If the image of quadratic function is known to pass through points (1,9), (2,4) and has only one intersection point with X axis, the analytic formula is obtained Please use the vertex formula y = a (X-H) 2 to solve it

If the image of quadratic function is known to pass through points (1,9), (2,4) and has only one intersection point with X axis, the analytic formula is obtained Please use the vertex formula y = a (X-H) 2 to solve it

Let the equation be y = a * (X-H) ^ 2
Then we can get two results
9=a*(1-h)^2
4=a*(2-h)^2
The results of simultaneous equations are as follows
3/(1-h)=2/(2-h)
The solution is as follows
h=4
We can see that:
a=1
Then the equation is as follows:
y=(x-4)^2

It is known that the image of the first-order function passes through the point (1,2), and the product of the abscissa of the intersection of the image and the x-axis and the ordinate of the focus of the y-axis is 9
Find the analytic expression of this first-order function

Let the first-order function be y = KX + B. It is known that: K + B = 2 ------ 1 is not the abscissa of the intersection of the image and the x-axis. Let the first-order function be y = 0, and find the abscissa as follows: (- B's Square divided by K, 0); let the first-order function be x = 0, and find the ordinate as follows: (0, b)

When the image of a given function passes through points (3,5) and (- 4, - 9), the coordinates of the intersection of the image of the function and the y-axis are obtained

Solution: let the analytic formula of this function be y = KX + B, and the two points are substituted into the function, which has 3K + B = 5 Formula 1 - 4K + B = - 9 formula 2. By subtracting formula 2 from Formula 1, we can get 3K + 4K = 5 + 9 formula 7K = 14 solution, k = 2 Formula 1, and 3 * 2 + B = 5 formula B = - 1, so the analytic formula of the first-order function is y = 2x-1

Given that the image of the first-order function passes through a point (1,2), and the product of the abscissa of the intersection of the image and the x-axis and the ordinate of the y-axis is 9, the first-order function can be obtained

For such a problem, we should first think of the undetermined coefficient method
That is to say, let the first-order function be y = KX + B
Then the equation or group of equations are solved by using the condition
In this problem, K + B = 2 can be obtained from the image passing through point (1,2)
From the latter condition, - B / k * b = 9 can be obtained
Make up the equations
Now I have solved it myself

If the image of a function of degree passes through points (3,5) and (- 4, - 9), then the coordinates of the intersection of the image of the function and the y-axis are______ ．

Because the image of a function passes through points (3,5) and (- 4, - 9), let the analytic expression of a function be y = KX + B, so 3K + B = 5 − 4K + B = − 9, and the solution is k = 2B = − 1, so the analytic expression of a function is y = 2x-1. When x = 0, y = - 1, so the coordinates of the intersection point of the image of the function and the Y axis are (0, - 1)

If the image of a function of degree passes through points (3,5) and (- 4, - 9), then the coordinates of the intersection of the image of the function and the y-axis are______ ．

Because the image of a function passes through points (3,5) and (- 4, - 9), let the analytic expression of a function be y = KX + B, so 3K + B = 5 − 4K + B = − 9, and the solution is k = 2B = − 1, so the analytic expression of a function is y = 2x-1. When x = 0, y = - 1, so the coordinates of the intersection point of the image of the function and the Y axis are (0, - 1)

When the inverse scale function y = K / X is known, the image passes through points a3,2
The image with known inverse scale function y = K / X passes through point a (3,2)
Determine the expression for this inverse scale image
Is point B (6,1) C (4 / 3,6) on the graph of this function

Because the image passes through points (3,2), we get the function y = 6 / X
Substituting two points B and C into the function formula, it is concluded that point B is on the function image and point C is not

If there is an intersection point (- 2,1) between the inverse scale function y = K / X and the linear function y = KX + m, then the coordinate of the other intersection point is ()?
This problem is so simple, but I suddenly feel like my head is flooded. Maybe I haven't been exposed to this problem for a long time
Please tell me the specific process of solving the problem,
I asked about the intersection point. I don't ask about K, M. I know how to calculate these two points!

Taking (- 2,1) into y = KX + m and y = K / X (- 2 into x, 1 into y), we can get - 2K + M = 1 and K / - 2 = 1. In this way, we can get k = - 2 by shifting K / - 2 = 1, and then we can get m = - 3 by substituting k = - 2 into - 2K + M = 1. We can get two analytic expressions y = - 2x-3 and y = - 2 / X. next (because I haven't learned inverse scale function), we can draw function image to solve it

If there is no common point between the image of the first-order function y = KX + 2 and the image of the inverse scale function y = 2 / x, the value range of K is obtained

The problem can be transformed into:
The equation KX + 2 = 2 / X has no real solution
That is KX ^ 2 + 2x-2 = 0, no solution or solution x = 0
Obviously, x = 0 is not the solution of the equation
So the equation KX ^ 2 + 2x-2 = 0 has no solution
△=4+8k

If the inverse scale function y = 2 / X and the first-order function y = KX + 1 have a common point, then the value range of K is

y=2/x=kx+1
kx^2+x-2=0
There is something in common
∴△=1-4×k×（-2）≥0
The solution is k ≥ - 1 / 8