Can we show these two cubic curves touch?

Can we show these two cubic curves touch?

Question

A curve is given by the equation

$$\begin{equation*} y = ax^3 - 6ax^2 + (12a + 12)x - (8a + 16), \tag{$*$} \end{equation*}$$

where

$a$

is a real number. Show that this curve touches the curve with equation

$$\begin{equation*} y = x^3 \tag{$*{*}$} \end{equation*}$$

at $(2,8)$. Determine the coordinates of any other point of intersection of the two curves.

1. Sketch on the same axes the curves $(1)$ and $(2)$ when $a = 2$.

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1. Sketch on the same axes the curves $(1)$ and $(2)$ when $a = 1$.

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1. Sketch on the same axes the curves $(1)$ and $(2)$ when $a = -2$.

Interactive graph

A curve is given by the equation

$$\begin{equation*} y = ax^3 - 6ax^2 + (12a + 12)x - (8a + 16), \tag{$*$} \end{equation*}$$

where

$a$

is a real number. Show that this curve touches the curve with equation

$$\begin{equation*} y = x^3 \tag{$*{*}$} \end{equation*}$$

at $(2,8)$.

You might like to use this applet to explore the behaviour of the curve $(eqi:1)$ as $a$ varies. The blue curve is $y=x^3$ $(eqi:2)$.

You can also zoom in or out if that helps to see the behaviour of the curves.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/NCk55VUa/width/700/height/500/border/888888/rc/false/ai/false/sdz/true/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

As $a$ varies, you might notice the following.

• The curves always touch at $(2,8)$ as the question suggests.

• What happens to the red curve when $a=0$?

• Are there any values of $a$ for which the curves do not intersect again?

Solution

A curve is given by the equation

$$\begin{equation*} y = ax^3 - 6ax^2 + (12a + 12)x - (8a + 16), \tag{$*$} \end{equation*}$$

where

$a$

is a real number. Show that this curve touches the curve with equation

$$\begin{equation*} y = x^3 \tag{$*{*}$} \end{equation*}$$

at $(2,8)$. Determine the coordinates of any other point of intersection of the two curves.

Let $f(x) = ax^3 - 6ax^2 + (12a + 12)x - (8a + 16)$ and $g(x) = x^3.$

To find the points of intersection, we set

$f(x) = g(x)$

to get

$$\begin{align*} & ax^3 - 6ax^2 + (12a + 12)x - (8a + 16) = x^3 \\ \iff & (a - 1)x^3 - 6ax^2 + (12a + 12)x - (8a + 16) = 0 \\ \iff & (x - 2)[(a - 1)x^2 - (4a + 2)x + (4a + 8)] = 0 \\ \iff & (x - 2)^2[(a - 1)x - (2a + 4)] = 0. \end{align*}$$

So the points of intersection have coordinates (2,8) and $\left(\frac{2a+4}{a-1},\left[\frac{2a+4}{a-1}\right]^3\right).$

The double root for $x = 2$ means that we have a touching there; the two curves are tangent to each other at $(2,8)$.

Alternatively…

To show that the curves $(1repeat)$ and $(2repeat)$ touch at $(2,8)$, we need to show that both curves pass through $(2,8)$ and that the gradients of the curves are equal at this point.

Now $f(2) = 8a - 24a + 2(12a + 12) - (8a + 16) = 8$ and $g(2) = 8,$ so both curves pass through the point $(2,8)$.

Also $f'(2) = 12a - 24a + (12a + 12) = 12$ and $g'(2) = 12,$ so the curves have the same gradient at the point $(2,8)$.

Therefore the curves $(1repeat)$ and $(2repeat)$ touch at $(2,8)$.

1. Sketch on the same axes the curves $(1repeat)$ and $(2repeat)$ when $a = 2$.

When $a = 2$, $(1repeat)$ becomes $y = 2x^3 - 12x^2 + 36x - 32.$ From above, we have that the two graphs intersect at $(8,512)$ and touch at $(2,8)$.

Now differentiating gives

$$\frac{dy}{dx} = 6x^2 - 24x + 36,$$

so stationary points occur when

$$\begin{align*} & 6x^2 - 24x + 36 = 0 \\ \iff & x^2 - 4x + 6 = 0 \\ \iff & (x - 2)^2 = -2. \end{align*}$$

This equation has no real solutions, so there are no stationary points.

Further, we can see from this equation that $y'>0$ for all $x$, and so $y$ is an increasing function.

Since $y<0$ when $x=1$ but $y>0$ when $x=2$, we know that it has a root (which must be the only root since $y$ is increasing) somewhere between $1$ and $2$.

We also know that $(1repeat)$ with $a = 2$ intersects the $y$ axis at $(0,-32)$.

So when $a = 2$, the graphs must have the following shape.

1. Sketch on the same axes the curves $(1repeat)$ and $(2repeat)$ when $a = 1$.

When $a = 1$, $(1repeat)$ becomes $y = x^3 - 6x^2 + 24x - 24.$ The two graphs touch at $(2,8)$ but do not intersect elsewhere because $\dfrac{2a+4}{a-1}$ is undefined when $a = 1$.

Differentiating gives

$$\frac{dy}{dx} = 3x^2 - 12x + 24,$$

so stationary points occur when

$$\begin{align*} & 3x^2 - 12x + 24 =0 \\ \iff & x^2 - 4x + 8 = 0 \\ \iff & (x - 2)^2 = -4. \end{align*}$$

This equation has no real solutions, so there are no stationary points.

Again, this means that $y'>0$ for all $x$, and so $y$ is an increasing function.

Since $y<0$ at $x=1$ but $y>0$ at $x=2$, we know that the graph of $y$ crosses the $x$-axis somewhere between $x=1$ and $x=2$, which should help our sketch.

We also know that $(1repeat)$ with $a = 1$ intersects the $y$ axis at $(0,-24)$.

So when $a=1$, the graphs have the following shape.

1. Sketch on the same axes the curves $(1repeat)$ and $(2repeat)$ when $a = -2$.

When $a = -2$, $(1repeat)$ becomes $y = -2x^3 + 12x^2 - 12x=-2x(x^2-6x+6).$ We can find that $y$ has roots at $x=0,3\pm\sqrt{3}$, or approximately $x=0, 1.3,4.7$.

The two graphs intersect at $(0,0)$ and touch at $(2,8)$.

Differentiating gives

$$\frac{dy}{dx} = -6x^2 + 24x - 12,$$

so stationary points occur when

$$\begin{align*} & -6x^2 + 24x - 12 = 0 \\ \iff & x^2 - 4x + 2 = 0 \\ \iff & x = 2 \pm \sqrt{2}. \end{align*}$$

So there are stationary points at $x \approx 3.4$ and $x \approx 0.6$.

So when $a=-2$, the graphs have the following shape.