If we know the first-order function y = kx-k + 1, the image does not pass through the second quadrant

If we know the first-order function y = kx-k + 1, the image does not pass through the second quadrant

Because the image of the function y = kx-k + 1 does not pass through the second quadrant
therefore
k>0
-k+11
therefore
k>1

If the image of a function y = KX + B passes through the first, second and fourth quadrants, then the image of a function y = bx-k does not pass through the () quadrant
A. One B. two C. three D. four

If the first-order function y = KX + B passes through quadrants 1, 2 and 4, the value of function y decreases with the increase of X, so K < 0; if the image intersects with the positive half axis of Y axis, then b > 0, so the coefficient of the first-order term of the first-order function y = bx-k b > 0, y increases with the increase of X. if the constant term - k > 0, then the function intersects with the positive half axis of Y axis, so it must pass through quadrants 1, 2 and 3, so the function does not pass through quadrants The fourth quadrant. So D

Given the first-order function y = kx-k + 2, what quadrants does the image pass through?

When k ＞ 0, - K + 2 ＞ 0, i.e. 0 ＜ K ＜ 2, the line passes through quadrant 1, 2 and 3; when k ＞ 0, - K + 2 ＜ 0, i.e. K ＞ 2, the line passes through quadrant 1, 3 and 4; when k = 2, the line passes through quadrant 1, 3 and the origin; when k ＜ 0, the line passes through quadrant 1, 2 and 4

What quadrant must the image of a linear function y = KX + (1 + k) (K ≠ 0) pass through
It must go through the second quadrant, through (- 1,1). I know that the answer should be detailed
Note: This is the problem of function in the first semester of the second year of junior high school. The more detailed the answer, the better the score

y=k(x+1)+1
therefore
Must pass x + 1 = 0
y=1
Namely
(- 1,1) point
And that's in quadrant 2, so
Go straight through the second quadrant

When the image of function y = KX (K ≠ 0) passes through the point (- 1 / 2,5), the analytic expression of the function is written, and the quadrants of the function image are explained

X = - 1 / 2, y = 5 is substituted into the positive proportion function equation
k(-1/2)=5
k=-10
The analytic expression of the function is y = - 10x
-10

1. When k > 0, which quadrant does the image of function y = KX pass through? When k > 0, b > 0, which quadrant does the image of function y = KX + B not pass through? When k > 0, b > 0, which quadrant does the image of function y = kx + B pass through

13,24,123,134.

When K0, the image whose function y = KX + B passes through the second step__ quadrant

When K0, the image whose function y = KX + B passes through the first, second and fourth quadrants

If the image of function y = KX + B passes through the fourth quadrant, then K0, B0,

k>0,b

If the image of the function y = KX passes through (2,8), then which quadrant does the image of y = (1-k) x pass through

2、 Four quadrants

If the line y = KX is parallel to the image of the linear function y = - 3x + 7, then k =?

K = - 3, two lines have the same slope