What is the function of this triangle?

1) Solution: (2,1) is substituted into y=mx-4 and y=nx.

1=2m-4， 2n=1

m=5/2 n=1/2

The primary function is y=5 2x-4, and the proportional function is y=1 2x2) solution: because the intersection point of the function y=3 2x-4 and the x-axis is (8 3,0), and both images pass through the point (2,-1), so the area of the triangle enclosed by the images of these two functions and the x-axis is: s=1 2 8 3 1=4 3

I hope to be selected as a satisfactory answer! Have a nice day!

Answer: 1) After the point (2, 1), there is x=2, y=1, and the two expressions y= and y= are obtained by substituting the two expressions respectively

2 The two functions and the x-axis are enclosed into a triangle, i.e.: y = with the intersection of the x-axis a(0, 0), y=with the x-axis intersection b(, 0) and the intersection of the two functions c(2, 1) The area of the triangle ABC is calculated as follows: from point C the perpendicular line of the x-axis intersects at point E to obtain the triangle ACE, the area of the triangle ABC = the area of the triangle ACE - the area of the triangle CBE =

<> this angle a is a simple trigonometric transformation.

When you find the area, you need to convert the area into two different expressions, so as to sort out an equation for the area, I think this will be better to solve the finger state, please see the specific process**.

1. Because f(-x)=[x 1 3+x (-1 3)] 5=-f(x), f(x) is an odd function.

Let the annihilation be defeated x 1 3=t, the original formula is (t-t one-part) divided by 5, x belongs to 0 to positive infinity, t belongs to 0 to 1, (t-t one-part) is the subtraction function, and f(x) is the subtraction function. Oak sails.

In the same way, x belongs to negative infinity, and to 0 is f(x) is an increasing function.

2. Algebraically f(4)-5f(2)g(2) and f(9)-5f(3)g(3) both = 0, so f(x 2)-5f(x)g(x) is equal to 0

Let cos(a-b)=u, belong to (-cosc,1],1 (sina) 2+1 (sinb) 2

(sina)^2+(sinb)^2]/(sinasinb)^2

2(2-cos2a-cos2b)/[cos(a-b)-cos(a+b)]^2

2[2-2cos(a+b)cos(a-b)]/(u+cosc)^2

4(1+ucosc) (u+cosc) 2, denoted f(u).

f'(u)=4{cosc/(u+cosc)^2-2(1+ucosc)/(u+cosc)^3]

4[(cosc) 2-2-ucosc] (u+cosc) 2<0, so f(u) is a subtraction function and the minimum is f(1)=4 (1+cosc)

The following studies v=1 (1+cosc)+1 (1+sinc).

2+sinc+cosc) (1+sinc+cosc+sinccosc), let t=sinc+cosc belong to (-1, 2], then sinccosc=(t 2-1) 2, v=(2+t) [1+t+(t 2-1) 2].

2(t+2)/(t^2+2t+1),2(t+2)/(t+1)^2

2[1 (t+1)+1 (t+1) 2], is a subtraction function, the minimum value of v = v( 2)=2( 2+2) ( 2+1) 2=2 2 ( 2-1)=4-2 2, so the minimum value of m = 4(4-2 2) = 16-8 2

Solution: y 1 2sin (x 4) cos(2x 2) sin2x

Minimum positive period: t2 |ω|2 2 The sine function is an odd function.

Select B [Formula].

cos2α＝1－2sin²α