Some time ago I read about a geometric application of polynomial division to finding the equations of tangents to points on a polynomial curve. This is obviously something which is more easily handled with basic differentiation, but still! I teach polynomial division, factor and remainder theorems about six weeks into my course and mention this application to my students. Occasionally I get asked to prove it but don’t have time in class. So here is my sketch of a proof for this claim.

Example: given the curve \[ f(x) = x^3 + x^2 -4x -6 \] We can find the equation of the tangent at the point \( x=2 \) by dividing \( f(x) \) by \( (x−2)^2 \).

Therefore, the equation of the tangent is \( y=12x−26 \).

The claim is that for any polynomial curve \( y=f(x) \) then the tangent line at the point where \( x=a \) can be found by finding the remainder after carrying out the division \( f(x)/(x−a)^2 \).

Here is my attempt at proving the result. First a picture of a function with a tangent to the point \( x=a \).

First, note that polynomial division results in a quotient and remainder. In this case: \[ \frac{f(x)}{(x-a)^2} = q(x) + \frac{r(x)}{(x-a)^2} \] which can be easily rearranged to give \[ f(x) = (x-a)^2 q(x) + r(x) \] A couple of things to note. First \( r(x) \) must be linear, since it is a degree less than \( (x−a)^2 \), which is quadratic.

Second, at the point \( x=a \) we find that \[ f(a) = r(a) \] In other words, \( f \) and \( r \) have the same value at \( x=a \).

The next bit of proof uses some differentiation. Applying the product rule we get: \[ f^{\prime}(x)=2q(x)(x−a)+q^{\prime}(x)(x−a)2+r^{\prime}(x) \] At the point \( x=a \) we note that \[ f^{\prime}(a)=r^{\prime}(a) \] In other words \( f^{\prime}(x) \) and \( r^{\prime}(x) \) have the same slope/gradient at \( x=a \). This is necessary if \( r(x) \) is a tangent line.

Since \( r(x) \) is linear, it has the form \[ r(x)=mx+c \] and we know straight lines are completely specified by knowing their gradient \(f^{\prime}(a) \) and a point \( (a,f(a)) \) on it.

Therefore \( y=r(x) \) must be a tangent line to the curve \( y=f(x) \) at the point \( x=a \).

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