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It is known that the three vertices of triangle ABC are all on the image of inverse scale function y = 1 / X. it is proved that its perpendicular center h is also on the image of this function
∵ a, B and C are all on y = 1 / x, let a, B and C coordinate in turn: (a, 1 / a), (B, 1 / b), (C, 1 /). Let h coordinate be (x, y). It is easy to get: the slope of AB = (1 / A-1 / b) / (a-b) = - 1 / (AB), the slope of BC = (1 / B-1 / C) / (B-C) = - 1 / (BC), ah
The inverse proportion function y = - 4 / X and the first order function y = - x-3 intersect at two points a and B. 1. Find the coordinates of two points ab. 2. Find the area of triangle AOB. S triangle AOB
-4/x = -x-3
X square + 3x - 4 = 0
（x-1）（x+4）=0
X = 1 or - 4
A（-4,1）B（1,-2）
The line connecting two points of AB is y = - 0.6x - 1.4
Intersection with X-axis at (- 7 / 3,0)
The bottom of the triangle is 7 / 3
The height is the sum of the absolute values of the two ordinates AB 3
Area = 1 / 2 * 7 / 3 * 3 = 4.5
As shown in the figure, it is known that the image of the first-order function Y1 = x + m (M is a constant) intersects with the image of the inverse scale function y2 = KX (k is a constant, K ≠ 0) at point a (1,3). (1) find the analytic expressions of the two functions and the coordinates of the other intersection B of the two images; (2) observe the image and write the value range of the independent variable X that makes the value of the first-order function greater than that of the inverse scale function
(1) Substituting a (1,3) into Y1 = x + m (M is a constant) and y2 = KX to get 1 + M = 3, k = 1 × 3, and solving M = 2, k = 3, so the analytic expressions of these two functions are y = x + 2, y = 3x; & nbsp; & nbsp; & nbsp; solving the equation y = x + 3Y = 3x to get x = 1y = 3 or x = - 3Y = - 1, so the coordinates of point B are (- 3
If the image of positive scale function y = (m-1) x passes through points a (1, Y1) and B (2, Y1 + Y2), then the value range of M is?
Y1, Y2 are constants and Y2 > 0
y2>0
So Y1 + Y2 > Y1
That is to say, the function value when x = 2 is greater than that when x = 1
So y increases as x increases
So the coefficient of X is greater than 0
m-1>0
m>1
RELATED INFORMATIONS
- 1. As shown in the figure, the images of inverse scale function Y1 = K1X and positive scale function y2 = k2x all pass through point a (- 1,2). If Y1 > Y2, then the value range of X is () A. - 1 ＜ x ＜ 0b. - 1 ＜ x ＜ 1C. X ＜ - 1 or 0 ＜ x ＜ 1D. - 1 ＜ x ＜ 0 or X ＞ 1
- 2. Given the linear function Y1 = - x + 3 and y2 = 3x-1, when x takes what value, Y1 > Y2? Y1 = Y2? Y1
- 3. Given that the image intersection point (- 1,5) of the linear function Y1 = 3x + A and y2 = - x + B, then when x satisfies what, Y1 = Y2
- 4. It is known that the intersection coordinates of the first-order function Y1 equal to 3x + 3 and Y2 equal to 2x + 8 in the same rectangular coordinate system are (1,6), then when Y1 is greater than Y2, the value range of X is (1,6)
- 5. Given the equation 3x-2y = 1, it is written that y is a first-order function of X
- 6. Using the pointer to the function as the function parameter, find the piecewise function. When x < 0, y = x + 2; when 0 ≤ x < 5, y = 3x-1; when x ≥ 5, y = x (x-1)
- 7. Given the image of the first-order function y = - 3x + 6, (1) x_ When y
- 8. The graph and Y-axis of the linear functions y = (m-2) x + 1 and y = (m-1) x + M & # 178; - 5 intersect point P and point Q respectively. If point P and point q are symmetric about x-axis, the value of M is obtained
- 9. Given the linear function Y1 = - 3x-4 and y2 = 5x-20, when x takes what value, the function Y1 is not less than Y2?
- 10. Make the graph of the following function of degree. (1) y = 4x-2 (2) y = - X-1 (3) y = 3x / 2 + 2 (4) y = - x + 2
- 11. Given that the point C is the inverse scale function, the perpendicular foot of the vertical line drawn from the point C to the coordinate axis is ab, and the area of the quadrilateral aobc is 6, the analytic solution of the inverse scale function is obtained
- 12. Take point P on the image of linear function y = - x + 3, make PA perpendicular to X axis A, Pb perpendicular to y axis B, and the quadrilateral OAPB is a square, then the P coordinate
- 13. The point P is hyperbolic y = - 4 / X (x)
- 14. The image with positive scale function y = KX (K ≠ 0) is a straight line passing through who and who; the image with linear function y = KX + B (K ≠ 0) passes through who and who
- 15. It is known that, as shown in the figure, the image of the linear function y = KX + B (K ≠ 0) passes through points a (0,3), B (4,6) 1) Finding the analytic expression of a function 2) Point P is a moving point on a linear function. Take P as the center of the circle and 5 as the radius to make a circle. If the length of the line segment cut by the circle P on the x-axis is 6, the coordinates of point P can be obtained Speed~
- 16. Given that the function y = log2 (x2-ax + 1) has a minimum value, then the value range of a is?
- 17. The known function f (x) = log (1-x) + loga (x + 3), where 0
- 18. Given that a ＞ 0 and a ≠ 1, x = loga (A3 + 1), y = loga (A2 + 1), try to compare the size of X and y
- 19. Let logac and Logbc be two of the equations x ^ 2-3x + 1 = 0, and find the value of loga / BC
- 20. Given f (x) = loga (a ^ x-1) (a > 0 and a ≠ 1), solve the equation f (2x) = loga (a ^ x + 1)