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As shown in the figure, a and B are any two points symmetrical about the origin on the graph of function y = 1 x, AC ∥ Y axis, BC ⊥ Y axis, then the area s of △ ABC=______ .
The area of △ ABC s = 4 × 12 ×| K | = 2
What is the image of y equal to the negative power of X
A. B is any two points on the image of the function y = which are symmetrical about the origin 0, AC is parallel to the Y axis, BC is parallel to the X axis, let the area of triangle ABC be s
Inverse scale function y = 1 / X to find the specific value of area s =? S
Suppose that the coordinates of point B fall in the first quadrant, and let the coordinates of point B be (x1, Y1). Because a and B are symmetrical about the origin, the coordinates of point a are (- x1, - Y1)
In the triangle ABC, | BC | = 2 | x1 |, | AC | = 2 | Y1 |, and two points a and B are on the function y = 1 / x, so we have Y1 = 1 / x1, and we get X1 * Y1 = 1
The area of triangle ABC is s = 1 / 2 * | BC | * | AC | = 2 | X1 * Y1 | = 2
This is a right triangle. You can draw
s=1/2*AC*BC
It should be like this.. ha-ha
The image of y = x two thirds power
When X & nbsp; = & nbsp; 0 & nbsp;, the function has the minimum value y = 0. Although & nbsp; X & nbsp; is located in the cubic root, there is x ^ 2 & nbsp;, & nbsp; under the cubic root, so the value of & nbsp; y & nbsp; must be greater than or equal to 0
Points a and B are inverse scale functions, y = 2 / X. if AC is parallel to X axis and BC is parallel to y axis, the area of △ ABC is equal to ()
Let a (1,2) be B (- 1, - 2), AC ∥ X axis, BC ∥ Y axis, so C (- 1,2), △ ABC is RT triangle. Angle ACB = 90 ° AC = 2, BC = 4, so s △ ABC = 1 / 2 AC.BC=1/2 ×2×4=4.
The cubic graph of y = (x-1)
In fact, y = x ^ 3 image is shifted one unit to the right
Given the quadratic function y = ax ^ 2 (a is greater than or equal to 1), the abscissa of bright spot a and B are - 1 and 2 respectively, and O is the origin of coordinates. If △ AOB is a right triangle, then the perimeter of △ AOB is
A (- 1, a) B (2, 4a) is vertical, so - A * 2A = - 1, a = (radical 2) / 2
OA = (radical 6) / 2 ob = 2 * (radical 3) AB = 3 * (radical 6) / 2 perimeter = 2 * (radical 6) + 2 (radical 3)
If the image of y = f (x) is symmetric to the x power of y = 2 about the Y axis, then f (3) =?
The function image is symmetric about the y-axis, that is, f (x) = 2 to the - x power, substituting 3, f (3) = 2 to the - 3 power = 1 / 8
Given the quadratic function y = AX2, the abscissa of two points a and B on the image are - 1,2,0 respectively, which are the origin. When the triangle OAB is a right triangle, the value of a is
A = two-thirds root sign, two mathematical symbols are not easy to type, sorry
If the image of the function y = f (x) is shifted two units to the right along the x-axis, the image obtained is C, and the image of C symmetrical about the x-axis is y = 2x, then the function expression of y = f (x) is ()
A. y=2x+2B. y=-2x+2C. y=-2x-2D. y=-log2（x+2）
If the image of C symmetric about X axis is y = 2x, then the image of C: y = - 2XY = f (x) can be obtained by translating 2 units to the right along X axis to get C, then translating 2 units to the left of C is f (x), so f (x) = - 2x + 2
RELATED INFORMATIONS
- 1. If we know that a and B are on the image of quadratic function y = AX2, the abscissa of these two points are - 2, 1 respectively, and the triangle AOB is a right triangle (o is the origin of coordinates), we can find the value of A
- 2. If the distance between the vertex A and the positive half axis of the graph a and B is the maximum distance on the positive half axis of the graph A. 32a−12aB. 32a+12aC. 62a−12aD. 62a+12a
- 3. Given that the right angle vertex a of RT triangle ABC moves on the straight line ρ cos θ = 9, C is the origin, and the angle ACB is 30 °, the polar coordinate equation of the trajectory of vertex B is obtained
- 4. Given that the length of the line segment l passing through point P (2,3) and cut by two parallel lines 3x + 4y-7 = 0 and 3x + 4Y + 8 = 0 is d 1), find the minimum value of D, 2) when the line L and 2) When the line L is parallel to the X axis, try to find the value of D
