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Axis of symmetry of inverse scale function Inverse proportion function of grade three Don't talk to me We need the formula The solution of symmetry axis
There are two axes of symmetry
One is the bisector of the angle between the first and third quadrants, that is, y = X
One is the bisector of the angle between the second and fourth quadrants, that is y = - X
Is the inverse scale function an axisymmetric figure
The image of inverse scale function is both axisymmetric and centrosymmetric. It has two axes of symmetry, and the center of symmetry is the origin of coordinates
Prove that the inverse proportion function is a special hyperbola!
According to the principle of invariance of symmetry center, the coordinate system can be reconstructed or the rotation transformation of coordinates can be used
Literacy: let the original coordinates be (x, y), the transformed coordinates be (x1, Y1), (x0, Y0) be the coordinates of the new origin, and a be the angle of the coordinate axis
1. Translation: x = X1 + x0, y = Y1 + Y0; X1 = x-x0, Y1 = y-y0;
2. Rotation: x = x1cosa-y1sina, y = x1sina + y1cosa; X1 = xcosa + ysina, Y1 = - xsina + ysina;
3. General transformation: x = x1cosa-y1sina + x0, y = x1sina + y1cosa + Y0
If you don't have specific proof, I can give it to you
It is proved that the inverse proportion function is hyperbola
The standard formula of hyperbola is as follows:
X^2/a^2 - Y^2/b^2 = 1(a>0,b>0)
The standard form of the inverse proportional function is xy = C (C 0)
But the inverse proportion function is really a hyperbolic function obtained by rotation, and the angle of rotation can be set as a (A0)
There are
X = xcosa + ysina
Y = xcosa - ysina
X^2 - Y^2 = (xcosa+ysina)^2 -(xcosa - ysina)^2
= 4xy(cosasina)
= 4c(cosasina)
therefore
X^2/4c(cosasina) - Y^2/4c(cosasina) = 1 (4c(cosasina)>0)
Y^2/(-4c(cosasina)) - X^2/(-4c(cosasina)) = 1 (4c(cosasina)
No, it's wrong. It should be x = xcosa + ysina, y = ycosa xsina
This simple point can be obtained by multiplying the rotation matrix: [(COSA, Sina); (- Sina, COSA)] by the vector
Complex points need to see the figure calculation, with a variety of trigonometric formula derivation
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