Recipe

Run quadratic-fit search on three bracket points

Use when: you have three points $x_1<x_2<x_3$ with known values $f_1,f_2,f_3$ from a bracket that traps the minimum, and want to exploit HOW MUCH lower one point is instead of only comparing which is smaller

Goal: a new candidate minimizer at the fitted parabola's vertex, replacing one of the three points for the next fit

Taught in Quadratic-fit search

Quick recipe

Fit the parabola through the three points and find its vertex
Evaluate $f$ at $x^*$ and confirm it lands inside the bracket
Discard the worst of the four points, keeping three for the next fit
Repeat until $x^*$ stops moving by more than the tolerance

1
Compute $x^*=x_2-\frac12\cdot \frac{(x_2-x_1)^2(f_2-f_3)-(x_2-x_3)^2(f_2-f_1)} {(x_2-x_1)(f_2-f_3)-(x_2-x_3)(f_2-f_1)}.$

Example

For $(x_1,x_2,x_3)=(0,2,5)$ with $(f_1,f_2,f_3)=(5,1,10)$ : $x_2-x_1=2$ , $x_2-x_3=-3$ , $f_2-f_3=-9$ , $f_2-f_1=-4$ . Numerator: $(2)^2(-9)-(-3)^2(-4)=-36+36=0$ , so $x^*=2-0=2$ regardless of the denominator.

Why: exactly one parabola passes through three points, and its vertex is that quadratic's unique closed-form minimum — no iteration is needed just to compute it.
2
Compute $f(x^*)$ , then check $x^*\in(x_1,x_3)$ before trusting it.

Example

For $(x_1,x_2,x_3)=(0,2,5)$ with $(f_1,f_2,f_3)=(4.6,1,11.35)$ : the fit gives $x^*\approx1.857$ (worked out below), which lies inside $(0,5)$ , so the fit is trusted and $f(1.857)\approx1.0203$ is evaluated.
- $x^*\in(x_1,x_3)$ — continue to Step 3 with the new point.
- $x^*\notin(x_1,x_3)$ — the fit isn't trustworthy yet; take one golden-section-search step instead (Section 6's recipe) and retry the fit once the bracket has updated.
Why: nothing in the vertex formula forces $x^*$ back inside the current bracket, especially early on when the three points are still far apart and $f$ doesn't yet look parabola-like between them.
3
Among $\{x_1,x_2,x_3,x^*\}$ , drop whichever point now has the highest $f$ -value, and sort the remaining three back into $x_1<x_2<x_3$ order.

Example

Continuing with $(x_1,x_2,x_3)=(0,2,5)$ , $(f_1,f_2,f_3)=(4.6,1,11.35)$ , and $x^*\approx1.857$ with $f(x^*)\approx1.0203$ : the highest value among all four is $f_3=11.35$ at $x_3=5$ , so it's discarded. Sorting the survivors gives the new triple $(0,\,1.857,\,2)$ with values $(4.6,\,1.0203,\,1)$ .

Why: keeping the three points closest to the emerging minimum is exactly the "discard the losing side" logic every bracketing method in this chapter uses, just applied to four candidates instead of two.
4
Refit the parabola through the new triple, compute a new $x^*$ , and repeat Steps 2-3. Stop once successive $x^*$ values differ by less than the required tolerance, and report the last $x^*$ as the estimated minimizer.

Why: convergence is superlinear once the points are close to the true minimum, so each refit typically buys more accuracy than the last — a numeric stopping rule on how much $x^*$ actually moved fits that better than a fixed iteration count.

Common mistakes

Trusting $x^*$ without checking it lands inside $(x_1,x_3)$ — the fitted parabola can extrapolate to a point outside the current bracket entirely.
Skipping the golden-section fallback step when $x^*$ falls outside the bracket, and using the out-of-range point anyway.
Discarding the wrong point — the survivors should be the three points closest to the emerging minimum, not simply "the two originals plus $x^*$ ."
Assuming a single fit always finds the exact minimum — only true when $f$ is genuinely quadratic; otherwise $x^*$ is one iterate in a sequence, not a final answer.