{Use of Tech} Elliptic curves The equation y² = x³ - ax + 3, w...

Question

{Use of Tech} Elliptic curves The equation y² = x³ - ax + 3, where a is a parameter, defines a well-known family of elliptic curves.

a. Plot a graph of the curve when a = 3.

Accepted Answer

Hello. In this video, we are going to be graphing the given classical curve using analytical methods. We will also be using implicit differentiation when needed. So, we are given the equation Y2 is equal to X cubed minus 2 X + 3. Now, before we start graphing this function, let's just go ahead and discuss the domain of the function. Now, our function is in the form of a polynomial, and the convenient thing about this being a polynomial, is that the domain of a polynomial is all real numbers. So our domain for the function is going to be all real numbers from negative infinity to positive infinity. Now let's go ahead and calculate the Y intercept of the function. In order to calculate the Y intercept, we will be plugging in the value of 0 for X. When we do this, we are left with Y2 is equal to 3. And if we take the square root of both sides of this equation, we have two Y intercepts. One is equal to square root of 3, and the other at negative square root of 3, and that tells us that the two Y intercepts are located at 0.2 root of 3 and 0. negative square root of 3. Now what we want to do is we want to go ahead and calculate the X intercept. Now, in order to calculate the X intercept, we will be setting Y equal to 0. But the problem with doing this is that the polynomial X cubed minus 2 X + 3 is not factorable by any methods. So instead, what we can go ahead and do is we can graph the function Y is equal to the square root of X cubed. Minus 2 X + 3, and by using this function, we can calculate that the value of X is approximately equal to -1.893. So the X intercept will be located at 1.893.0. Now that we have the X and Y intercepts in the domain, let's go ahead and talk about the symmetry of the graph. Now, what we will go ahead and do is determine whether this is an even odd function, and if it has symmetry about Y is equal to X. So, what we will go ahead and do is determine if this graph has symmetry about the y axis. In order to determine if this graph has symmetry about the y axis, we will go ahead and plug in the value negative X into our function. That will allow us to rewrite the function as Y2 is equal to X cubed minus 2 multiplied by negative X plus 3. By simplifying this function, we are left with Y2 is equal to negative X cubed plus 2 X + 3. Since this is not the original function, we can conclude that there is no symmetry about the y axis. Now, what about symmetry about the X-axis? In order to determine if this graph has symmetry about the X-axis, we will go ahead and plug in the value negative Y into our function. When we do this, we get negative Y2 is equal to X cubed minus 2 X + 3, and negative Y2 will simplify to be Y2d, and this will give us the original function from the problem. Since this is the original function, this graph will have symmetry about the X-axis. Now we need to determine if there is going to be symmetry about the line, Y is equal to X. But the problem with this is that in order to determine this, we have to plug in negative X and negative Y. When we plug in negative Y, we will get the original left-hand equation, but the X version on the right hand side will not be the original. Therefore, we can conclude that there is no symmetry about the line Y is equal to X. Now that we've talked about symmetry, let's go ahead and calculate the critical points. In order to calculate the critical points, we are going to need to take the derivative of the given function. But since this is a function of the variables X and Y, we will need to take the derivative using implicit differentiation. So The the function given to us is Y2 equal to X cubed minus 2 X + 3. And we will go ahead and take the derivative with respect to X. On the left-hand side, the derivative of Y2d will be 2 Y and through implicit differentiation, we will generate a DYDX term. On the right hand side, the derivative of X cubed will be 3X2d. The derivative of -2 X will be -2, and the derivative of 3 will be 0. And finally, in order to solve for the DYDX derivative, we will go ahead and divide both sides of this equation by 2 Y, leaving us with DYDX is equal to 3X2 minus 2 divided by 2 Y. Now what we want to do is we want to find the values of the derivative that are 0. In order to do this, we will go ahead and take our numerator, set it equal to 0, and solve. That will be 3 X2 minus 2 equal to 0, and when we solve for the value of X, we will get 2 values of X. X is equal to positive and negative, square root, 2 divided by 3. Now, when we plug in this value into our function, we are going to get 4 different points. If we plug in positive square root of 2 into the original function, we get square root 2 divided by 3, approximation of 1.382. And we also get square root 2 divided by 3, 1.382. When we plug in the value of negative square root 2 divided by 3 into the original function, we get square root 2 divided by 3.2.022. And we get square root 2 divided by 3, 2.022. So, now that we have the locations of the critical points, we have the X intercepts and the Y intercepts, and we've talked about the domain. All we need to do now is just pick the graph that accurately represents this information. That graph in this case. This is going to be option C, and this is going to be the solution to this problem. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

{Use of Tech} Elliptic curves The equation y² = x³ - ax + 3, where a is a parameter, defines a well-known family of elliptic curves.a. Plot a graph of the curve when a = 3.

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{Use of Tech} Elliptic curves The equation y² = x³ - ax + 3, where a is a parameter, defines a well-known family of elliptic curves.

a. Plot a graph of the curve when a = 3.