- 5. It is known that the integer part of 5 - √ 5 is a, and the decimal part is B. find a & # 178; + AB + B & # 178;
- 6. 6274 times the 5th power of (1 + x) equals 15031. Find out how much x equals! Keep 4 decimal places 1090 times the 5th power of (1 + x) is equal to = 3276. Find out how much x is equal to! Keep 4 decimal places 4233 times the 5th power of (1 + x) is equal to = 10496. Find out how much x is equal to! Keep 4 decimal places 3093 times the 5th power of (1 + x) is equal to = 5503. Find out how much x is equal to! Keep 4 decimal places 11597 times the 5th power of (1 + x) is equal to = 34089. Find out how much x is equal to! Keep 4 decimal places If there is a way, it would be better! 3Q
- 7. Given that the cube of a + the cube of B = 27, the square of a, the square of b-ab = - 6, find (the cube of B - the cube of a) + (the square of a, the square of b-3ab) -2 (cube of B - square of a) for detailed explanation, each step of explanation
- 8. What is the cube of the square of (a + b) (a-b) - (B-A) (a + b)
- 9. If a is a real number not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- If a is not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- 1) = 1 / 2. If A1 = - 1 / 3, A2 is the difference reciprocal of A1, A3 is the difference reciprocal of A2, A4 is the difference reciprocal of A3, and so on, then a2013=_______ .
- 10. How to determine the number of real number and imaginary number roots of m-degree equation For example, the number of roots of x ^ 7 + x ^ 3 = 12?
- 11. What is the area of the triangle ABC when the line y = four-thirds + 4 intersects with X at a, intersects with y at B, and O is the origin There are a few more. Give me 20 more ① What is the image line that the image of a linear function y = - 2x + 3 does not pass through ② It is known that the image of positive scale function y = K1 x intersects with the image of y = k2-9 at point P (3, - 6) If the first-order function y = k2x-9 intersects the x-axis at point a, the coordinates of a can be obtained,
- 12. Given that the line L passes through the point P (2,1), and intersects with the positive half axis of X axis and Y axis at two points a and B respectively, and O is the coordinate origin, then the minimum area of triangle OAB is? Why is 2 / A + 1 / b greater than or equal to 2 √ 2 / AB
- 13. When the inverse scale function y = K / X (k > 0) is known, the image passes through point a (2, m), where ab ⊥ X axis passes through point B, and the area of △ AOB is 3 [1] Find the direct [2] point C [x, y] of K and m on the image of inverse scale function y = K / x, find the value range of function value y when 1 ≤ x ≤ 3, take d [1,0] on the y-axis to connect BD, make a square with BD as the edge in the first quadrant, bdef asks whether e, f fall on the image of y = K / X
- 14. If the line y = KX + 1 and the ellipse X225 + y2m = 1 always have a common point, then the value range of the real number m is______ .
- 15. If the line y = K (x + 1) + 1 and the ellipse x2 / 5 + Y2 / M = 1 always have a common point, and the focus of the ellipse is on the x-axis, then the value range of M is
- 16. 1. It is known that for K ∈ R, the line y = KX + 1 and the ellipse (x ^ 2) / 5 + (y ^ 2) / M = 1 always have a common point, then the value range of the real number m is_______ . How can I figure it out wrong A (0,1) B (0,5) C [1,5] ∪ (5, positive infinity) d [1,5] I figured it out. It's wrong for me
- 17. How to calculate the distance from parabolic focus to hyperbolic asymptote?
- 18. It is known that the left and right focal points of hyperbola are F1F2, the eccentricity is following sign 2 and passing through point (4, - following sign 10) 1 to solve the equation of hyperbola
- 19. It is known that the asymptote equation of hyperbola C is y = ± 3x, and the distance from the right focus f (C, 0) to the asymptote is 3. (1) find the equation of hyperbola C; (2) make a straight line with slope k through F, l intersects the hyperbola at two points a and B, and the central perpendicular of line AB intersects the X axis at D
- 20. A focal coordinate is F1 (0, - 13), and the absolute value of the difference between a point P and two focal distances on the hyperbola is 24