The analytical formula of the function in the second year of junior high school, and the method of f

Let y kx+b

Because a, b are on a function.

So substitute the coordinates of a and b.

Get 2 3k+b

6＝-k+b

Get k 2b -4

So y 2x-4

When x 2a, y 2*2a-4 4a-4

So p(2a,4a -4) is on the image of this primary function.

Let the analytic expression of this function be y=kx+b

Because x=3, y=2

x=-1,y=-6

3k+b=2

k+b=-6

k=2,b=-4

So the analytic expression of this function is y=2x-4

When x=2a, y=4a-4

So the point p(2a,4a -4) is on the analytic formula of this function.

Let the analytic formula of the function be: y=cx+b

The image of a function passes through the points a(3,2),b(-1,-6).

Substitution: 3c+b=2

c+b=-6

The solution gives b=-4 c=2 y=2x-4

When x=2a, y=4a-4

So p(2a,4a -4) is on the image of this primary function.

Establish. y=kx+b

When x=3, y=2

When x=-1, y=-6

This results in the analytic formula of the function.

Then when x=2a, we can find y

If y=, then the point p(2a,4a -4) is on the function graph.

If it is not equal to 4a-4, it is not on the image.

Proportional function.

y=kx（k≠0）

As long as you know the value of a pair of x and y or the coordinates of a point, you can find k after substitution, so as to get the analytic formula.

A one-time function. y=kx+b（k≠0）

As long as you know the values of two pairs of x and y or the coordinates of two points, you can find k and b after substitution, so as to get the analytic formula.

Inverse proportional function.

y=k/x（k≠0）

As long as you know the value of a pair of x and y or the coordinates of a point, you can find k after substitution, so as to get the analytic formula.

Quadratic functions. General form: y=ax +bx+c(a≠0).

You need to know the values of three pairs of x and y or the coordinates of three points, and then you can find a, b, and c after substituting them, so as to get the analytic formula.

Vertex formula: y=a(x-h) +k,(a≠0).

If vertex coordinates.

For (h,k), then use the above formula to set the analytic formula, and then know the coordinates of a point to determine a.

Intersection formula: y=a(x-x1)(x-x2),(a≠0).

Here x1 and x2 are the coordinates of the intersection point of the quadratic function and the x-axis on the x-axis, if you know such conditions, use the intersection formula to set the analytic formula, and then use other points to determine a. This saves you the trouble of understanding the system of equations.

2)ac=15,cd bisected aco,so dry rock od da=oc ac=9 15=3 5,od=oa*3 8=12*3 stove friend 8=9 2,3) as ef cd in f,ec=ef, so cf=fd,ocd=90°- fce= cef, so ocd fec,cd ec=od fc,fc=cd 2=(1 2) (oc 2+od 2)=(9 4) no defense 5, ec=cd*fc od=2*81 16*5*2 9=45 4,e(45 4,9), let de:y=kx+b, then.

9k/2+b=0,①

45k/4+b=9., get 27k 4=9, k=4 3, substitut , get b=-6

So de:y=4x 3-6

y=kx+b (primary function) y=kx proportional function.

y=k x y=kx -1 xy=k inverse proportional function.

y=ax 2+bx+c quadratic function.

y=a(x-x1)(x-x2) quadratic function intersection.

y=a(x-k) 2+h quadratic function vertex formula.

Use the definition of the inverse bilius function to find the analytic formula:

There are three expressions of inverse proportional functions: (1) y=k x; (2)y=kx-';(3) xy=k, where k is a constant, and k≠0(The second form is that y is equal to the product of k and x to the minus 1 power), pay special attention to k≠0,1, solution: from m-10=a1, the solution is m= 3, and m=a3 when k=(m+3)=0, m=3, then k=m+3=6, and the analytic formula of the inverse proportional function is y=6 x

2. Solution: From 3m+m-5=a1, the solution obtains m=1 or m=a4 3, and m=1, k=m-1=0, m=a4 3, then m-1=7 9, so the inverse proportional function is parsed by y=7 (9x).

Information on capital expansion

Use the properties of the inverse proportional function to find the analytic formula:

From the concept of inverse proportional function, in question 3 n+2n-9=1, since the inverse proportional function decreases with the increase of x in each quadrant, n+3 is a positive number; Question 4: m-5=-1, and because the image of the inverse proportional function increases with the increase of the value of x in each quadrant, m is a negative value.

Solution: From the meaning of the question, n+2n-9=a1, the solution is n=a4 or n=2, because the image y decreases with the increase of x value in each quadrant, so n+3>0, n=2, then n+3=5, so the inverse proportional function image is y=5 x

Solution: From the meaning of the question, m-5=a1, solution m= 2, and because the image y increases with the increase of x value in each quadrant, m=a2, so the analytic formula of the inverse proportional function is y=a2 x

Hello classmates! In middle school, I learned the following four types of functions:

Proportional function: y=kx (k≠0).

