Average Rates For Quadratic Functions

From an Interval to Its Midpoint

A quadratic function can curve, so its rate of change is not generally constant across an interval. Yet for every quadratic, the average rate over an interval has a particularly simple connection to the function’s instantaneous slope: they are equal at the interval’s midpoint.

This result links two different ways of describing change. The average rate compares the outputs at the interval’s endpoints. The instantaneous slope describes how the function is changing at one input. For a quadratic, the midpoint is exactly where those two rates agree.

Average Rate and Instantaneous Slope

Let \(f\) be a function, and let \(u\) and \(v\) be distinct inputs in its domain. The average rate of change of \(f\) from \(u\) to \(v\) is the change in output divided by the change in input. Here we take \(u<v\), so the interval is \([u,v]\).

Geometrically, this quotient is the slope of the secant line through the graph points \((u,f(u))\) and \((v,f(v))\).

The instantaneous slope at an input \(x\), when it exists, is the limit of the average rate over intervals from \(x\) to nearby inputs. For the quadratic functions considered here, this limit exists at every real input. It is the derivative \(f'(x)\).

We will consider a quadratic in the standard form \(f(x)=ax^2+bx+c\), where \(a\), \(b\), and \(c\) are real numbers. If \(a\ne 0\), its graph is a parabola; the calculation below also works when \(a=0\), in which case the function is linear.

The Average-Rate Formula for a Quadratic

For a quadratic, the endpoint changes can be factored in a way that removes the interval length from the denominator. That cancellation reveals why the midpoint appears.

In particular, the average rate depends on the interval’s midpoint \(m\), not on its width. The width still matters for specifying the interval, but changing the width while keeping the same midpoint does not change this quadratic’s average rate.

Why the Formula Works

First calculate the average rate directly. Since \(f(x)=ax^2+bx+c\), substituting the two endpoints gives

The third line uses the difference-of-squares identity \(v^2-u^2=(v-u)(v+u)\). The last cancellation is valid because \(u\ne v\). Since \(u+v=2m\), the result is \(2am+b\).

To compare this with the instantaneous slope, use the derivative’s limit definition at \(m\). For \(h\ne 0\), the average rate from \(m\) to \(m+h\) is

As \(h\) approaches zero, the term \(ah\) approaches zero. Therefore the limit exists and gives the instantaneous slope at \(m\): \(f'(m)=2am+b\). This is exactly the average rate already calculated.

Worked Examples

A Common Error: Forgetting to Divide by the Input Change

The average rate is not just the change in output. It is the change in output divided by the change in input. For example, on \([1,5]\), the function \(p(x)=x^2\) has endpoint values \(p(1)=1\) and \(p(5)=25\). A mistaken calculation might stop at \(25-1=24\) and call \(24\) the average rate.

That number is the output change, not the average rate. The input change is \(5-1=4\), so the correction is

The midpoint is \(3\), and \(p'(x)=2x\), so \(p'(3)=6\), as required. Remember to include the denominator before comparing an average rate with a slope.

What the Midpoint Result Tells Us

The equality is useful because it identifies a particular input where the instantaneous slope represents the average change across the whole interval. To find it, calculate the midpoint of the inputs—not the midpoint of the function’s endpoint outputs—and evaluate the quadratic’s derivative there.

The result relies on the quadratic’s algebraic form. Its proof depends on factoring \(v^2-u^2\), which lets the interval factor \(v-u\) cancel. The conclusion is therefore a specific property of quadratic functions, rather than a rule to apply automatically to every curved graph.

For any interval \([u,v]\) with \(u<v\), the key calculation can be organized as follows:

The central relationship is concise: for a quadratic, the average rate on an interval equals the instantaneous slope at its midpoint. The algebra proves the relationship, and the examples show how to verify it with actual values.