A curve-sketching masterclass

An implicit differentiation question dealt with $y^4 – 2x^2 + 8xy^2 + 9 = 0$. Differentiating it is easy enough for a competent A-level student – but what does the curve look like? That requires a bit more thought.

My usual approach to sketching a function uses a structure I call DATAS:

  • Domain: where is the function defined?
  • Axes: where does the function cross them?
  • Turning points: where are the function's critical points?
  • Asymptotes: what happens when $x$ or $y$ get really big?
  • Shape: what does the graph look like overall?

As it turns out, this approach is almost entirely useless here: we're not even looking at a function! It's not at all clear from first inspection what $x$ values are consistent with $y$, setting $x=0$ reveals that the curve doesn't cross the $y$-axis (although $y=0$ leads to $x^2 = \frac{9}{2}$, which at least gives us a couple of points on the curve.) Critical points? The gradient is never zero.

Asymptotes are a bit more helpful: as $y$ and $x$ get enormous, the curve approaches $y^4 = x^2$, which looks like $y=\sqrt{x}$, but reflected in both axes.

But still, there's not really anything to build a shape on. Instead, I'm going to have to do some Actual Maths.

The first thing I notice is that I've got $y^4$ and $y^2$ in there, which suggests I might be able to complete the square to get it into a nicer form. Indeed, it sorts itself out fairly nicely as $(y^2 + 4x)^2 = 18x^2 – 9$, with a little bit of work. But how to sketch that?

Let's play with the variables a bit. If I call the bracket on the left $z$, just to see what happens, I get $z^2 = 18x^2 -9$. That's a hyperbola, crossing the $x$-axis when $x = \pm \frac{\sqrt{2}}{2}$, and approaching the lines $z = \pm 3x\sqrt{2}$ as $x$ gets large. Now it's just a case of transforming.

Now, if I let $w = y^2$, I've got $z = w + 4x$ – so I can transform from $z$-space to $w$-space by shifting the graph "down" by $4x$, which I suppose is a shear. The easiest way for me to picture this is to consider how the asymptotes move: the one that was $z = 3x\sqrt{2}$ becomes $w = (3\sqrt{2}-4)x$, which remains a positive gradient, but only just. The other, $z=-3x\sqrt{2}$, becomes much steeper, $w = (3\sqrt{2}+4)x$.

The overall effect on the hyperbola is to push it downwards on the right and upwards on the left – with the crest on the left above the $x$-axis and the one on the right below.

One last transformation! We need to replace $w$ with $y^2$. Clearly, $y^2$ only takes on positive values, so we can ignore anything below the $x$-axis. On the right, things are simple: instead of approaching the straight line, the 'hyperbola' now approaches a pair of square root curves. It's symmetrical about the $x$-axis, and is vertical when it crosses (at $x = \frac 32 \sqrt{2}$, as previously mentioned.)

On the left is where things get interesting! The part of the $w$-space curve above the $x$-axis doubles back on itself – which means the $y$-space curve does the same thing, but now with a reflection below the axis. The curve is vertical when $x = -\frac{\sqrt{2}}{2}$ and when $x = -\frac32 \sqrt{2}$ (as it crosses the axis), making a sort of 3 shape to the left of the $y$-axis.

And we're done! I think it looks like a fish smoking a cigar, but others suggest less innocent interpretations.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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