The inverse proportional function y = k (k≠0) of x

Primary function y=kx+b (k≠0).

Quadratic function y=ax +bx+c (a≠0).

Proportional function: y=kx (k≠0) Inverse proportional function y=k (k≠0) of x

The primary function y=kx+b (k≠0) and the quadratic function y=ax +bx+c (a≠0).

We only learned these four types of functions in junior high school.

Useful for learning? In the future, I won't be the same person who can only connect to wifi and play with my mobile phone.

y=- +3 is not less x ah (1) no less : the straight line y=- +3 intersects with the y axis at the point q then the coordinates of the q point are (0, and the coordinates of p are symmetrical to his x axis and the coordinates of p are (0, at this time the image of the function y=kx+b passes through the point (-2,5), and it is okay to bring these two points into the equation k=15 4,b= (2) :less :

Then q(0,3) then p(0,-3), and in the same way bring these two points into k=-4 , b=-3

y=- x+3 intersects the y-axis at the point q, indicating x=0, and the solution is y=3, q(0,3).

The point q is symmetrical with p with respect to the x-axis, p(0,-3).

The image of the primary function y=kx+b intersects the y-axis at the point p, where x=0,y=b, p(0,-3).

b = -3 for a one-time function y = kx + b image through the point (-2,5), put the point (-2,5), b = -3

Substituting yields 5=-2k+3, k=-1

So, y=-x+3

The image of the primary function y=kx+b passes through the point (-2,5) The point (-2,5) satisfies the image of the function y=kx+b When x=-2, y=5

5=-2k+b

The line y=- +3 intersects the y-axis at the point q, and the point q is symmetrical with p with respect to the x-axis The line y=, then the coordinates of the q point are: (0, and the coordinates of the point p are: (0, when x=0, y=

At that time, the analytic expression of the k= primary function was: y=,5

This is something I do very seriously and I hope you succeed.

Bringing the point into the primary function gives -2k+b=5, and the y-axis intersects the point p, then p(0,b) the straight line y = -1, 2x+3 intersects the y-axis at the point q(0,3) and p, and the q point is symmetrical with respect to the x-axis, then b=-3, and -2k+b=5, then k=-4 is analytically is: y=-4x-3

Solution: Passing through point B to make the high intersection of the triangle OAB on the M point, so s=1 2*OA*MB=3y=3(8-x)=24-3y

2).There is a question to know the value of the second question, so it can be found by making a mark.

The title does not say the range in which the value of y varies with the change of x, so it should be a primary function (there seems to be no quadratic function in the second year of the first month).

And because the value of y increases with the increase of x, the auspicious coefficient of x should be positive. Let y=kx+b, put (1.-1) Bring in and get.

1=k+b, and then take a random fixed value according to k must be greater than zero (can be determined and can not be zero). For example, upstairs, he takes k to 1, so the formula is.

1=1+b, and then you can make a noise to get b=-2.

I hope you understand my explanation.

The analytic formula for the bisector of the third quadrant angle is: y=-x

Let p (slag a,a) be on a straight line, and the point p is also coarse on an inverse proportional function image, so a=k a, i.e., aa=k

And because the distance from the point p to the origin is 4, the beam oak.

aa+aa=16, so k=8, and the analytic formula is y=8 x

Solution: Let y=kx+b, when x=0, y=b, i.e., ob=b, when y=0, x=-b k, i.e., oa==-b k, get: b+-b k=12

And because the image passes through the point p(3,2), it can be solved by substituting the equation 3k+b=2, and the two squares start to take the equation or form a system of equations.

The general expression of the primary function is y=kx+b

In order to determine the analytic expression of a function, two conditions are required to find the values of k and b, which may be the coordinates of two points, or the corresponding values of two sets of x and y, or two indirect conditions. Replace these two conditions to obtain two binary systems of linear equations about k and b, solve this system of equations to obtain the values of k and b, and then substitute them into the general equation to obtain the analytical formula of the satisfying one-time function.

It is not difficult to function once, pay attention to the following knowledge points: two variables, an x value determines a y value. Representation:

Image method, relational formula, ** method. The relationship between the primary function and the proportional function: the primary function y=kx+b(k≠0), the proportional function y=kx(k≠0), the intersection of the primary function and the y-axis (0,b),

Pass the point (2,0).

0=-4+m

m=4 because the straight line l is parallel to y=3x.

So k=3 so the analytic formula for the straight line l is.

y-0=3x(x-2)

y=3x-62 put a

The two points b are brought to the two analytic formulas, the column equation and the system of equations, which should be easy to solve.

One, m=0-2x2=-4

Solution: Let y=3x+b

Bring a(2,0) in, b = -6

y=3x-6

Two, y=k x, bring a(1,3) into y=3 xThen bring b(n,1) in, n = 3

i.e. point b is (3,1).

Then bring the two points a and b into a functional relation, and solve y=4x-1, which I solved with dark circles! Give some face.