Given sin (π + x) = 4 / 5, X ∈ (π / 2,3 π / 2), then TaNx=

Given sin (π + x) = 4 / 5, X ∈ (π / 2,3 π / 2), then TaNx=

Since sin (π + x) = 4 / 5, SiNx = - 4 / 5
X ∈ (π / 2,3 π / 2) so cosx

If the image of a function is known to pass through (- 4,9) and (6,3), the analytic expression of the function is
Linear function

How easy it is
Let y = KX + B
Countable equations {9 = - 4K + B
3=6k+b
The solution is {k = - 3 / 5
b=33/5

Given that the graph of a function of degree passes through points (3,5) and (- 4, - 9), the analytic expression of the function of degree is obtained

Let the linear function be y = KX + B (K ≠ 0), (1 point) because its image passes through (3,5), (- 4, - 9), so 3K + B = 5 − 4K + B = − 9. The solution is: k = 2B = − 1, (3 points), so the linear function is y = 2x-1. (5 points)

The image of a given function passes through (2,3) and (1, - 3) 1. Find the analytic formula 2. Judge whether the point (- 1,1) is on the image of a given function

Let the function be y = ax + B
Substituting (2,3) and (1, - 3) into the above formula has
2a+b=3
a+b=-3
The solution is: a = 6
b=-9
The equation is y = 6x-9
When (- 1,1) is substituted into the equation, 6 × (- 1) - 9 is not equal to 1, so the point is not on the function image

Given the image of a function of degree, through a (2,4) B (- 2, - 8). (1) find the analytic expression of a function of degree (2) draw the function image
Do not copy and paste, please look at the coordinate point, it is too late

The analytic formula of a function is y = KX + B
The image of a linear function passes through a (2,4) B (- 2, - 8)
We obtain {4 = 2K + B}
-8=-2k+b
The solution is k = 3, B = - 2
The analytic expression of the first-order function is y = 3x-2
Drawing a straight line through a (2,4) B (- 2, - 8) is an image of y = 3x-2

The image passes through a (- 1,0), B (3,0), and the function has a maximum value of - 8. Find the analytic formula of the function

Let y = a (x + 1) (x-3) = a (X & # 178; - 2x-3) = a (x-1) &# 178; - 4A
The maximum value is - 4A = - 8
A = 2
So y = 2 (x + 1) (x-3)
No, it should be the minimum

In the following functions, part of the image is shown in Figure 1-8
The intersection of a y = sin (x + π / 6) b y = sin (2x - π / 6) c y = cos (4x - π / 3) d y = cos (2x - π / 6) and X axis is (- π / 6,0) a is 1, and another known point is (π / 12,1) w = 2. It is mainly about how to distinguish it from cosine function or sine function

If y = sin (@ x + &), then @ * (- π / 6) + & = 0, @ * π / 12 + & = π / 2
The solution is: D sin (2x + π / 3) = cos (2x - π / 6)

If the image of a function y = KX + 4 passes through a point (- 3, - 2), then (1) find the expression of this function; (2) judge whether (- 5,3) is on the image of this function

(1) Substituting (- 3, - 2) into the analytic expression, we can get - 3K + 4 = - 2, the solution is k = 2, the analytic expression is y = 2x + 4; (2) substituting (- 5,3) into the analytic expression, it does not satisfy the analytic expression of the function, so the point is not on the image of the function

It is known that the image of a function of degree passes through two points m (1,3), n (- 2,12). Try to judge whether the point P (2a, - 6A + 8) is on the image of the function and explain the reason

y＝kx+b
3＝k+b
12＝-2k+b
k＝-3 b＝6
So y = - 3x + 6
When x = 2A
Y = - 6A + 6 is not equal to - 6A + 8
So p is not here
If you have any help, please take it. Thank you

If we know that the image of a linear function is parallel to the line y = - x + 1 and passes through the point (8,2), then the analytic expression of the linear function is ()
A. y=-x-2B. y=-x-6C. y=-x+10D. y=-x-1

From the meaning of the problem, we can get the equation system k = - 18K + B = 2, the solution is k = - 1B = 10, then the analytic formula of this linear function is y = - x + 10