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

Next page : y^3+y^2=(x^4)y+x^2

Let us look at

Here we can make *x* the subject, but just for the sake of it, let’s not do it. Instead we will analyse it in a similar way as the previous example. It will also be a case where we can test our methods without the need for some tool that can do implicit plots. Any calculator will do.

**Zeroes**

If *x* is 0 then we get

And thus that *y*=0 or *y*=−1.

For *y*=0 we get

And thus *x*=0 or *x*=−1.

**Other obvious solutions**

The point (1, 1) is quite obviously on the graph.

**Asymptotes**

If *x* goes toward infinity *y* will go toward infinity, since in both cases *x*^{2}+*x* will be positive and *y*^{3}+*y*^{2} will be negative for negative values of *y*. For large x, is dominated by *x*^{2} and *y*^{2} is dominated by *y*^{3}, so we get that the graph tend to

or

close to 0. Technically you might not get an asymptote using the trick of discarding term that grows much slower than the dominant term, but you will at least get something that behaves somewhat the same.

If *x* and *y* is close to 0 we get that our relation is close to

Since *y*^{3} and *x*^{2} goes toward 0 much faster than *y*^{2} and *x*. We should thus expect the graph to behave like

close to the origin.

**Symmetries**

We have two zeroes at −1 and 1. It can be worthwhile to try to shift the relation 1/2 step in the positive direction to see if we get a symmetric relation, and indeed

We can now see that we have a symmetry about the *y*-axis, and thus around the *y*=−1/2 for the original relation. That means that we know that the points (−1,−1) and (−2,1) will be on the graph too.

**Points with horizontal tangents**

To find these we find *y*´=*dy/dx* of the relation using implicit derivatives. We get

For *y*´=0 we get

Or *x*=−1/2, i.e. at the line of symmetry. It might be easiest to use the symmetric version with *x*=0.

This gives us that y is about −1.18.

**Points with vertical tangents**

We can now find x´=*dx/dy* using implicit derivatives, or take the reciprocal of y´. We have

or

So to find points where *x*´=0 we can find where

I.e. at *y*=0 and *y*=−2/3

For *y*=0 we already have that *x*=−1 or *x*=0, so we have a vertical slope at those two points.

For y=−2/3 we get

This gives, using an ordinary calculator *x*=0.13 and *x*=−1.13 (rounded to two decimals).

**Slopes**

We can use

To find the slope at the other points we have found, in particular (0,-1) and (1,1)

For (0,−1) we get y´=1 and thus, by symmetry, −1 at (−1,−1).

For (1,1) we get *y’*´=3/5 and thus, by symmetry, −3/5 at (−2,1).

**Putin it all together**

So let us see what we have:

It seems rather obvious how to connect these, and indeed we have the graph:

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

Next page : y^3+y^2=(x^4)y+x^2Last modified: