If the line L and the line y = 2x-1 are symmetric about the X axis, find the functional relation of the line L and solve it by the method of function

If the line L and the line y = 2x-1 are symmetric about the X axis, find the functional relation of the line L and solve it by the method of function

If the point (x0, Y0) is on the line y = 2x - 1, then the point (x0, - Y0) is on the line L
Because: the point (x0, Y0) is on the line y = 2x - 1, so: Y0 = 2x0 - 1
It is - Y0 = - 2x0 + 1
That is (- Y0) = - 2x0 + 1
So the line L is y = - 2 x + 1

Given that the line y = 2x + 1 intersects with the y-axis at points a and B, and that point C and point a are symmetrical about the y-axis, the functional relationship expressed by the line passing through points B and C is obtained

It's obvious that a is the intersection of X and a
Let y = 0,2x + 1 = 0
x=-1/2
Let x = 0, y = 1
Then a (- 1 / 2,0), B (0,1)
According to the meaning C (1 / 2,0)
So the function of the line BC is
y=kx+1
Substituting (1 / 2,0)
1/2k+1=0
k=-2
y=-2x+1

If the image of a linear function is parallel to the straight line y = 2x and intersects the X axis with a point (- 3,0), then the relation of this function is__

Because the solution is parallel to y = 2x
Then the slope of the straight line k = 2
And because the point (- 3,0) intersects with the x-axis, then the relation y = KX + B of the first-order function
We get y = 2x + 6

The graph of a function passes through the point P (- 2.3) and is parallel to the line y = - 1 / 2x

The linear function is parallel to the straight line y = - 1 / 2x, let its function relation be y = - 1 / 2x + B, because it passes through the point P (- 2.3), and substituting into 3 = 1 + B, B = 2, so the function relation is y = - 1 / 2x + 2

It is known that in the plane rectangular coordinate system, O is the coordinate origin, a and B are two points on the x-axis, point a is on the left side of point B, the image of quadratic function y = AX2 + BX + C (a ≠ 0) passes through points a and B, and intersects with the y-axis at point C. (1) as shown in the figure, what is the relationship between the symbols of a and C? (2) If the length of line OC is the middle of the proportion of the length of line OA and ob, it is proved that a and C are reciprocal to each other; (3) under the condition of (2), if B = - 4 and ab = 43, the values of a and C are obtained

(1) It can be seen from the figure that when the opening of the parabola is downward, i.e. a < 0, C < 0; when the opening of the parabola is upward, i.e. a > 0, C > 0; therefore, a and C have the same sign. (2) let a (m, 0), B (n, 0), in the analytical formula of the parabola y = AX2 + BX + C, let y = 0, we get: AX2 + BX + C = 0, so OA · ob = Mn = CA; and oc2 = C

Drawing steps of Ln (1-x) image

Find the derivative of Ln (l-x) and draw the derivative image. The derivative is the slope, which can be used to draw the approximate image of Ln (l-x)

Drawing hyperbola with Geometer's Sketchpad

The mathematical principle of "new function" in Geometer's Sketchpad is: the definition of function - the value of Y is uniquely determined by the value of independent variable x, that is, it is enough to input a variable. The hyperbola you said is a standard hyperbolic equation for solving several equations. Because it is not a function, it cannot be drawn directly with "new function"

Geometer's Sketchpad for hyperbola
Is to use conic to do, such as ellipse, with two concentric circles at the intersection of the vertical line, and then track the point of animation
How to make hyperbola

There are many ways to do this. I'll just talk about one. If there are other ways, you can ask:
1. Draw a line AB and draw a point C on the line ab
2. Draw two points F1 and F2 again to make | F1F2 | > | ab|
3. The intersection of circle F1 and F2 is m, M & # 180;
4. Select points c and m to construct the track (or track point m is also OK), select points c and m to construct the track (or track point m ')

How to draw any curve in Geometer's Sketchpad

If you know the equation of a curve, it's simple, just like what you said on the first floor: first click "data → new function", input your function, confirm and then select the function you get, and then click "drawing → drawing function". If the curve is arbitrary and irregular, first use the pen in the toolbar to draw the curve you want, but

How to make a circle a straight line in the Geometer's Sketchpad

Roll a circle into a straight line or use parameters to draw point trajectory change! Please describe clearly, I can achieve