Extreme values of quadratic function with errors

Question

Suppose I have this quadratic function (a least-square fitting result) with errors in its coefficients: $$ y = 0.06(\pm0.16)x^2-0.65(\pm0.04)x+1.2(\pm0.001) $$ The numbers after $\pm$ sign represent the errors of each coefficient. I'm interested in the error band of this function. I'm not pretty sure how I can find its boundary. Intuitively, the maximum value seems like happen when all the coefficients are positive: $$ y_{max} = 0.06\cdot(0.16)x^2+0.65\cdot(0.04)x+1.2\cdot(0.001) $$ and minimum $y_{min}$ otherwise (all negatives). However, I'm not sure if that's correct (for all intervals). Is it possible the extreme value happens when the errors are somewhere in the middle of $\pm0.16...$? Thanks:)

Answer 1

Let $y=ax^2+bx+c\;\;(a>0).\quad$ Then, by completing the square, $$y\geq c-\frac{b^2}{4a}.$$

Therefore—assuming that you meant the first coefficient to be $(0.06\pm0.016)$ instead of $(0.06\pm0.16)$—as $x$ varies, the given function has a minimum value at $$(1.2\pm0.001)-\frac{(-0.65\pm0.04)^2}{4(0.06\pm0.016)}.$$

As the coefficients' errors vary, the smallest such minimum is $$(1.2-0.001)-\frac{(-0.65-0.04)^2}{4(0.06-0.016)}=-1.50611\overline{36},$$ while the biggest such minimum is $$(1.2+0.001)-\frac{(-0.65+0.04)^2}{4(0.06+0.016)}=-0.02301\overline{315789473684210526}.$$

Hence, as the coefficients' errors and $x$ vary, the function's least possible value is $-1.50611\overline{36},$ with the absolute error being $1.483\overline{100478468899521531}$ upwards of this value.