How to draw x ^ 3 + y ^ 3 = 1 curve with Geometer's Sketchpad, 4 To the nth power

How to draw x ^ 3 + y ^ 3 = 1 curve with Geometer's Sketchpad, 4 To the nth power

1. Chart ------ new parameter, a = 3
2. Chart - draw new function, y = (1-x ^ a) ^ (1 / a), finish
3. You can double-click the parameter box to change the parameters to get 4 times, 5 times, etc
4. The image is a whole graph with odd number for a and part with even number for a. the other part is made according to the characteristics of even function

How to draw 75 (x-1) (Y-1) = x + y with Geometer's Sketchpad?

The Geometer's Sketchpad can only draw functions, not equations. Such equations need to be converted into the form of functions. But if you use other software, such as Yingren's Sketchpad, you can draw directly. The image looks like hyperbola

Can the students with Geometer's Sketchpad help me draw the image of y = 2 M2 + 4 m - 1 (M is x)

sure

The value range of y = | SiNx + sin | x | is But the former range is {0,1} and the latter is {- 1,1} The answer is {0,2} Use images be deeply grateful

The answer {0,2} is correct

The value of | sin | is | sin | x | be deeply grateful

0≤|sinx| ≤1
-1≤sin|x| ≤1
When 0 ≤ sin | x | ≤ 1, | SiNx | = sin | x | 0 ≤ SiN x | + sin | x | ≤ 2
When - 1 ≤ sin | x | ≤ 0, | SiNx | = - sin | x | SiN x | + sin | x | = 0
Therefore, the value range of y = | SiNx + sin | x | is 0 ≤ y ≤ 2

The graph and properties of the following functions are studied: ysin (arcsinx); y  arcsin (SiNx)

y=sin(arcsinx)
arcsiny=arcsinx
y=x x∈[-1,1]
z=arcsin(sinx)
sinz=sinx
z=x x∈R
You can see that the first curve is part of the second curve

The function y = SiNx and y = arcsinx are inverse functions. Why is it wrong?

Pay attention to their domain of definition and value
If functions are inverse functions of each other, then the definition domain of the original function is the value range of the inverse function
Definition domain of inverse function
We will use this to examine these two functions
Y = SiNx, the definition domain is (negative infinity, positive infinity), and the range is [- 1,1]
Y = arcsinx defines the domain as [- 1,1], and the range of values is [- pi / 2, PI / 2]
Obviously not

Image features and functional properties of y = arcsinx Please answer separately

In this paper, we discuss the properties of sin-2, odd value domain

The function y = sin (x-1 / 3) enlarges the abscissa of the image by 2 times and the ordinate remains unchanged The function y = sin (x-1 / 3) enlarges the abscissa of the image by 2 times, the ordinate remains unchanged, and then moves 1 / 3 pie to the left

When the abscissa of function y is expanded by two times
y=sin(x/2-π/3)
Then shift π / 3 units to the left
y=sin(x/2-π/6)

In order to get the function y = 3 (1 / 3) ^ x, how can the image of function y = (1 / 3) ^ X be moved? There has to be a process·· I know the answer is to shift three units to the right, but I don't understand

If you know that the answer is to shift right, then you have written a wrong question
The answer to your original question should be that the abscissa does not move, and the ordinate is expanded three times
According to your answer, the question should be y = (1 / 3) ^ (x-3)