Consider the equation x(x2-4) = y(y2-1). Its graph is represented in the Figure I.
For the equation x(x2-4) = y(y2-1), the number 4 is changed with the numbers 5, 3, 2, 1, 1.1, 0.9, and -3 orderly. Now let us see the change of the graph between the interval [-10,10] y using a slider. The relevant video of the graph of the equation x(x2-n) = y(y2-1) is attached here. Please see the video of the graph.
While the number n is changing, an unusual event is observable. When n is 1, the equation is x(x2-1) = y(y2-1), and the graph becomes as it is seen in the Figure III below. To see teh vide of the equation, please see the video of the graph.
In the figure above, it can be noticed that when n is equal to 1, an ellipse and a line which is passing through the origin are obtained. For the equation x(x2-1) = y(y2-1), we have x3-x = y3-y.
Then it is x3-y3= x-y. Also, x3-y3= (x-y)(x2-y2+xy).Thus, we have (x-y)(x2-y2+xy)= x-y. Then (x-y)(x2-y2+xy)-(x-y)= 0 and (x-y)(x2-y2+xy-1)= 0. In this situation, x-y=0 or (x2-y2+xy-1)=0, so x=y or (x2-y2+xy-1)=0. As it is seen in the figure, the line x=y and the ellipse (x2-y2+xy-1)=0 are obtained two are intersecting in two points.
What happens if a constant is added to one side of the equation such that x(x2-4)= y(y2-1)+1?
The graph for the equaiton is given below in the Figure IV.
Now, I use Graphing Calculator to create a graph of a 3-D surface with the equation: x(x2-4)-y(y2-1)= z. Please look through the Figure V and see the video of the graph.
Let make up linear functions f(x) and g(x) and explore, with different pairs of f(x) and g(x) the graphs for
i. h(x) = f(x) + g(x
ii. h(x) = f(x).g(x)
iii. h(x) = f(x)/g(x)
iv. h(x) = f(g(x))
The graphs of the difrent pairs of the f(x) and g(x) are represented in the Figure VI and the following video. Please see the video of the graphs for i,ii, iii, and iv.
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