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 \(\eqref{eq:1repeat}\) and \(\eqref{eq:2repeat}\) 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 \(\eqref{eq:1repeat}\) and \(\eqref{eq:2repeat}\) touch at \((2,8)\).
- Sketch on the same axes the curves \(\eqref{eq:1repeat}\) and \(\eqref{eq:2repeat}\) when \(a = 2\).
When \(a = 2\), \(\eqref{eq:1repeat}\) 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 \(\eqref{eq:1repeat}\) with \(a = 2\) intersects the \(y\) axis at \((0,-32)\).
So when \(a = 2\), the graphs must have the following shape.
- Sketch on the same axes the curves \(\eqref{eq:1repeat}\) and \(\eqref{eq:2repeat}\) when \(a = 1\).
When \(a = 1\), \(\eqref{eq:1repeat}\) 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 \(\eqref{eq:1repeat}\) with \(a = 1\) intersects the \(y\) axis at \((0,-24)\).
So when \(a=1\), the graphs have the following shape.
- Sketch on the same axes the curves \(\eqref{eq:1repeat}\) and \(\eqref{eq:2repeat}\) when \(a = -2\).
When \(a = -2\), \(\eqref{eq:1repeat}\) 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.
