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If the image of a function y = KX + B passes through points a (0,2), B (1,0), then B+___ ,k=_____
Take a and B into the equation
2=0k+b
B = 2, k = - 2
0=1k+b
RELATED INFORMATIONS
- 1. If the image of a function y = kx-6 passes through a point (- 1,5), then K=______ ．
- 2. The image with positive scale function y = KX (K ≠ 0) is a straight line passing through point () and point (1,)
- 3. The image with positive scale function y = KX (K ≠ 0) is processed by___ It's a straight line The image of a linear function y = KX + B (K ≠ 0) is a point (0___ ),(____ , 0)_____ .
- 4. The image of an inverse scale function is shown in the figure. Point a is the upper point on the image, ab ⊥ X axis, AC ⊥ Y axis, and the perpendicular feet are B and C respectively. If the area of the rectangular aboc is 8, then the analytical expression of the inverse scale function is______ ．
- 5. As shown in the figure, if point a is on the image of function y = 2x (x > 0), then the area of rectangle aboc is______ A unit of square
- 6. How to calculate the area of inverse scale function image?
- 7. As shown in the figure, point a is a point of the inverse scale function in the first quadrant, AB and AC are perpendicular to the XY axis, B and C are perpendicular feet, and the area of rectangular aboc is 3 Mark as - 2, the line ad intersects the y-axis at point E, point E is on line OC, and Ce: od = 1:2 (1) find the analytic formula of inverse scale function (2) find the coordinate of point a (3) find the analytic formula of line ad
- 8. As shown in the figure, point a is a point of the inverse scale function in the first quadrant image, AB and AC are perpendicular to the X axis and Y axis respectively, B and C are perpendicular feet, and rectangular aboc The area of the line ad is 3, and the coordinate of point D is (- 2,0). The line ad intersects the Y axis at e, and the point E is on the line OC, and CE = OE. (1) find the analytic formula of the inverse proportion function; (2) find the coordinate of point a; (3) find the analytic formula of the line ad
- 9. Take the point P on the image of the linear function y = - x + 3, make PA perpendicular to the x-axis and Pb perpendicular to the y-axis. If the s rectangle paob = 9,4, how many points can be found It is helpful for the responder to give an accurate answer
- 10. Take the point P on the image of the linear function y = - x + 3, make PA perpendicular to the X axis and Pb perpendicular to the Y axis. If the s rectangle paob = 2, how many points can be found
- 11. It is known that the image of the first-order function y = KX + B (k > 0) passes through the point P (3,2), and the area of the triangle formed by it and the two coordinate axes is equal to 4. It is better to find several more similar problems in solving the analytic formula of the function~
- 12. If the area of the triangle formed by the image of the first-order function y = KX + 3 and the two coordinate axes is 9, then the value of K is?
- 13. When the image of the first-order function y = k2x-k + 5 passes through the point (- 2,4), the inverse scale function y = K / X in the quadrant where the image is located, y increases with the increase of X
- 14. If y = log2 (x ^ 2 - ax + 1) has the minimum value, then the value range of a is 2 after log is the base
- 15. Given the function f (x) = loga (AX-1) (a > 0, a ≠ 1); (1) find the domain of definition of function f (x); (2) discuss the monotonicity of function f (x)
- 16. Given the function f (x) = loga (a-ax) (a > 0, and a ≠ 1), find the domain of definition and range of value
- 17. Finding the domain of definition of function y = LG (xsquare-4x-5) Write the process
- 18. F (x) = log2 (x + 1) g (x) = 2log2 (2x + 4) Q to find the domain and minimum value of G-F
- 19. Given that the function f (x) is the domain of definition loga (x + 1) (a > 0, a ≠ 1), the maximum value in the interval [1,7] is 1 / 2 larger than the minimum value, the value of a is obtained
- 20. Find the definition field (1) y = loga (X & # 178; + 1) (a > 0 and a ≠ 1) (2) loga absolute value (x-1) + loga (x + 1) (a > 0 and a ≠ 1) (1) Y = loga (X & # 178; + 1) (a > 0 and a ≠ 1) (2) Absolute value of loga (x-1) + loga (x + 1) (a > 0 and a ≠ 1) (3) F (x) = 1 / radical (1-x) + LG (3 + x)