(2014. The first mock exam in Zhangzhou) when alpha {-1, 1 When 2,1,3}, the quadrant that the power function y = x α cannot pass through is () A. Second quadrant B. The third quadrant C. Fourth quadrant D. The second and fourth quadrants

(2014. The first mock exam in Zhangzhou) when alpha {-1, 1 When 2,1,3}, the quadrant that the power function y = x α cannot pass through is () A. Second quadrant B. The third quadrant C. Fourth quadrant D. The second and fourth quadrants

Draw α = - 1, 1
The graph of power function y = x α at 2,1,3,
As shown in the figure,
According to the image, the quadrant that the image of the above function cannot pass through is the second and fourth quadrant
Therefore, D

Drawing with Mathematica I want to draw a cycloid, Please send out the specific operation steps, I'm a rookie, only copy and paste~ Besides, if I want to draw another tangent line of the cycloid, what should I do? Cycloid x = RA rcosa y = - R + rsina A is a parameter angle and R is a constant How can the tangent line be drawn? Also, the image coordinates are not suitable, too flat, how to modify it? It's the tangent line at any point. I'll use the auxiliary diagram to illustrate the correctness of the formula.

The code is as follows: r = 1; parametricplot [{r a - R cos [a], - R + R sin [a]}, {a, - 2 pi, 2 pi}] should not be wrong case

How does Mathematica draw function images Basically, how to draw f (x) = SiNx in rectangular coordinates, polar coordinates and space coordinates?

Plot[sinx,{x,-111,111}]
Space coordinate system plot3d
It's in the documentation

Mathematica the command to draw multiple three-dimensional function images in a graph First: x ^ 2 + y ^ 2 = 3Z ^ 2 The second: x + y + Z = 2A (a is an unknown constant) The third: z = XY Notice that it's in the same picture,

…… Students, look for an introduction book, or the next Chinese version of 8.0, the help is very easy to use:
a = 1; ContourPlot3D[{x^2 + y^2 == 3 z^2,
x + y + z == 2 a,
z == x y},{x,-1,1},{y,-1,1},{z,-1,1}]
A I give it at will. Change it if you want

Please use Mathematica or other software to draw a simple function graph α = α 2F2 + α 1 (1-f2), where α 1 and α 2 are constants, F2 is an independent variable and α is a dependent variable,

Manipulate[
Plot[a2 f2 + a1 (1 - f2), {f2, 1, 0}, PlotRange -> {0, 15}], {a1, 1,
10}, {a2, 1, 10}]

How to draw two function images at the same time with Mathematica One is f [x] = x ^ 5, and the other is g [x] = 5 ^ x, which requires that there is no scale on the coordinate axis ～

Show[Plot[x^5,{x,-10,10}],Plot[5^x,{x,-10,10}]]

The amplitude, period and initial phase of the function are written directly without drawing a picture, and how the function can be obtained by changing the sine curve is explained Y = 8sin (x / 2-pie / 4)

When the amplitude A = 8, the period T = 2 π / (1 / 2) = 4 π, the initial phase is - π / 4, the sine curve can first shift π / 4 units to the right; then keep the ordinate unchanged, the abscissa elongation is 2 times of the original; finally, keep the abscissa unchanged, the ordinate elongation is 8 times of the original, the image of function y = 8sin (x / 2 - π / 4) can be obtained

y=x*（1-1/e）,

This is the straight line passing through the origin with a slope of 1 - (1 / E) = (E-1) / e. it's easy to draw!

1. Draw the image of the function y = x? 2. Observe from the image that when x < 0, y increases with the increase of X or

For the parabola with y = x? 2 opening upward, the vertex is at the origin, and Y axis is the symmetry axis. When x < 0, y decreases with the increase of X

1. Draw the image of function y = 1 / 2 (x + 3) ^ 2 in rectangular coordinate system The function y = 1 / 2 (x + 3) 2 is drawn in rectangular coordinate system (1) The symmetry axis and vertex coordinates of the function image are pointed out; (2) According to the answer of the image, when x takes what value, y decreases with the increase of X? When x takes what value, y increases with the increase of X? When x takes what value, y takes the maximum or minimum value? (3) How to translate the image of function y = 1 / 2x 2 to get the image of function y = 1 / 2 (x + 3) 2? 2. It is known that the vertex of the parabola y = 1 / 5 (X-5) 2 is a parabola intersecting with y axis at point B as a parallel line of X axis, and the parabola intersects with another point C (1) Find the coordinates of a, B, C three points; (2) Find the area of △ ABC; (3) Try to judge the shape of △ ABC and explain the reason If there is a solution, if you can put the above picture better, hope to answer out in two hours, the math is good to enter!

One
(1) Axis of symmetry x = - 3; (- 3,0)
(2) When x = - 3, y is the smallest;
(3) Move three units to the left
Two
（1）A（5,0）；B（0,5）；C（10,5）；
（2）25
(3) Isosceles triangle;
I'm sorry to give you the answer