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Given a positive proportional function and a first-order function, their images pass through the point P (- 3,3), and the image of the first-order function intersects with the Y axis at the point Q (0, - 2), the analytic expressions of the two functions are obtained
Let the analytic expression of the positive proportional function be y = K1X, ∵ it passes through the point P (- 3,3), ∵ 3k1 = 3, and the solution is K1 = - 1, ∵ the analytic expression of the positive proportional function is y = - x; let the analytic expression of the primary function be y = K2 + B, ∵ it passes through the point P (- 3,3), and the point Q (0, - 2), ∵ 3k2 + B = 3B = - 2, and the solution is K2 = - 53B = - 2
Given that the image of inverse scale function y = k-2x is located in the first and third quadrants, then the value range of K is ()
A. k＞2B. k≥2C. k≤2D. k＜2
The image of ∵ y = k-2x is located in the first and third quadrants, ∵ K-2 ＞ 0, K ＞ 2
Given that there is no intersection point between the image of the first-order function y = - x + 4 and the inverse scale function y = KX in the same rectangular coordinate system, the value range of K is ()
A. k＞0B. k＜4C. k＞-4D. k＞4
By solving the equations y = − x + 4, ① y = KX, ②, and substituting ① into ②, we get - x + 4 = KX. After sorting out, we get x2-4x + k = 0, ∵ there is no intersection point between the image of the first-order function y = - x + 4 and the inverse proportion function y = KX in the same rectangular coordinate system, ∵ Δ= 16-4k < 0, ∵ k > 4
Given that there is no intersection point between the image of the first-order function y = - x + 4 and the inverse scale function y = KX in the same rectangular coordinate system, the value range of K is ()
A. k＞0B. k＜4C. k＞-4D. k＞4
By solving the equations y = − x + 4, ① y = KX, ②, and substituting ① into ②, we get - x + 4 = KX. After sorting out, we get x2-4x + k = 0, ∵ there is no intersection point between the image of the first-order function y = - x + 4 and the inverse proportion function y = KX in the same rectangular coordinate system, ∵ Δ= 16-4k < 0, ∵ k > 4
RELATED INFORMATIONS
- 1. Let the function y = (m-2) x * M-4 be an inverse scale function. (1) find the value of M. (2) draw the image of this function and point out which quadrant is its image located? Let y = (m-2) x * M-4 be an inverse proportional function (1) Finding the value of M (2) Draw an image of this function and point out which quadrants its image is in? (3) If the line y = KX intersects with the image of this function at two points a and B, and the ordinate of point a is 2, it is the solution set of the inequality (m-2) x * M-4 > KX about X
- 2. Given the function y = (M-3) x ^ (m ^ 2-10), when what is the value of M, it is an inverse scale function? In which quadrant is his image located
- 3. If the inverse scale function y = KX passes through (- 3,2), then the image______ Quadrant
- 4. When the image with positive scale function y = 5x passes through the first quadrant and the second quadrant, point (0,) and point (1,) are on the image, y The image with positive scale function y = 5x passes through the second image Point (0,) and point (1,) are in their image Y increases with the increase of X?
- 5. The expression of positive scale function of image passing through (1,2) is______ ．
- 6. Given that the image of positive scale function y = KX passes through (2,3a) and (a, - 6) (a is greater than 0), the value of X when y = 3 is obtained
- 7. If the images of positive scale function and linear function pass through point m (3,4) If the images of the positive scale function and the first-order function pass through the point m (3,4), and the area between the images of the positive scale function and the first-order function and the Y axis is equal to 15 / 2, the analytic expressions of the two functions are obtained
- 8. A problem of positive proportion function in the second grade of junior high school, It is known that y + m is proportional to x-n, and when x = 2, y = 3; when x = 1, y = - 5. Find the value of y when x = - 1 I need it tonight
- 9. Proportion is divided into positive proportion and negative proportion
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- 11. A function image translation problem How is the image of F (x) translated from the image of F (x-1)? Let's take an example,
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- 15. Why can the vector translation point be directly moved, while the function needs to be done by undetermined coefficient method?
- 16. What is the principle of inverse scale function translation left plus right minus up plus down minus
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- 20. Why is left and right translation of a function image left plus right minus? I didn't understand when I was studying, but now it involves quadratic function