# In quadratic interpolation of 3 points - is the minimum guaranteed to be within the left and right points?

I have 3 ordered points, $$(x_1,y_1),(x_2,y_2),(x_3,y_3),$$ with $$x_1, of which I calculate the interpolationg quadratic function. In my case, $$(x_2,y_2)$$ is a local minimum in or maximum in this list (i.e. ($$y_2>y_1$$ and $$y_2>y_3$$) or ($$y_2 and $$y_2). My question: is the parabola's minimum or maximum guaranteed to be within the support of these points, so that $$x_1?

I thought about just working out the formula for $$x_\text{min}$$, but surely there is a more easy argument?

• Of course not. Take a parabola whose minimum is at $x=17$, and take $x_1,x_2,x_3$ all bigger than (or all smaller than) $17$. – Gerry Myerson Feb 28 at 2:59
• But that one does not satisfy the ordering of the y-values as either $y_3>y_2>y_1$ or $y_3<y_2<y_1$ in that case. In my case $y_2$ is a 'bump' in the list of points, as indicated in the question. – Ewoud Feb 28 at 3:12
• OK. Well, a parabola has exactly one turning point, so if the value at an intermediate point is bigger than (resp., or smaller than) the value at two outer points, then it must have a maximum (resp., minimum) between the two outer points. – Gerry Myerson Feb 28 at 5:40
• Seems so obvious now... Thanks anyway :) – Ewoud Feb 28 at 6:03

Assume $$x_\text{min} < x_1$$. Then if $$x_1, $$y_1,y_2,y_3$$ is a monotonic sequence because the parabola is monotonic after $$x_\text{min}$$. Similarly for $$x_\text{min}>x_3$$. Hence, $$x_1.