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Let us look at
We can immediately see that the point (0, 0) is on the graph.
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).
Say y goes toward infinity then we basically have
So we should expect that y=x2 and y=−x2 to be asymptotes.
What if x goes toward infinity? This is quite less obvious, at least to start with, since we have the term x4y. But if x and y are very small we get that
will behave as
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.
Intersection between then asymptotes and the relation
Since we have an x2 term in the relation it might be an idea to see if the substitution
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
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
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 then, 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, derivatives
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