2-2x1*x2=(-b/a)^2-2c/a=10 (2)
f(0)=c=3 (3)
The solution is a = 1, B = - 4, C = 3
f(x)=x^2-4x+3
To solve the equations 2x + 3Y = 1 ① 3x-6y = 7 ②, the substitution elimination method is used
2x+3y=1①
3x-6y=7②
From ①, y = (1-2x) / 3
Substituting into 2, we get 3x-6 * (1-2x) / 3 = 7
The solution is: x = 9 / 7
Substituting into the solution of (2), y = - 11 / 21
Therefore, the solution of the original equations is:
x=9/7
y=-11/21
It is found that (1) y = (2) / X
Substituting 2: 3 (1-3y) / 2-6y = 7
3-9y-12y=14
-21y=11
y=-11/21
So x = (1 + 11 / 7) / 2 = 9 / 7
The solution is x = 9 / 7, y = - 11 / 21
Given that the image of quadratic function y = f (x) passes through the point (0,3), and the sum of squares of the two equations f (x) = 0 is 10, for any x, f (1 + x) = f (1-x), then f (x)
The image of quadratic function y = f (x) passes through the point (0,3), so C = 3. For any x, there is f (1 + x) = f (1-x), so the axis of symmetry is x = (1 + X + 1-x) / 2 = 1. So let f (x) = ax & # 178; - 2aX + 3 and ax & # 178; - 2aX + 3 = 0. The sum of squares of the two is 10x1 + x2 = 2x1x2 = 3 / ax1 & # 178; + x2 & # 178; = (x1 + X
Please use the global substitution method to solve the equations 2x-3y-2 = 0 ① 6y-4x + 3 / 7 = 2Y + 1 ②
2x-3y-2=0 ① 6y-4x+3/7=2y+1 ②
3y-2x=-2 ③
6y-4x+3/7=2y+1
2(3x-2y)+3/7=2y+1
-4+3/7=2y+1
y=-16/7
x=-23/14
[I hope my answer can help you]
[if you don't understand, please ask, if you are satisfied, please adopt]
In duplicate: 2x-3y = 2 2x = 2 + 3Y
Two formulas: 6y-4x + 3 = 14y + 7, 8y + 4x = - 4
By substituting 2x = 2 + 3Y into 8y + 4x = - 4, we can get the following results
8y+2（2+3y）=-4
14y=-8
y=-4/7
By substituting y = - 4 / 7 into 2x-3y = 2, we can get the following result:
... unfold
In duplicate: 2x-3y = 2 2x = 2 + 3Y
Two formulas: 6y-4x + 3 = 14y + 7, 8y + 4x = - 4
By substituting 2x = 2 + 3Y into 8y + 4x = - 4, we can get the following results
8y+2（2+3y）=-4
14y=-8
y=-4/7
By substituting y = - 4 / 7 into 2x-3y = 2, we can get the following result:
2x-3（-4/7）=2
2x=2-12/7
2x=2/7
X1 / 7 * Stow
① X 2 + 2
-4+3/7=2y+1
2y=32/7
y=16/7 ③
By substituting (3) into (1), we get
2x-3×（16/7）-2=0
2x=-23/7
x=-23/14
2X-3Y=2 (1)
4X-4Y=-4/7,X-Y=-1/7, （2）
2x-2y = - 2 / 7 is substituted into Formula 1
-2 / 7-y = 2, y = - 16 / 7 into 2
X=-17/7
②：-2（2x-3y）+3/7=2y+1
①：2x-3y=2
Substituting ① into ②: - 2 * 2 + 3 / 7 = 2Y + 1
2y=3/7-4-1
y=-16/7
Substituting y into (1): 2x-3 * (- 16 / 7) = 2
x=-17/7
2x-3y = 2 is substituted into the formula - 2 (2x-3y) + 3 / 7 = 2Y + 1 & # 8658; - 4 + 3 / 7 = 2Y + 1 & # 8658; y = - 4 / 3 is substituted into the formula to get x = - 1. Your 3 / 7 should be 7 / 3, right? Seven out of three, right? If not, please explain it again according to the appeal method. Follow up: 6x-4x + 3 out of 7
Let the quadratic function f (x) satisfy that f (x + 2) = f (2-x), and the sum of the two squares of F (x) = 0 is 10. If the image passes through the point (0,3), the analytic expression of the function f (x) is obtained
Ask for advice!
