A, the first quadrant b, the second quadrant c, the third quadrant d and the fourth quadrant
Solution: ∫y = through (1, -2), ∴ -2=, ∴ k=-2.
∴ linear function y=kx+ 1=-2x+ 1, whose image passes through the second, fourth and first quadrants.
∴ C was chosen without going through the third quadrant.
2. Known: K
Solution: ∫k < 0, ∴ y 1= in the second and fourth quadrants.
0 in the second and fourth quadrants of k<, y2=kx.
Choose B.
3. It is known that Y is inversely proportional to x=4 and y=-, then the functional relationship between Y and X is ().
a、y=- B、y=-
c、y=- D、y=-
Solution: ∫y is inversely proportional to and will be used as an independent variable. Let the analytical formula be y=.
When x=4 and y=-,
∴ - = ,∴ k=2×(- )=-，
Y = =, choose one.
Investigation definition: given that two inverse proportional variables are y and y, the analytical formula should be set to y=, not y=.
4. The function value y of inverse proportional function y=(k+ 1) decreases with the increase of x, so the value of k is ().
A, -2 B, 0 C, -2 or 0 D,-1
Solution: ∫ inverse proportional function y=(k+ 1),
k2+2k- 1=- 1，k2+2k=0，
K 1=0 or k2=-2.
∵ y decreases with the increase of x value, ∴ k+1>; 0,∴k & gt； - 1。
Choose B. k = 0
The above four cases focus on the basic content of the concept and nature of inverse proportional function, which is the key to in-depth study and should be mastered carefully.
Example 2. Given the function y=(m2+m-6), when asked what the value of m is, the function is an inverse proportional function, and the image is in the second and fourth quadrants.
Solution: The ∫ function is an inverse proportional function.
∴ m2-3m+ 1=- 1, the solution is m= 1 or m=2.
The images are in the second and fourth quadrants.
Substituting m= 1 into m2+m-6 to get 12+ 1-6 < 0, which meets the requirements.
When m=2 is substituted into m2+m-6=0, the function is not an inverse proportional function.
Note: 1. Inverse proportional function y=, the degree of independent variable x is-1, and the coefficient k≠0, when k
Example 3. It is known that y=y 1+y2, y 1 is directly proportional to X, and y2 is inversely proportional to X. When x= 1 and x=2, the value of y is 6. When x=-4, find the value of y.
Solution: ∫y 1 is proportional to x, ∴ y1= k1x.
∫y2 is inversely proportional to X, ∴ y2=
∴ y=k 1x+
And ∵ x= 1, y=6, x=2, y=6.
According to the meaning of the problem, there is a solution.
∴ y 1=2x, y2=, that is, y=2x+
When x=-4, y=2×(-4)+ =-8- 1=-9.
Note: In the same topic, multiple function relationships should be expressed by different undetermined coefficients k 1, k2…… ...; Although k is a constant, the constants are not necessarily equal under different relations.
Example 4. As shown in the figure, the images of inverse proportional function y=- and linear function y=-x+2 intersect at point A and point B. Find the coordinates of (1)A and B.
(2) the area of △ AOB.
Analysis: The intersection of two functions' images on the image should satisfy the analytical expressions of the two functions at the same time, so the solution of the equations formed by the simultaneous analytical expressions of the two functions is the coordinates of the intersection. Triangle ABC is not a right triangle, three sides can be found, but the height is difficult to find, and there is a rectangular coordinate system in the drawing, so it is much easier to decompose the drawing into the sum of the areas of several right triangles with ready-made right angles.
Solution: Simultaneous solution of (1) equations.
solve
So the coordinates of point A are (-2,4), and the coordinates of point B are (4,2).
(2) Let a straight line y=-x+2 intersect the X axis at m and the Y axis at n, then it is easy to get m (2 2,0) and n (0 0,2).
∴
=
= =6
Note: To find the area of a graph in a rectangular coordinate system, the graph is usually divided into the sum of the areas of several triangles. The principle of splitting is to take the line segment on the coordinate axis as one side of a small triangle as far as possible, that is, to split complex graphics with the coordinate axis as the boundary, so it is easy to find the bottom and height of the triangle. The idea of decomposing complex graphics into simple graphics is the most basic idea to solve the triangle area, and it can also be used here.
S△AOB=S△AOM+S△BOM= ×2×4+ ×2×2=6 to get the result. The topics related to S- algebra and geometry are very important, with many knowledge points and many changes, which are the focus of the senior high school entrance examination.