1. If the image of the function y = KX + 1 passes through the point (2,5), then k = () 2. Once the function y = 3x-2 moves up 4 units, a straight line ()

1. If the image of the function y = KX + 1 passes through the point (2,5), then k = () 2. Once the function y = 3x-2 moves up 4 units, a straight line ()

1. If the image of function y = KX + 1 passes through point (2,5), then k = (2)
2. A straight line (y = 3x + 2) is obtained after the linear function y = 3x-2 is translated upward by 4 units

Does the image of a function shift to the left and the domain of definition change?

The domain also follows the translation
So change

How does the vertical translation of function image affect the function range

The left-right translation has no effect on the function image range
There are two kinds of up and down translation: ① when the function is translated up, as long as the translation distance is added to the original function range
② When a function is translated downward, it only needs to subtract the translated distance from the original function range

It is known that the image of the inverse scale function y = KX (K ≠ 0) passes through the point (- 2, - 12). (1) find the analytic expression of this function. If the point a (m, 1) is a point on the image of the inverse scale function, find the value of M; (2) by using the result of (1), do there exist a point P on the coordinate axis, so that the triangle with a, O, P as the vertex is a right triangle? If it exists, please write the coordinates of point P directly; if not, please explain the reason

(1) ∵ the image of inverse scale function y = KX passes through (- 2, - 12), ∵ k = - 2 × (- 12) = 1, ∵ y = LX, ∵ point a (m, 1) is the point on the image of inverse scale function, ∵ 1 = 1m, ∵ M = 1; (2) from (1), a (1,1), connecting OA, passing through a as p3p4 ⊥ OA, AP1 ⊥ X axis, ap2 ⊥ Y axis, ∵ a (1,1), ∵ AP1 = ap2 = OP1, namely P1 (1,0); P2 (0,1); ∵ p2p4 = P 1p3 = 1, OP3 = OP4 = 2, namely P3 (2,0), P4 (0,2), then the coordinates of the subject point P are (1,0) or (0,1) or (2,0) or (0,2)

What are the characteristics of the two four quadrant inverse proportion function

y=k/x
The characteristic of two four quadrant inverse proportion function is k ≠ 0 and ＜ 0

When n takes what value, the (n + n-1) power of y = (n + 2n) x is an inverse scale function? In which quadrant is its image?

N + n-1 = - 1, n = 0 or n = - 1 when n = 0, N + 2n = 0 is not appropriate, when n = - 1, N + 2n = - 1

If the image with inverse scale function y = (m-1) x ^ (3-m ^ 2) is in the second and fourth quadrant, then M =?

The inverse scale function, then the degree of X is - 1
3-m²=-1
m²=4
m=±2
The image is in the second and fourth quadrants
So the coefficient M-1

How many quadrants does the image of linear function y = - 5x-6 not pass through

If the slope is negative, go through quadrants 2 and 4, and the y-axis intercept is negative, you can know that it is only one quadrant

The quadrant that the image of the square of the linear function y = 5x = (B-1) must not pass through is

Fourth quadrant

As shown in the figure, given that the point (1,3) is on the image of the function y = KX (x > 0), the edge BC of the square ABCD is on the X axis, the point E is the midpoint of the diagonal BD, and the image of the function y = KX (x > 0) passes through two points a and E, then the abscissa of the point E is___ ．

Substituting (1, 3) into y = KX to get: k = 3, so the analytic formula of the function is y = 3x, let a (a, 3a) (a > 0), according to the image and the meaning of the problem, we can see that point E (a + 32a, 32a), because the image of y = 3x passes through e, so substituting e into the analytic formula of the function to get: 32A (a + 32a) = 3, that is, A2 = 32, to get: a = 62 or a = -