If every student can draw the image of the function correctly, it will be relatively intuitive and simple for us to solve this problem, but for middle school students, many students can't master the image of the function well, so this kind of problem is easy to make mistakes. It is also the easiest place for students to lose points. I will introduce such problems one by one in the following situations:
1, inverse proportional function y = x/k (k > 0), when x BK. Because the inverse proportional function y= x/k, when K > 0, y decreases with the increase of x, for example, the function y=x/2, and when x >- 1, the value range of y is y 1.
2. Inverse proportional function y = x/k (k < 0). When x A/K or Y < B/K, the inverse proportional function y= xk increases with the decrease of x because K < 0. For example, the function y=x/2? When x >- 1, the value range of y is y > 2；; When x < 2, the value range of y is y.
3. Inverse proportional function y= x/k(k≠0). When a < x < b and a > yn. For example, given the function y=x/2, point A (1, Y 1), B (2 1, Y2) and C (2, Y3) on the image of the function, find the size relationship of Y 1, Y2 and Y3. Since 2 1 < 1 < 2, Y2 > Y 1 > Y3 can be easily obtained according to the above method.
2. the inverse proportional function y = x/k (k < 0), point A 1 (X 1, Y 1), A2 (X2, Y2) an (xn, yn) are all on the image of the inverse proportional function, and x1is known. In this problem, we directly use the property of inverse proportional function (when k < 0, y increases with the increase of x), and it is easy to get y 1 < y2 < y3 < yn. For example: known function y=x/2? , points A (1, Y 1), B (2 1, Y2), C (2, Y3) on the image of the function, find the size relationship of Y 1, Y2, Y3. Since 2 1 < 1 < 2, Y2 < Y 1 < Y3 can be easily obtained according to the above method.
3. the inverse proportional function y = x/k (k > 0), points a 1 (x 1, y 1), a2 (x2, y2) an (xn, yn) are all on the image of the inverse proportional function, and x1is known.
So we always have the y value corresponding to Ak+ 1An greater than the y value corresponding to A 1Ak. The size of Y 1 and Y2Yk is easy to judge: y1> y2 > > yk; The order of YK+ 1 and YK+2yn is YK+ 1 > YK+2 > > YN. To sum up, we get the relationship between Y 1, Y2 and Y3Yn: yk+1> yk+2 > yn > y1> y2 > > yk; If we don't consider so much, it can be summarized in a simplified sentence: when the inverse proportional function y= x/k, k > 0, the y value of any point on the image with positive abscissa is greater than the y value of the point with negative abscissa. If the signs of abscissa are the same, we can compare them according to the properties of inverse proportional function. For example, given the function y=x/2, point A (- 1, Y 1), b (-2 1, Y2), c (2, Y3), D(2.5, Y4) is on the image of the function, and find y/. Analysis: k=2 is greater than zero, and the abscissas of a, b, c and d are positive and negative. The y value of the point with positive abscissa is greater than that of the point with negative abscissa, so there must be Y3 and Y4 greater than Y 1, Y2. When k > 0, y decreases with the increase of x, so there is Y4.