It is known that the coefficient of quadratic term of quadratic function f (x) is a, and f (0) = 3 (1) If the solution set of the inequality f (x) + 2x < 0 is (1,3), find the analytic expression of F (x) (2) Under the condition of (1), if the real number m satisfies f (m) + 2 > 0, the value range of M is obtained

Let f (x) = ax & sup2; + BX + C, f (0) = 3
C=3
∴f(x)=ax²+bx+3
∴ax²+bx+3+2x0
(m-3)²>4
The solution is m ∈ {M5}
Two minus five is a root of x square minus four, x plus C equal to zero. What is the other root of the equation?
Solution, according to Veda's theorem
X1+X2=-b/a=4
It is known that X1 = 2 - √ 5
So x2 = 2 + √ 5
It is known that the coefficient of quadratic term of quadratic function y = f (x) is a, and the solution set of unequal trial f (x) > 2 is (1,3) (1) the equation f (x) + 6A = 0 has two equal roots, so we can find the analytic trial of f (x); (2) if the maximum value of F (x) is a positive number, we can find the value range of A
Let: y = ax ^ 2 + BX + C (a, B, C are constants, a ≠ 0), the vertex is (1.3, - 2) solved by (- B / 2a, (4ac-b ^ 2) / 4A), ax ^ 2 + BX + C + 6A = 0
To solve the equation: * the square minus 2 times the root sign 2 minus 1 equals 0
x^2-2√2-1=0
(2*2^(1/2) + 1)^(1/2)
-(2*2^(1/2) + 1)^(1/2)
x²-2=0
x=±2
x²-2√2x-1=0
x²-2√2x+2=3
(x-√2)²=3
x-√2=±√3
X = {2 +} 3 or x = {2 -} 3
Positive and negative root sign 2
As shown in the figure, it is known that the vertex m of the image of the quadratic function y = AX2 + 2x + C (a > 0) is on the inverse scale function y = 3x and intersects with the X axis at two points ab
As shown in the figure, it is known that the vertex m of the image of the quadratic function y = AX2 + 2x + C (a > 0) is in the inverse scale function y = 3x
And it intersects with X axis at two points ab
(1) If the symmetry axis of quadratic function is x = - 1 / 2, try to find the value of a and C;
(2) Find the length of AB under the condition of (1);
(3) If the intersection of the symmetric axis and the X axis of the quadratic function is n, when no + Mn is the minimum, try to find the analytic expression of the quadratic function
(1) According to the symmetry axis X = - B / (2a) = - 1 / 2, a in the quadratic function y = AX2 + 2x + C (a > 0) is obtained,
Then, according to the vertex on the inverse scale function y = 3 / x, C can be obtained;
(2) The coordinate of the intersection of the parabola and x-axis is obtained, and then the abscissa of point a is subtracted from the abscissa of point B
(3) The coordinates of points m and N can be expressed by the formula containing a, that is, the value of a can be obtained, and then the analytical formula can be obtained
(1) The symmetry axis of quadratic function is x = - 1 / 2,
∴-2/(2a)=-1/2 ,
The solution is a = 2,
∵ quadratic function y = AX2 + 2x + C (a > 0) the vertex m of the image is on the inverse scale function y = 3 / X,
The vertex is (- 1 / 2, C-1 / 2),
∴ 1/2（c﹣1/2 ）=﹣3,
The solution is C = - 11 / 2,
The analytic formula of quadratic function is y = 2x2 + 2x-11 / 2;
(2) The analytic formula of quadratic function is y = 2x2 + 2x-11 / 2;
Let y = 0,2x2 + 2x-11 / 2 = 0;
The solution is x = (- 1 ± 2 √ 2) / 2
∴AB=(-1+2√2)/2-(-1-2√2)/2 =2√3 ；
(3) According to the symmetry axis X = - 1 / A, when x = - 1 / A, y = - 3a,
When 3A = 1 / A, no + Mn is the smallest,
When 3A * 2 = 1, a = √ 3 / 3,
The analytic expression of quadratic function is y = x2 + 2x + 3
The analytic expression of quadratic function is y = √ 3 / 3 x ^ 2 + 2x + 3 √ 3
a=2，c=-（5/2），
If the intersection of the symmetry axis and X axis of the quadratic function is n, when no + Mn takes the minimum value, try to find the quadratic function ∵ y = ax + 2x + C2 √ 3A = 2, 3A - √ 3 = CA = √ 3 / 3, C = 0
Given that the straight line passes through P (- 3, - radical 3), find the locus of the midpoint m of the chord where the point P intersects the circle x ^ 2 + y ^ 2 = 9
Let the center of the circle be o and the coordinates of the m point be (x, y)
In RT △ OMP, by using Pythagorean theorem,
OM²+PM²=OP²
（x²+y²）+【x-（-3）】²+【y-（-√3）】²=（-3）²+（-√3）²
arrangement
x²+y²+3x+√3y=0
（x+3/2）²+（y+√3/2）²=3
Let m (x, y) be at the origin (0,0), then OMP forms a right triangle. Om ^ 2 + MP ^ 2 = OP ^ 2, that is, x ^ 2 + y ^ 2 + (- 3-x) ^ 2 + (- √ 3-y) ^ 2 = 9 + 3, then we can get (x + 1.5) ^ 2 + (y + + √ 3 / 2) ^ 2 = 3, which is the trajectory equation of M.
