minor fix

documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex

@@ -150,7 +150,7 @@ Otherwise, there is only one solution $x_1 = 0$.
 
 \textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
 \begin{align}
-  d_{P,{f_2}}(x)  &= \sqrt{(x-z)^2 + (f(x)-w)^2}\\
+  d_{P,{f_2}}(x)  &= \sqrt{(x-z)^2 + (f_2(x)-w)^2}\\
     &= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 + w^2)}\\
     &= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 + w^2}\\
   0 &\stackrel{!}{=} \Big(\big(d_{P, {f_2}}(x)\big)^2\Big)' \\