How to find the global equation of a dual curve?

How do I find the equation of a tangent line to a curve?

• I'm given $x^2+2x-4$ at $x=2$ and I have to find the tangent line to this curve at that point...

• Answer:

  Here's an algebraic approach that avoids the explicit use of derivatives. We are given a quadratic function $f(x) = x^2 + 2x -4$, and we want to find the equation of the tangent to the parabola $y = f(x)$ at the point $(2, 4)$. (Note that $f(2) = 2^2 + 2 \cdot 2 - 4 = 4$.) Assume that it is given by the equation $$ y = m(x-2) + 4, \tag{$\ast$} $$ where $m$ is its slope. Let's consider the intersection of the parabola with the tangent; this is given by the system of equations $$ \begin{cases} y &=& x^2 + 2x - 4, \\ y &=& m(x-2)+4. \end{cases} $$ In other words, to find the intersection, we should solve the quadratic equation $ x^2 + 2x - 4 = m(x-2)+4$, or $$ x^2 + (2-m)x+(2m-8) = 0. \tag{$\ast\ast$} $$ Pictorially it is clear that the tangent meets the parabola meets in exactly one point, so we want $(\ast \ast)$ to have a unique solution. This implies that the discriminant of $(\ast \ast)$ vanishes: $$ \begin{array}{crcl} &(2-m)^2 - 4 \cdot (2m-8) &=& 0 \\ \implies \qquad & m^2 - 12m + 36 &=& 0. \end{array} $$ Conveniently (although this is not a numerical coincidence), this equation has a unique solution $m=6$: this is the solution we are after. Plugging $m=6$ in $(\ast)$, we get the equation of the tangent at $(2, 4)$ to be $y = 6x-8$.