Quadratic Functions¶
Recall (doc/postdoc). Not a first course.
Job: \(f(x)=ax^2+bx+c\) as Morse critical points / projective conics in \(1\)D; vertex = completing square.
Core¶
\[
f(x)=ax^2+bx+c=a\Bigl(x+\frac{b}{2a}\Bigr)^2+\frac{4ac-b^2}{4a},\quad a\neq0
\]
| Object | Formula |
|---|---|
| Vertex | \(\bigl(-\frac{b}{2a},\,f(-\frac{b}{2a})\bigr)\) |
| Axis | \(x=-b/(2a)\) |
| Range (\(\mathbb{R}\)) | \([f_{\min},\infty)\) if \(a>0\); \((-\infty,f_{\max}]\) if \(a<0\) |
| Graph | parabola; opens up/down by \(\operatorname{sign}a\) |
| Roots | zeros of \(f\); discriminant of \(ax^2+bx+c\) |
Vertex form: \(f(x)=a(x-h)^2+k\) — translation of \(a x^2\).
Second derivative test: \(f''=2a\); extremum at critical point of \(f'=2ax+b=0\).
Used later for¶
- Calculus: local max/min, Taylor degree \(2\)
- Optimization: quadratic programming
- Geometry: conic sections
- Probability: Gaussian \(\propto e^{-x^2/(2\sigma^2)}\)
Check¶
Is the vertex always a global min or max on \(\mathbb{R}\)?
Yes for genuine quadratics (\(a\neq0\)) — \(f\) is strictly convex/concave; unique critical point is global extremum.