At the same time, it has the following properties: ① the minimum positive period is π; ② the image with respect to the straight line x = π 3 symmetry; 3 6，π 3) One of the functions is () A. y=sin（x 2+π 6） B. y=cos（x 2−π 6） C. y=cos（2x+π 3） D. y=sin（2x-π 6）

At the same time, it has the following properties: ① the minimum positive period is π; ② the image with respect to the straight line x = π 3 symmetry; 3 6，π 3) One of the functions is () A. y=sin（x 2+π 6） B. y=cos（x 2−π 6） C. y=cos（2x+π 3） D. y=sin（2x-π 6）

∵ the minimum positive period of the function is π,
∴2π
When ω = π, we get ω = 2. The answer should be selected from C and D, excluding a and B
∵ at (- π)
6，π
3) It's an increasing function
ν when x = - π
At 6, the function has a minimum value when x = π
At 3, the function has a maximum value
For C, f (- π)
6）=cos（-π
3+π
3) = 1 is the maximum value, which does not conform to the meaning of the question;
And for D, it's exactly f (- π)
6）=sin（-π
2) = - 1 is the minimum value, f (π)
3）=sinπ
2 = 1 is the maximum
And x = π
At 3, y = sin (2x - π)
6) There is a maximum, so x = π with respect to the straight line
3 symmetry, 2 also holds
Therefore, D is selected

The image of positive scale function y = 2x and the image of linear function y = - 3x + K intersect at point P (1, m), and the value of (1) K (2) is defined by two straight lines and X axis The image with positive scaling function y = 2x intersects with the image of linear function y = - 3x + k at point P (1, m), Find: (1) the value of K (2) The area of a triangle bounded by two straight lines and the x-axis

1、
Put P in y = 2x
m=2×1=2
Replace P (1,2) with y = - 3x + K
2=-3+k
K=5
2、
The intersection point of y = 2x and X axis is (0,0)
y=-3x+5=0,x=5/3
So the intersection point with the X axis (5 / 3,0)
So the bottom of the triangle = | 5 / 3-0| = 5 / 3
The height is the distance from P to x cycles = | m | = 2
So area = 5 / 3 × 2 △ 2 = 5 / 3

The image of the first order function y = 3x-5 is parallel to the image of the positive scale function () and intersects with the point ()

The image of the first order function y = 3x-5 is parallel to the image of positive scale function (y = 3x) and intersects with the point (0, - 5) along the y-axis

As shown in the figure, the image of the first order function y = KX + B intersects the X axis at a (- 6,0), and the image with positive proportion function y = MX is in the second quadrant at point B, and its abscissa is - 4, area 15

Take - 4 into two equations
-4m=-4k+b
-6k+b=0
So 2K = - 4m
k=-2m
6b=30
B=5
-4m=8m+5
-12m=5
m=-5/12
k=5/6
So y = 5x / 6 + 5
y=-5x/12

It is known that the image of primary function intersects point a (6,0) with x-axis, and intersects point B with image of positive proportional function. Point B is in the first quadrant and abscissa is 4. If the area of △ AOB (o is the origin of coordinates) is 15, the functional relationship between this function and positive proportional function is obtained

As shown in the figure, make BC ⊥ OA in C, ∵ s ⊥ OAB = 12oa · BC,

As shown in the figure, it shows the image of a positive proportional function and a linear function. They intersect at point a (4,3). The image of the first-order function intersects with the y-axis at point B, and OA = ob. Find the analytic expressions of these two functions and the area of the triangle formed by two straight lines and x-axis

Through a, make AC ⊥ X axis at point C
Then AC = 3, OC = 4, so OA = 5 = ob
Then B (0, - 5) (1 point)
Let the line Ao: y = NX pass through a (4, 3)
Then 3 = 4N, n = 0.75 (2 points)
So y = 0.75x (3 points)
Let AB: y = KX + B pass through a (4, 3), B (0, - 5)
Then:
b＝−5
4k+b＝3 ．
The solution is as follows:
b＝−5
K = 2. (4 points)
So: y = 2x-5 (5 points)
Let y = 0, x = 2.5
Then d (2.5, 0) (6 points)
The area of AOD is 2.5 × 3 △ 2 = 3.75 (7 points)

We know that the image of the positive scale function y = KX and the first order function y = KX + B intersect at (8.6), the first-order function and X-axis intersect at point B, and ob = 3oa It is known that the image of the positive scaling function y = KX and the first order function y = KX + B intersects at (8.6), the first-order function intersects with the x-axis at point B, and the first-order function intersects with the x-axis at point B, and ob = 3oa

If there is no graph, it is not easy to do. There are two functions with slope of K. the lines of function graph are parallel and cannot intersect. Unless B = 0, the two lines coincide. The problem is

As shown in the figure, the image of the first-order function y = ax + B and the positive proportional function y = KX intersect point a in the third quadrant, and the y-axis intersects at point B (0, - 4), and the area of OA = Ba and △ AOB is 6. Find the analytic formula of the two functions

Make ad ⊥ Y-axis on D, ∵ OA = Ba,

If the image of a positive scale function passes through point a (3, - 1), then the analytic expression of the positive scale function is

Let the positive proportional function be y = KX
Because it goes through point (3, - 1)
Then, it is substituted into the equation
-1=k*3
k=-1/3
So the analytic formula is y = - X / 3

It is known that: the image of positive scale function is over (- 4,8) (1). If points 9 (a, - 1) and (2, - b) are on the image, find the values of a and B (2) Through a point P on the image as the vertical line of the Y axis, the foot of the perpendicular is Q (0, - 8) to find the area of the triangle OPQ

Suppose that the positive proportional function is y = KX, if (- 4,8) is brought in, k = - 2, so this function is y = - 2x (1) by substituting (a, - 1) and (2, - b) into y = - 2x respectively, a = 1 / 2, B = 4 (2) through a point P on the image as the vertical line of Y axis, the perpendicular foot is Q (0, - 8), the ordinate of point P is - 8, the abscissa sign is - 8 / (- 2) = 4, and the triangle OPQ is a right triangle