Given that the image of the quadratic function y = x2 + BX + C passes through point a (C, 0) and is symmetrical about the straight line x = 2, the analytical formula of the quadratic function may be______ Just write a possible analytical expression

Given that the image of the quadratic function y = x2 + BX + C passes through point a (C, 0) and is symmetrical about the straight line x = 2, the analytical formula of the quadratic function may be______ Just write a possible analytical expression

C2 + BC + C = 0 (1), B = - 4A = - 4 (2)
(1) (2) the solution of simultaneous equations is b = - 4, C = 0 or 3
Then the analytical formula of quadratic function is y = x2-4x or y = x2-4x + 3

In the quadratic function ax quadratic + BX + C (how to find C?) Keep it simple Is the constant term of a quadratic function

When x = 0, y = C
C is the ordinate of the intersection of the parabola and the Y axis

Given the quadratic function f (x) = AX2 + BX + C, f (2) = 0, f (- 5) = 0, f (0) = 1, find this quadratic function

∵ y = AX2 + BX + C satisfies that f (2) = 0, f (- 5) = 0, f (0) = 1,
That is, after three points a (2,0), B (- 5,0) and C (0,1),

4a+2b+c=0
25a−5b+c=0
c=1 ,
Solution:
a=−1
ten
b=−3
ten
c=1 ,
Therefore, the analytical formula of this quadratic function is f (x) = − 1
10x2−3
10x+1.

Given the quadratic function f (x) = AX2 + BX + C, f (2) = 0, f (- 5) = 0, f (0) = 1, find this quadratic function

∵ y = AX2 + BX + C satisfies that f (2) = 0, f (- 5) = 0, f (0) = 1,
That is, after three points a (2,0), B (- 5,0) and C (0,1),

4a+2b+c=0
25a−5b+c=0
c=1 ,
Solution:
a=−1
ten
b=−3
ten
c=1 ,
Therefore, the analytical formula of this quadratic function is f (x) = − 1
10x2−3
10x+1.

In triangle ABC, a = root 5, B = root 15, a = 30, then edge C =?

By the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2bc * cosa
Bring in the solution and get C = root 5 or 2 root 5

In triangle ABC, given a = 2, B = 2 (root 2), ∠ C = 15 degrees, find ∠ a ()

From the double angle formula: cos2x = 2cos ^ X-1, it can be obtained that: cos ^ x = (1 + cos2x) / 2, so: cos ^ 15 ° = (1 + cos30 °) / 2 = (1 + √ 3 / 2) / 2 = 1 / 2 + √ 3 / 4 = (1 / 8) * (4 + 2 √ 3) = (1 / 8) * (1 + 2 √ 3 + 3) = (1 / 8) * (1 + √ 3) ^∵ cos15 ° > 0, ‡: cos15 ° = √ (1 / 8) * √ (1 + √ 3