Previous page : y^3+y^3=x^2+x

Next page : y^3+x^2=y/x

Let us look at

We can immediately see that the point (0, 0) is on the graph.

**Zeroes**

If *y*=0 we get

So we have, the already known, zero at *x*=0

If *x*=0 we get

So we have *y*-intercepts at *y*=0 and *y*=−1.

So we have the points (0, 0), again, and (0, −1).

**Asymptotes**

Say *y* goes toward infinity then we basically have

or

and thus

So we should expect that *y*=*x*^{2} and *y*= –*x*^{2} to be asymptotes.

What if *x* goes toward infinity? This is quite less obvious, at least to start with, since we have the term *x*^{4}*y*. But if x and y are very small we get that

will behave as

or

We could thus expect the graph to cross itself at the origin, and that the two parts will have slopes of 1 and –1 respectively.

**The intersection between the asymptotes and the relation**

Since we have an *x*^{2} term in the relation it might be an idea to see if the substitution

We get

or

So we now know we have points at (±1, ±1)

**Horizontal tangent points, i.e. stationary points
**

We may find these by finding where *y*´=0. So we take the implicit derivative of our relation to get

Here we needed both the chain rule and the product rule. For *y*´=0 this gives us

This is a bit problematic since we have both x and y in the equation. We could make y the subject and substitute that into the original equation, but that would give us an eight-degree polynomial to solve, so let us instead make y´ the subject to see what slopes we have at the known points. We get

Or

And thus

For the point (0, −1) it gives us

For the point (0, 0) gives us that the derivative is undefined – as is expected.

For the vertical tangent points, we get into the same problem as with the horizontal slopes. We get an expression for the slope that is dependent on both *x* and *y*. We get that

Could it be that any of our known points satisfies this? We know (0,0) is a point where the graph intersects itself, and that the graph is horizontal at (0,-1). But how about the (pm1,pm1). The sign of the *x*-coordinate cannot matter, since we take that to the power of four. So, let us test

For this to be true *y*=−1. So we have that the graph is vertical at (1, −1) and (−1,−1).

We could also have made* y* the subject of

To get

So any possible points where the graph is vertical could be along the two lines given by the above equation. We could calculate this for a few points and connect the points. For x=0 we get y=0 and y=-2/3, for x=1 we get y=1/3 and y=-1 (as expected). For x=1/2 we get

I.e. just slightly above 0 or just slightly below −2/3.

Putting it all together (using paint, and the mouse (implies ugly)). I get

The dark lines are either asymptotes, or, if they have vertical lines on them, they are places where one could possibly have vertical stretches of the graph. So, next, I tried to draw some kind of graph that could fulfil this (it would look rather much better on a piece of paper).

So, how does it compare to the actual graph?

Not too bad.

Up a level : Differentiation, derivativesPrevious page : y^3+y^3=x^2+x

Next page : y^3+x^2=y/xLast modified: