Derive and use quadratic rate patterns

Try

Consider the synthetic model Q(x)=−x²+6x+2 on the restricted domain 0≤x≤5. Inputs and outputs are numerical model units. Compute Q at the six integer inputs and list the changes over each unit interval. The values are 2,7,10,11,10,7. The changes 5,3,1,−1,−3 decrease steadily even though the outputs initially increase.

Predict where the model’s largest output occurs. Also predict whether Q(5)=7 is the smallest output just because the curve is falling near its right endpoint. A local direction of change does not by itself compare every point in the stated domain.

Predict

What happens to those interval rates if the starting input moves right by one? What if the interval width changes from one to two? Record both predictions. The phrase “the rate at x” is incomplete here: our rate belongs to two endpoints, x and x+h, with positive width h.

Do not extend a contextual model beyond its stated domain merely because a polynomial can be evaluated there. We will distinguish the unrestricted quadratic’s end behavior from the extrema of the restricted model.

Experiment

Interval	Output change	Average rate
[0,1]	7−2=5	5
[1,2]	10−7=3	3
[2,3]	11−10=1	1
[3,4]	10−11=−1	−1
[4,5]	7−10=−3	−3

Now calculate on [1,3] and [2,4], both with width two. The rates are (11−7)/2=2 and (10−10)/2=0. A zero average on [2,4] does not say the function is constant: it rises to 11 and falls back to 10. Equal endpoint outputs can hide a turning point.

Understand

Let q(x)=ax²+bx+c, and let h>0 with both endpoints in the model domain. Expand before dividing:

q(x+h)−q(x)=a(x²+2xh+h²)+b(x+h)+c−(ax²+bx+c)=2axh+ah²+bh.

Every term contains h. Dividing by the positive interval width gives

[q(x+h)−q(x)]/h=a(2x+h)+b.

The constant c disappears because a vertical translation changes neither endpoint difference nor rate. For fixed h, moving the starting input by Δx changes this rate by 2aΔx. Thus rates across equally wide intervals increase when a>0, decrease when a<0, and stay constant when a=0. When a=0 the formula describes a linear function rather than a genuine quadratic. This derivation uses finite interval algebra; no instantaneous-rate claim is needed.

A quadratic with a>0 is concave up: its graph lies below its secant chords between their endpoints. With a<0 it is concave down and lies above those chords. Choose input endpoints u<v in the domain and a fraction s with 0≤s≤1. The corresponding input is (1−s)u+sv, and the chord height is (1−s)q(u)+sq(v). The chord-minus-curve difference there is a·s(1−s)(v−u)², so its sign follows directly from a. In this worked model a=−1, making Q concave down throughout its domain even on the part where Q is increasing. Concavity describes how rates and chord positions behave, not the sign of the output or of every interval rate.

For Q, a=−1 and b=6, so the rate is 6−2x−h. With h=1 this becomes 5−2x, reproducing the table. With h=2 it becomes 4−2x, giving 2 on [1,3] and zero on [2,4]. Changing h also changes the interval’s midpoint. Indeed a(2x+h)+b=2a(x+h/2)+b, so quadratic secant rates depend on that midpoint. Equal midpoints give equal secant slopes even for different widths, provided all endpoints are allowed.

Complete the square: Q(x)=11−(x−3)². The squared term is never negative, so the largest output is 11, attained at x=3. On [0,5], the farthest allowed input from 3 is 0, at distance 3; hence the minimum is Q(0)=2. This argument compares the whole interval, not just the six measured-looking dots.

For any u<v≤3 in the domain, Q(v)−Q(u)=(v−u)(6−u−v)>0. For 3≤u<v, the second factor is negative and the output decreases. The vertex separates these trends. Notice that a downward-opening quadratic can have positive interval rates to the left: decreasing rates do not mean all rates are negative.

Polynomial turning and inflection without differentiation

A graph can have a local extremum, a change of concavity, or very flat nearby secants. These are not the same property. Exact comparisons let us study them without using derivative rules.

For p(x)=x³−3x, the identities

• p(x)−p(−1)=(x+1)²(x−2),
• p(x)−p(1)=(x−1)²(x+2)

show a strict local maximum p(−1)=2 and a strict local minimum p(1)=−2: use a neighborhood in which the last factor has the indicated sign. To check curvature, expand the equally spaced values:

[p(m+h)+p(m−h)]/2−p(m)=3mh².

On an interval to the left of zero, this is negative: the midpoint lies above the chord. To the right it is positive: the midpoint lies below the chord. Midpoint inequalities imply the full chord inequalities for a continuous function: repeatedly bisect a segment to get every dyadic fraction, then approximate an arbitrary fraction by dyadic ones and use continuity. Thus p is concave down on (−∞,0) and concave up on (0,∞), with an inflection at (0,0). Zero is not an extremum: near it p is positive on the left and negative on the right.

For x³, every secant from a to b>a has slope a²+ab+b²>0, so the function is strictly increasing. But its symmetric secant from −h to h has slope h², tending to zero. It has the same change of concavity at zero and no extremum there. Horizontal limiting secants do not by themselves prove a turn.

Why two distinct polynomial zeros force an intervening extremum

Let a nonconstant real polynomial P have zeros a<b. A nonzero polynomial has only finitely many zeros, by successive factor division, so some c strictly between them has P(c)≠0. Continuity and the compact-interval maximum/minimum property give the result: if P(c)>0, a positive maximum on [a,b] cannot be at either zero endpoint and is an interior local maximum. If P(c)<0, use the negative minimum. This proves existence, not a method locating the point, nor uniqueness.

Here the compact-interval property need not be a calculus black box. A polynomial is bounded on a bounded interval by bounding each coefficient term, so its values have a finite supremum M. Bisect [a,b]; at least one closed half has the same supremum M, since the supremum on a finite union is the larger of the two suprema. Repeat, obtaining nested closed intervals whose lengths tend to zero and whose common point z exists by completeness of the reals. If P(z)<M, continuity would make every value in a sufficiently short retained interval less than (P(z)+M)/2, contradicting its supremum M. Thus P(z)=M. Apply this argument to −P for the minimum. This supplies the attainment step without a derivative or an unstated optimization rule.

Changed proof task. Analyze q(x)=(x−2)³−12(x−2) on [−1,5]. Use factor comparisons, not a plotted guess, to identify local and absolute extrema and the inflection. Check the included endpoints separately: on a restricted domain an endpoint can be a one-sided local extremum without being the absolute one. The registered task gives the exact comparison factors and rubric. In an unfamiliar graph question, report only the precision supported by the supplied graph; finite samples alone do not certify unsampled extrema or curvature.

Apply

Restrict the same formula to [1,2]. Identify its maximum and minimum without using the unrestricted vertex as an allowed point. Then restrict it to [4,5]. Explain how a changed domain changes the answer even though the formula is identical.

Check the domain-dependent extrema

On [1,2], Q increases, so the minimum is Q(1)=7 and maximum Q(2)=10. The vertex at 3 is outside this domain. On [4,5], Q decreases, so the maximum is Q(4)=10 and minimum Q(5)=7. Saying “the maximum is always 11” ignores the allowed inputs. A graph construction must mark its domain endpoints as well as its turning point.

If Q is instead defined on all real inputs, its end behavior is written lim_x→∞ Q(x)=−∞ and lim_x→−∞ Q(x)=−∞. This means outputs become arbitrarily negative in both distant directions; infinity is not an input or an attained output. The completed-square formula justifies the statement because (x−3)² grows without bound. The original restricted model on [0,5] has no inputs tending to positive or negative infinity, so those end-limit claims describe an unrestricted extension, not its permitted operating range.

Continue

In practice, identify a,b,c, the positive interval width and the allowed domain. Derive or use the finite rate identity, interpret both its sign and its trend, and justify extrema by a square-completion or endpoint comparison. Keep sampled evidence, exact model assumptions and unrestricted end behavior separate.