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Knowledge points of linear function in junior high school mathematics
Linear function: the image and properties of linear function are one of the compulsory contents in the senior high school entrance examination. The score in the middle school exam is about 10. The issues are diverse, flexible and comprehensive. There are even existential inquiry topics. The main contents are as follows: ① You can draw the image of the function at one time and master its properties. ② According to the known conditions, the analytic formula of the linear function is determined by the undetermined coefficient method. We can use a linear function to solve practical problems. ④ Investigate the relationship among ic function, binary linear equations and univariate linear inequality. Breakthrough methods: ① correctly understand and master the concept, image and properties of linear function. ② Using the idea of mathematical combination to solve the problems related to linear function images. (3) Master the method of spheroidizing a resolution function with undetermined coefficients. ④ Do some comprehensive exercises to improve the ability of analyzing problems.

Function properties:

The change value of 1.y is directly proportional to the corresponding change value of x, and the ratio is k, that is, Y = KX+B (where k and b are constants, k≠0), ∫ when x increases m, K (x+m)+B = Y+km, km/m = K.

2. When x=0, b is the point of the function on the Y axis, and the coordinate is (0, b).

When b=0 (y=kx), the image of a linear function becomes a proportional function, which is a special linear function.

4. In two linear function expressions:

When k and b in the expression of quadratic function are the same, the images of quadratic function overlap; When k is the same and b is different in the expression of quadratic function, the images of quadratic function are parallel; When k and b in two linear function expressions are different, the images of two linear functions intersect; When k is different and b is the same in quadratic function expressions, the images of quadratic functions intersect at the same point (0, b) on the Y axis. If the relationship between two variables x and y can be expressed as Y=KX+b(k, b is constant, and k is not equal to 0), then y is said to be a linear function image property of x.

1. Practice and graphics: Through the following three steps: (1) list.

(2) tracking points; [Generally, two points are taken. According to the principle of "two points determine a straight line", it can also be called "two-point method". The general image of y=kx+b(k≠0) can be drawn as a straight line passing through (0, b) and (-b/k, 0).

The image of the proportional function y=kx(k≠0) is a straight line passing through the coordinate origin, and generally takes two points (0,0) and (1, k). (3) Connecting lines can make images of linear functions-straight lines. So the image of a function only needs to know two points and connect them into a straight line. Usually, the intersections of the function image with the X axis and the Y axis are -k points B and 0, 0 and B, respectively.

2. Nature:

Any point P(x, y) on the (1) linear function satisfies the equation: y=kx+b(k≠0).

(2) The coordinates of the linear function intersecting with the Y axis are always (0, b), and the images of the proportional function intersecting with the X axis at (-b/k, 0) are all at the origin.

3. Function is not a number, it refers to the relationship between two variables in a certain change process.

4. Quadrant where K, B and function images are located:

When y=kx (that is, b is equal to 0 and y is proportional to x):

When k>0, the straight line must pass through the first and third quadrants, and Y increases with the increase of X; When k < 0, the straight line must pass through the second and fourth quadrants, and y decreases with the increase of x.

When y=kx+b:

When k>0, b>0, when the image of this function passes through the first, second and third quadrants; When k>0, b<0, when the image of this function passes through the first, third and fourth image limits; When k0, the image of this function passes through the first, second and fourth quadrants; When k < 0, b0, the straight line must pass through the first and second quadrants; When b<0, the straight line must pass through the third and fourth quadrants.

In particular, when b=0, the straight line passing through the origin o (0 0,0) represents the image of the proportional function. At this time, when k>0, the straight line only passes through the first and third quadrants, not through the second and fourth quadrants. When k < 0, the straight line only passes through the second and fourth quadrants and does not pass through the first and third quadrants.

4. Special positional relationship:

When two straight lines in the plane rectangular coordinate system are parallel, the k value of the resolution function (that is, the coefficient of the first term) is equal. When two straight lines in the plane rectangular coordinate system are perpendicular, the resolution function K is negative reciprocal (that is, the product of two K values is-1).