Let m (x, y)
OM² + MP² = OP²
(x - 0)² + (y - 0)² + (x + 3)² + (y + √3)² = (-3 - 0)² + (-√3 - 0)²
x² + y² + x² + 6x + 9 + y² + 2√3y + 3 = 9 + 3
x² + 3x + y² + √3y = 0
(x + 3/2)² + (y + √3/2)² = 3
It is known that the vertex m of the image of the quadratic function y = ax + 2x + C (a > 0) is on the inverse scale function y = 3 / X and intersects with the x-axis and points a and B
(1) The symmetry axis of quadratic function is x = 1 / 2. Try to find the value of a and C (2). Under the condition of (1), find the length of AB (3). If the intersection of symmetry axis and X axis of quadratic function is n, when no + Mn is the minimum, try to find the analytic formula of quadratic function
We must change the known a & gt; 0 to a & lt; 0. (1) - 2 / (2a) = 1 / 2, a = - 2, m (1 / 2, C + 1 / 2) on y = 3 / x, so c + 1 / 2 = 6, C = 11 / 2. (2) - 2x ^ 2 + 2x + 11 / 2 = 0, that is 4x ^ 2-4x-11 = 0, let two be x1, X2, then X1 + x2 = 1, x1x2 = - 11 / 4, ab = | x2-x1 | = √ [(x1 + x2) ^ 2-4x2] = 2 √ 3
P (- 3, - 3 times the root sign 3) finds the locus of the midpoint m of the chord where the point P intersects the circle x ^ 2 + y ^ 2 = 9
Idea: let the center of the circle be o and the coordinates of m point be (x, y)
In RT △ OMP, we can use Pythagorean theorem to sort out
If 1 / A, 2 / B and 3 / C are equal difference sequence and equal ratio sequence, then the two real roots of the equation AX ^ 2 + bx-c = 0 are
4/b=1/a+3/c=(c+3a)/ac
4/b^2=3/ac,
4/b^2=(c+3a)^2/4(ac)^2=3/ac
(c+3a)^2=12ac
(c-3a)^2=0
c=3a
b=2a,
ax^2+bx-c=0
turn
ax^2+2ax-3ax=0
a(x-1)(x+3)=0
x1=1,x2=-3
Simple method: let the middle be K, then K ^ 3 = (k-d) (K + D) k, so K ^ 2 = k ^ 2-D ^ 2, then d = 0, so q = 1. We get a; B; C = 1:2:3, so X1 = 1, X2 = - 3
It is known that P (1,2) is a certain point in the circle x2 + y2 = 9. If two mutually perpendicular arbitrary chords intersect at points B and C through P, then the trajectory equation of point m in BC is______ .
Let m (x, y), B (x, x, Y 1, X (x, Y 1, Y 1, X (x, Y 1, Y 1, y 2), then x 1 + x 2 = 2x, Y 1 + y 2 = 2x, Y 1 + y 2 = 2X, x 1 + y 12 = 2x2, y 2 = x (x, x, Y 1, Y 1, Y 1, x 1, Y 1, y 2) 2 = x 12 + x 2 (x 1, y 2) 2 = y 2 = 2x, Y 1 + 2 = 2x2 = 2x2, and x 12 + y 2 = 9, x 12 + 12 + 12 = 9, x 12 + 12 + 12 = 9, x 12 + 12 + 12 + 2x2 + 2x2 + 2x2 + 2x2 + 2x2 + 2y22-2-2 + 2x2-2-2-2x2-2-2-2-2-2 + 2-2-2-2-2-2-2-2-2-2-2-2-2-2-1, we get x1x2 + y1y2-2 (Y1 + Y2) - (x1 + x2) ）+5 = 0, substituting into 2x2 + 2y2-9-4y-2x + 5 = 0, so the M trajectory equation is x2 + y2-x-2y-2 = 0

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