Let f (x) = ax ^ 2 + BX + C
If x = 0, f (x) = 3 is substituted by (0,3), then f (0) = 3 = C
So C = 3
f(x)=ax^2+bx+3
f(x+2)=f(2-x)
arcsinx-x
a(x+2)^2+b(x+2)+3=a(2-x)^2+b(2-x)+3
It is concluded that 2A + B = 0
b=-4a
f(x)=ax^2-4ax+3
The roots of F (x) = 0 are X1 and X2 respectively
x1^2+x2^2=(x1+x2)^2-2x1x2=4^2-2*3/a=10
The solution is a = 1
So f (x) = x ^ 2-4x + 3
X minus y parts x times x plus y parts y square minus x quartic minus y parts x quartic y divided by X twice plus y parts x twice
Original formula = [x / (X-Y)] * [y ^ 2 / (x + y)] - [x ^ 4Y / (x ^ 2 + y ^ 2) (x ^ 2-y ^ 2)] / [x ^ 2 / (x ^ 2 + y ^ 2)]
=xy^2/(x^2-y^2)-x^2y/(x^2-y^2)
=xy(y-x)/(x^2-y^2)
=-xy/(x+y).
Let the quadratic function satisfy: F (2-x) = f (2 + x), and the sum of the squares of the two real roots of F (x) = 0 is 10. The analytic expression of F (x) can be obtained through the point (0,3) of the function image
How to solve it with undetermined coefficient method?
If f (x) = ax ^ 2 + BX + C, let the two real roots of F (x) = 0 be x1, then x2 has X1 + x2 = - B / ax1x2 = C / a (x1) ^ 2 + (x2) ^ 2 = (x1 + x2) ^ 2-2x1x2 = (b ^ 2 / A ^ 2) - 2C / a = 10F (2-x) = f (2 + x), so the axis of symmetry of F (x) is x = 2, and x = - B / 2A = 2. Because of its passing point (0,3), f (0) = C = 3, and C = 3-B / 2A =
Let f (x) = ax ^ 2 + BX + C
Because the function image passes through the point (0, 3), that is, f (0) = 3
So C = 3,
Then f (x) = ax ^ 2 + BX + 3
Because f (2-x) = f (2 + x), f (2-x) = f (2 + x),
Then the axis of symmetry of F (x) is x = 2
So - B / (2a) = 2
Then B = - 4A
So f (x) = ax ^ 2-4ax + 3
And the square sum of the two real roots of F (x) = 0 is 10
From the root formula, we can get another expansion of
Let f (x) = ax ^ 2 + BX + C
Because the function image passes through the point (0, 3), that is, f (0) = 3
So C = 3,
Then f (x) = ax ^ 2 + BX + 3
Because f (2-x) = f (2 + x), f (2-x) = f (2 + x),
Then the axis of symmetry of F (x) is x = 2
So - B / (2a) = 2
Then B = - 4A
So f (x) = ax ^ 2-4ax + 3
And the square sum of the two real roots of F (x) = 0 is 10
From the root formula, we can get an equation with only a, then we can solve a
It's easy to understand. Put it away
Solve the equation (1) 2x + 3Y = 40x − y = − 5 (2) x + 13 − y + 24 = 0x − 34 − y − 33 = 112
(1) 2X + 3Y = 40, ① x − y = − 5, ②, from ②, y = x + 5, ③ is substituted into ①, 2x + 3 (x + 5) = 40, the solution is x = 5, substituting x = 5 into ③, y = 5 + 5 = 10, so the solution of the equations is x = 5Y = 10; (2) the equations can be reduced to 4x − 3Y = 2, ① 3x − 4Y = − 2, ② × 4, 16x-12y = 8, ③, ② × 3, 9x-12y = - 6, ④, ③ - ④, 7x = 14, the solution is x = 2, substituting x = 2 into ① So, the solution of the equations is x = 2Y = 2