Can we sketch $y = x^2(2-x)$, and from this, $y^2 = x^2(2-x)$?

Can we sketch $y = x^2(2-x)$, and from this, $y^2 = x^2(2-x)$?

Question

Draw a sketch-graph of the curve whose equation is

$$\begin{equation*} y = x^2(2-x). \end{equation*}$$

Hence, or otherwise, draw a sketch-graph of the curve whose equation is

$$\begin{equation*} y^2 = x^2(2-x) \end{equation*}$$

indicating briefly how the form of the curve has been derived.

Solution

The polynomial on the right-hand side has a double zero at the origin and a zero at $x = 2$.

Further,

$$\begin{equation*} \frac{dy}{dx} = -3x^2 + 4x = x(4-3x) \end{equation*}$$

so the curve has two stationary points, at $x = 0$ and $x = \dfrac{4}{3}$.

Since the leading term of the polynomial is $-x^3$, $y \to \infty$ as $x \to -\infty$, and $y \to -\infty$ as $x \to \infty$.

Consequently, the first stationary point must be a local minimum and the second stationary point must be a local maximum.

This information, when put together, leads to the following sketch.

Firstly, note that this curve does not exist where the right hand side is negative, that is when $x>2$. This corresponds to where the curve in the first part of the question is below the $x$-axis.

For other values of $x$, the curve has two branches, $y=\pm x\sqrt{2-x}$.

The curve cuts or touches the $x$-axis at $x=0$ and at $x=2$.

So we have some kind of loop above and below the $x$-axis between $x=0$ and $x=2$, while for $x<0$ the curve is in two branches, one increasing and the other decreasing.

In order to discover the nature of the axis crossings, we could think about how $y=\sqrt{2-x}$ behaves and what happens when we multiply it by $x$. Alternatively, we can do some calculus.

Differentiating implicitly

we find

$$\begin{align*} y^2 &=2x^2-x^3 \\ \implies\quad 2y\frac{dy}{dx} &= 4x-3x^2 \\ \iff\quad \frac{dy}{dx} &=\frac{4x-3x^2}{2y} \end{align*}$$

As $x$ approaches $2$, $y\to 0$ and $\frac{dy}{dx}\to\frac{8-12}{2y}\to\pm\infty$. So a tangent to the curve at $x=2$ would be vertical.

We can’t evaluate $\frac{dy}{dx}$ at the origin, but nearby $x$ is very small so $y^2\approx2x^2$. This means that $y\approx\pm\sqrt{2}x$ and we have $\frac{dy}{dx}\approx\frac{4x-3x^2}{\pm 2\sqrt{2}x} \approx \pm\sqrt{2}.$ Hence the curve passes through the origin with gradient $\pm\sqrt{2}$.

The curve whose equation is

$$\begin{equation*} y^2 = x^2(2-x) \end{equation*}$$

has two parts, each the reflection of the other in the $x$-axis:

$$\begin{equation*} y = \pm \sqrt{x^2(2-x)} \quad\text{where }x \le 2\text{.} \end{equation*}$$

The restriction that $x \le 2$ is necessary, since $x^2(2-x) \ge 0$ if and only if $x \le 2$. We have a graph of the function $y = x^2(2-x)$, so we need to think how a graph is transformed by the map $y \mapsto \sqrt{y}$. The three points where $y = 1$ are invariant; both the old curve and the new curve go through all three points. The new curve will be ABOVE the old curve for $0

1$, since in this region,$y^2 > y$. But the map$x \mapsto \sqrt{x}$is reasonably well-behaved; as$x$increases,$\sqrt{x}$increases, and as$x$decreases,$\sqrt{x}$ decreases. In particular, the stationary points remain stationary points as, by the

,

$$\begin{equation*} \frac{d}{dx} \left( \sqrt{f(x)} \right) = \frac{f'(x)}{2\sqrt{f(x)}} \end{equation*}$$

when $f(x)$ is a function such that $f(x) > 0$.

We are led, then, to the following sketch.

** Both curves are plotted together below.