From a Polynomial Rule to an Interval Rate
A polynomial can rise on one part of its graph and fall on another. To describe what happens across a whole interval, we compare the polynomial’s outputs at the two endpoints. The resulting average rate tells us the net output change per unit of input change over that interval.
The calculation is direct, but careful bookkeeping matters. Each endpoint must be substituted into the polynomial correctly, and the output difference and input difference must be taken in the same order. A graph provides a useful check: if the output at the right endpoint is lower than at the left endpoint, the average rate from left to right must be negative.
Definition and Endpoint Method
For a polynomial model, \(f(a)\) and \(f(b)\) are found by substituting the endpoint values into the polynomial. After simplifying those two outputs, subtract in the order \(f(b)-f(a)\), find \(b-a\), and divide. The interval’s endpoints must be distinct so that \(b-a\ne 0\).
The value \(f(b)-f(a)\) is the net output change over the interval. It records the difference between the ending and starting outputs, not every change that may occur between them. If this difference is positive, the graph’s endpoint at \(b\) is higher than its endpoint at \(a\); if it is negative, that endpoint is lower.
Why the Formula Gives the Secant Slope
The two points on the graph corresponding to the interval endpoints are \((a,f(a))\) and \((b,f(b))\). The straight line through these points is called a secant line. The slope between two points is vertical change divided by horizontal change, so here the vertical change is \(f(b)-f(a)\) and the horizontal change is \(b-a\).
This establishes why the average rate is the slope of the secant line: its two points are precisely the graph’s endpoint values. The line summarizes the graph’s overall endpoint-to-endpoint change. It does not claim that the polynomial follows that straight line between the endpoints.
A Reliable Calculation Sequence
Writing down the stages separately helps prevent sign and substitution errors. In particular, keep the endpoint order visible until the final division.
Identify the starting input \(a\) and ending input \(b\), with \(a\ne b\).
Substitute each input into every term of the polynomial, using parentheses for negative inputs.
Calculate \(f(b)-f(a)\) and \(b-a\) in the same starting-to-ending order.
Compute the average rate, then compare its sign with the endpoint rise or fall on the graph.
Worked Polynomial Examples
Worked Example: A Cubic with a Positive Average Rate
Find the average rate of \(f(x)=2x^3-x^2+x+1\) from \(x=0\) to \(x=2\).
First evaluate the polynomial at the starting input:
Now evaluate it at the ending input:
The output change is \(15-1=14\), and the input change is \(2-0=2\). Therefore, the average rate is
Check: The endpoint output rises from \(1\) to \(15\), a net increase of \(14\) over an input increase of \(2\). The rate \(7\) is positive, and \(7\cdot2=14\), recovering the output change.
Worked Example: A Cubic with a Negative Average Rate
Find the average rate of \(g(x)=x^3-3x^2+2\) from \(x=0\) to \(x=2\).
At the starting input,
At the ending input,
The endpoint output change is \(-2-2=-4\), while the input change is \(2-0=2\). Thus,
Check: The graph’s endpoint output falls from \(2\) to \(-2\), so the net output change is negative. The rate is negative as expected; multiplying \(-2\) by the input change \(2\) gives \(-4\), the endpoint output change. This endpoint comparison does not by itself describe every point between \(0\) and \(2\).
Worked Example: A Quartic Across a Wider Interval
Find the average rate of \(p(x)=x^4-2x^2+x\) from \(x=-1\) to \(x=2\). Because the starting input is negative, substitute \(-1\) in parentheses.
At the ending input,
The output change is \(10-(-2)=12\). The input change is \(2-(-1)=3\). Therefore,
Check: From the left endpoint to the right endpoint, the output rises from \(-2\) to \(10\), giving a net increase of \(12\) across \(3\) input units. The positive rate \(4\) matches this rise, and \(4\cdot3=12\).
Reading the Result Against the Graph
When the interval is written from left to right, \(a<b\), the denominator \(b-a\) is positive. Therefore the average rate has the same sign as the endpoint output change \(f(b)-f(a)\). This gives a simple visual check: a higher right endpoint means a positive rate, a lower right endpoint means a negative rate, and equal endpoint heights mean a zero rate.
A zero average rate does not necessarily mean the polynomial’s graph is flat throughout the interval. It means only that the endpoint outputs are equal. Likewise, a positive average rate does not require the graph to rise at every point in between. The average rate summarizes the net change across the interval, not the local behavior at each input.
This distinction is useful when interpreting a polynomial graph. A curve may bend, rise, or fall between the endpoints, while the average rate still depends only on the two endpoint values. The secant line gives a compact measure of that overall change, but it is not a claim that the curve itself is straight.
Common Sign and Substitution Pitfalls
A frequent error is to reverse the denominator while leaving the numerator in its original order. For the second example, the correct changes from \(0\) to \(2\) are \(g(2)-g(0)=-4\) and \(2-0=2\), giving \(-2\). If only the denominator is reversed, the incorrect calculation is
This incorrectly gives a positive rate even though the graph’s right endpoint is lower. The correction is to use matching orders. Reversing both differences would give the same correct rate, but reversing just one changes its sign.
Another common mistake occurs when substituting a negative endpoint. For example, \((-1)^2=1\), not \(-1\). Parentheses make clear that the entire negative input is being squared. Finally, do not confuse a negative average rate with negative endpoint outputs: in the cubic example, the starting output is \(2\), the ending output is \(-2\), and the average rate is negative because the output falls overall.
Why Endpoint Checks Matter
Polynomial calculations can involve several powers and signs, so a final number should be checked against the original meaning of the rate. First compare the two endpoint outputs. Then check that the rate’s sign agrees with their net rise or fall. If the rate is \(r\) across an input change of \(b-a\), the output change should be \(r(b-a)\). These checks catch many arithmetic and ordering errors.
For instance, if a calculation reports a positive rate over a left-to-right interval whose right endpoint is lower, something is inconsistent. Recheck the endpoint substitutions, the subtraction order, and the division. The graph is not a replacement for the calculation, but it is an effective way to test whether the result makes sense.
Check Your Understanding
For each question, keep the endpoint order consistent and use the endpoint values to check the sign.
- For \(h(x)=x^3+x\), find the average rate from \(x=1\) to \(x=3\). What are the two endpoint outputs?
- For \(q(x)=x^3-2x^2\), calculate the average rate from \(x=0\) to \(x=2\). Does the result indicate a net rise or fall?
- If a polynomial has the same output at both ends of an interval with distinct endpoints, what is its average rate on that interval?
- Why must the output difference and input difference be written in matching endpoint order?
- Does a positive average rate guarantee that the polynomial rises at every point in the interval? Explain using the meaning of endpoint change.