③ Point inclination y-y 1=k(x-x 1)(k is the slope of the straight line, (x 1, y 1) is the point where the straight line passes).

④ Two-point formula (y-Y 1)/(y2-y1) = (x-X 1)/(X2-x1) (on a known straight line, (x1,y/kloc).

1. Find the k value of the function image: (y 1-y2)/(x 1-x2) 2. Find the midpoint of the line segment parallel to the X axis: | x1-x2 |/2 3. Find the midpoint of the line segment parallel to the Y axis: | Y 1.

4. Find the length of any line segment: √ (x 1-x2) 2+(y 1-y2) 2 (note: the sum of squares of (x1-x2) and (y1-y2) under the root sign) 5. Find two ones.

Two linear functions y 1 = k1x+y1= y2 = k2x+B2 make y 1x+b 1 = k2x+b2 replace the solution value of x=x0 back to y1=

6. Find the midpoint coordinates of a line segment connected by any two points: [(x 1+x2)/2, (y 1+y2)/2].

7. Find the first resolution function of any two points: (x-x1)/(x1-x2) = (y-y1)/(y1-y2) (where denominator is 0, then numerator is 0) x y+. Negative) in the third quadrant+,-(positive and negative) in the fourth quadrant 8. If two straight lines y1= k1x+b1∑ y2 = k2x+B2, then k 1=k2, b1.Then k1× k2 =-/kloc-. Y = k (x-n)+b is a translation of n units to the right.

4. Quadrant where K, B and function images are located:

When y=kx (that is, b is equal to 0 and y is proportional to x):

When k>0, the straight line must pass through the first and third quadrants, and Y increases with the increase of X; When k < 0, the straight line must pass through the second and fourth quadrants, and y decreases with the increase of x.

When y=kx+b:

When k>0, b>0, when the image of this function passes through the first, second and third quadrants; When k>0, b<0, when the image of this function passes through the first, third and fourth image limits; When k0, the image of this function passes through the first, second and fourth quadrants; When k < 0, b0, the straight line must pass through the first and second quadrants; When b<0, the straight line must pass through the third and fourth quadrants.

Particularly, when b=0, the image of the proportional function is represented by a straight line of the origin o (0 0,0). At this time, when k>0, the straight line only passes through the first and third quadrants, not through the second and fourth quadrants. When k < 0, the straight line only passes through the second and fourth quadrants and does not pass through the first and third quadrants.

4. Special positional relationship:

When two straight lines in the plane rectangular coordinate system are parallel, the k value of the resolution function (that is, the coefficient of the first term) is equal. When two straight lines in the plane rectangular coordinate system are perpendicular, the resolution function K is negative reciprocal (that is, the product of two K values is-1).

③ Point inclination y-y 1=k(x-x 1)(k is the slope of the straight line, (x 1, y 1) is the point where the straight line passes).

④ Two-point formula (y-Y 1)/(y2-y1) = (x-X 1)/(X2-x1) (on a known straight line, (x1,y/kloc).

1. Find the k value of the function image: (y 1-y2)/(x 1-x2) 2. Find the midpoint of the line segment parallel to the X axis: | x1-x2 |/2 3. Find the midpoint of the line segment parallel to the Y axis: | Y 1.

4. Find the length of any line segment: √ (x 1-x2) 2+(y 1-y2) 2 (note: the sum of squares of (x1-x2) and (y1-y2) under the root sign) 5. Find two ones.

Two linear functions y 1 = k1x+y1= y2 = k2x+B2 make y 1x+b 1 = k2x+b2 replace the solution value of x=x0 back to y1=

6. Find the midpoint coordinates of a line segment connected by any two points: [(x 1+x2)/2, (y 1+y2)/2].

7. Find the first resolution function of any two points: (x-x1)/(x1-x2) = (y-y1)/(y1-y2) (where denominator is 0, then numerator is 0) x y+. Negative) in the third quadrant+,-(positive and negative) in the fourth quadrant 8. If two straight lines y1= k1x+b1∑ y2 = k2x+B2, then k 1=k2, b1.Then k1× k2 =-/kloc-. Y = k (x-n)+b is a translation of n units to the right.