Quadratic function problems in the third grade of Mathematics In the plane rectangular coordinate system, OA = 12 cm and ob = 6 cm are known. Point P starts from point O and moves along the edge of OA to point a It moves at a speed of 1 cm / S; point Q moves from B along Bo to o at a speed of 1 cm / s P. Q starts at the same time, and t (second) represents the moving time (0 ≤ T)

Quadratic function problems in the third grade of Mathematics In the plane rectangular coordinate system, OA = 12 cm and ob = 6 cm are known. Point P starts from point O and moves along the edge of OA to point a It moves at a speed of 1 cm / S; point Q moves from B along Bo to o at a speed of 1 cm / s P. Q starts at the same time, and t (second) represents the moving time (0 ≤ T)

Analysis: (1) according to the speed of P and Q, the length of OQ and op can be expressed by time t, and the functional relationship of Y and t can be obtained through the area formula of triangle; (2) according to the functional formula of (1), the length of OQ and op can be obtained by calculating the value of X when y is the maximum, and then the coordinates of point C and the analytic formula of line AB can be obtained

Give me a few questions about quadratic function, not too simple, not too level

It is known that the length of the hypotenuse ab of the right triangle ABC is 10cm, and the sine values of its two acute angles are the two roots of the equation m (x ^ 2-2x) + 5 (x ^ 2 + x) + 12 = 0

On mathematical quadratic function
Y = 1.5x + 1 when is x bigger and Y bigger? = = 0 OK?

X belongs to any real number, and Y increases with the increase of X. y can not be equal to 0, and X can be equal to 0

It is known that the parabola y = X2 - (4m + 1) x + 2m-1 intersects the x-axis at two points. If the abscissa of one intersection is greater than 2, the abscissa of the other intersection is less than 2, and the intersection of the parabola and the y-axis is below the point (0, − 12), then the value range of M is ()
A. 16 ＜ m ＜ 14b. M ＜ 16C. M ＞ 14d. All real numbers

According to the theme, Let f (x) = X2 - (4m + 1) x + 2m-1, ∵ parabola y = X2 - (4m + 1) x + 2m-1 and X axis have an intersection abscissa greater than 2, another intersection abscissa less than 2, and parabola opening upward, ∵ f (2) < 0, that is, 4-2 (4m + 1) + 2m-1 < 0, the solution is: M > 16, and ∵ parabola and Y ∵ parabola

Quadratic function
When x = - 1, the maximum value of the function is 2

It can be done in vertex form
Let y = a (x + 1) & sup2; + 2
Over point (- 2 - 1)
∴-1=a（-2+1）²+2
-1=a+2
a=-3
The analytic formula is y = - 3 (x + 1) & sup2; + 2
=-3x²-6x-1
It can also be done in general
Let y = ax ^ 2 + BX = C
When x = - 1, the maximum value of the function is 2

The title of quadratic function
It is known that the intersection of the parabola y = ax square and the straight line y = KX + 3 is (x1,9 / 2) and (x2,2), where X1 and X2 (x1 is less than x2) are two roots of the equation x square-x-6 = 0, and the analytical expressions of the parabola and the straight line are obtained
It is known that the intersection of the parabola y = ax square and the straight line y = KX + 3 is (x1,9 / 2) and (x2,2), where X1 and X2 (x1 is greater than x2) are two roots of the equation x square-x-6 = 0

x^2-x-6=0
(x-3)(x+2)=0
x1>x2
So X1 = 3, X2 = - 2
(3,9/2)
Substituting parabola and straight line
9/2=A*3^2=9A
A=1/2
(-2,2)
Substituting y = KX + 3
2=-2k+3
k=-1/2
So y = x ^ 2 / 2
y=-x/2+3

On the subject of quadratic function
If ob = OC = 1 / 2 OA, then the value of B is ()
Please write the process
The answer is (- 1 / 2)

Suppose a is to the left of the origin and B,
Let ob = m, then the parabola y = ax ^ 2 + BX + C passes through B (m, 0), a (- 2m, 0), C (0, m)
Substituting y = ax ^ 2 + BX + C, we get b = - 0.5
When a is on the right side of B, B = 0.5

The problem of quadratic function
As shown in the figure, the parabola y = ax * 2 + BX + C, X-axis at two points a and B, intersection Y-axis at point C (0, root 3), vertex D (1, - 4 √ 3 / 3)
1) Find the coordinates of a, B and C
2) The quadrilateral aebc is obtained by rotating △ ABC 180 ° around the midpoint m of ab
① Find the coordinates of point E
② Try to judge the shape of quadrilateral aebc and explain the reason
3) Try to explore: whether there is a point P on the straight line BC, which makes the perimeter of △ pad minimum. If there is, ask for the coordinates of point p; if not, please explain the reason?

It's a typical senior high school problem. I've forgotten it for a long time, but try to solve it. Let's talk about the way of thinking. You'd better calculate it yourself. 1) intersection y is 0, root 3, so C = root 3, vertex D is simply brought in. First of all, there is a + B + √ 3 = - 4 √ 3 / 3, so a + B = - 7 √ 3 / 3. If I'm not wrong, the formula of quadratic function is a (x +...)

Quadratic function problems
It is known that the parabola y = a (x + m) 2 + K has the same vertex as the parabola y = (x + 1) 2 + 3 and passes through a (0,1)
1. Find the analytic expression of quadratic function and the coordinate of P
2. Find the coordinates of point a (0,1) about the axis of symmetry and the area of triangle APB

Question 1: y = - 2 (x + 1) 2 + 3
But where's your p-spot

Find the analytic expression of quadratic function
It is known that the function y = f (x) is not only an even function, but also a periodic function with period 6. If f (x) = - x ^ 2 + 2x + 4 when x belongs to [0,3], then f (x) = - x ^ 2 + 2x + 4 when x belongs to [3,6]=_____
I'm a little confused
Help me to say it clearly

When x belongs to [0,3], f (x) = - x ^ 2 + 2x + 4
If the function y = f (x) is even, then if x belongs to [- 3,0], f (x) = - x ^ 2-2x + 4
It is also a periodic function with period 6
Then, if x belongs to [3,6],
f(x)=-(x-6)^2-2(x-6)+4
=-x^2+10x-20