Dear Uncle Colin,

I’m told that the graphs of the functions $f(x) = x^3 + (a+b)x^2 + 3x - 4$ and $g(x) = (x-3)^3 + 1$ touch, and I have to determine $a$ in terms of $b$. Where would I even start?

- Touching A New Graph Except Numerically Troubling

Hi, TANGENT, and thanks for your message!

For two graphs to touch, there needs to be a point that lies on both curves, and for the derivatives of the two curves at that point to be equal.

### The naive way

A naive approach, then, is to differentiate and find out where the gradients are the same, then use that information to find the common point.

Differentiating gives $f’(x) = 3x^2 + 2(a+b)x + 3$ and $g’(x) = 3(x-3)^2 = 3x^2 - 18x + 27$.

Those match when $3x^2 + 2(a+b)x + 3 = 3x^2 - 18x + 27$, or $(a+b+9)x = 12$ (*).

The graphs intersect where $f(x)=g(x)$, or $x^3 +(a+b)x^2 + 3x -4 = x^3 - 9x^2 + 27x -26$, which is to say $(a+b+9)x^2 - 24x + 22 = 0$. (**)

Substituting in the value of $x$ from (*) gives $\frac{144}{a+b+ 9} - \frac{288}{a+b+9} + 22 = 0$, so $11(a+b+9) = 72$, and the relationship between $a$ and $b$ can be determined from there.

### More elegantly

A more elegant way of saying the same thing is to note that subtracting the curves would lead to a point on the $x$-axis where the gradient is zero. We’ve already (effectively) worked that bit out: it’s equation (**).

In this context, that’s a quadratic that has a single real root! That means $(-24)^2 - 88(a+b+9) = 0$. That gives $11(a+b+9) = 72$ directly!

No, I haven’t written $a$ in terms of $b$. It’s not difficult to rearrange ($a = -\frac{27}{11}-b$), but I think I like it best in the form $11a + 11b + 27=0$.