Given the function y = (K + 1) x + k-1, when k -, it is a linear function, when k -, it is a positive proportional function

Given the function y = (K + 1) x + k-1, when k -, it is a linear function, when k -, it is a positive proportional function

Given the function y = (K + 1) x + k-1, when k is not equal to - 1, it is a linear function, and when k = 1, it is a positive proportional function

Given the positive proportional function y = (K-3) x, if y increases with the increase of X, then the value range of K is

The positive scaling function y = (K-3) x, y increases with the increase of X
k-3＞0
k＞3

1. Given the positive proportional function y = (A-1) x, if the value of Y increases with the increase of X, then the value range of a is____
2. When the image with positive scale function y = KX (K ≠ 0) passes through the point (1 / 3, - 1 / 2), the image passes through the point____ Quadrant, K=______
3. The image of positive scale function y = KX (K ≠ 0) and the image of y = 1 / 5x are symmetric about the Y axis, k = 0=_____

1∵a-1＞0 ∴a>1
2. Four quadrants 3. K = - 1 / 5

Is y = (3-K) x-2k square + 18 a positive proportional function?

Y = (3-K) x-2k square + 18 is a positive proportional function
Then: - 2K & # 178; + 18 = 0
k²=9
k=±3
K = 3, rounding off
The function relation is y = 6x

If y = (K-2) x + (the square of k-2k) is a positive proportional function of X, find K
Know how to say, do not know, please leave

According to the positive proportion function of X, we can know that k-2k = 0, so k = 0 or 2, but we should ensure that this is a function, so K-2 ≠ 0, then K ≠ 2, so k = 0

Proof: no matter what the value of K is, the image of the first-order function (2k-1) x - (K + 3) y - (k-11) = 0 is always over a point
The answer must be detailed!
A little more detail

It is proved that when 2k-1 = 0, that is, k = 1 / 2,
-(1/2 +3)y-(1/2 -11)=0,y=3
When K + 3 = 0, i.e. k = - 3,
（-3*2 -1）x -(-3-11)=0,x=2
So a straight line must pass (2,3)

No matter what the value of K is, the image of the linear function (K + 1) x + (2k + 5) y + 3 = 0 is always over a point

（k+1）x+（2k+5）y+3=0
k(x+2y)+(x+5y+3)=0
x+2y=0 (1)
x+5y+3=0 (2)
(1)-(2)
y=-1
So x = 2
So the image of a function of degree (K + 1) x + (2k + 5) y + 3 = 0 is always over a point (2, - 1)

No matter K is any value of the inequality-3 ＜ x ＜ 1 / 2, the graph of the linear function y = (2k-1) x / (K + 3) + (11-k) / (K + 3) is constant over a certain point
Find the coordinates of this point

The image of a linear function is a straight line, so you can look at the equation which is a straight line and transform it into a straight line
(k+3)y=(2k-1)x+11-k
(2k-1)x-(k+3)y+11-k=0
k(2x-y-1)-(x+3y-11)=0
The line passes through constantly
2x-y-1 = 0 and - (x + 3y-11) = 0, that is, the intersection of X + 3y-11 = 0
Equations
2x-y-1=0
x+3y-11=0
The solution of x = 2, y = 3
So the constant passes through the fixed point (2,3)
The conclusion is correct after examination
In addition, for any real number k, the conclusion is true
Another way of thinking
y=[(2x-1)k+11-x]/(k+3)
If it is a fixed value, then
(2x-1)/1=(11-x)/3
The solution is: x = 2, substituting into the original linear equation to get y = 3,
So over the fixed point (2,3)

It is known that the image of quadratic function has and only has one intersection a (2,0) with X axis, and the intersection B (0,4) with y axis, and its symmetry axis is parallel to y axis
Q: take point m on the image where the quadratic function is located between two points a and B, pass through point m as the vertical line segment of x-axis and y-axis respectively, and the vertical feet are points c and D respectively. Calculate the minimum perimeter of the rectangular MCOD and the M coordinate of the point where the perimeter of the rectangular MCOD is the minimum
The first question is to find the analytic expression of the function, which I have worked out

Since there is only one intersection point between the parabola and the x-axis, let the analytical expression of the parabola be y = a (X-2) &# 178;, and substitute the B coordinate into y = (X-2) &# 178; = x & # 178; - 4x + 4. Let m (P, q), because PC ⊥ x-axis, so PC = q, PD ⊥ y-axis, PD = P, (0 ﹤ P ﹤ 2), (0 ﹤ Q ﹤ 4), because m is on the parabola, so q = P & #

The symmetry axis of the image of a quadratic function is a straight line passing through the point (4,0) and parallel to the Y axis. The abscissa of its two intersections with the X axis and the ordinate of the Y axis are the same
The area of the triangle with three points of intersection on the coordinate axis is 3

If the coordinates of a, B and C are all integers and the area of △ ABC is 3.. there are two possible cases: 1, ab = 4, OC = 3; 2, ab = 6, OC = 1. So there are a (2,0), B (6,0), C (0,3), or C (0, - 3); or a (1,0), B (7,0), C (, 0,1), or C (0, - 1). Y = ± 1 / 4 (X & # - 178; - 8x + 12), Or y = ± 1 / 7 (X & # 178; - 8x + 7)