AP Precalculus · Lesson 1

Rates of change and polynomial zeros

Know Know it

The average rate of change of ff from x=ax=a to x=bx=b is f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}. It is the slope of the line through the two points, and its units are output units per input unit.

A function is increasing on an interval when larger inputs give larger outputs there. It is concave up when its rate of change is increasing, and concave down when its rate of change is decreasing. A point of inflection is where the concavity changes.

In a table with equally spaced inputs, work out the rate over each step. Positive rates suggest the function is increasing. Rates that grow suggest concave up; rates that shrink suggest concave down.

A polynomial of degree nn has at most nn real zeros and at most n−1n-1 local extrema. Counting multiplicity and nonreal zeros, it has exactly nn complex zeros, and with real coefficients the nonreal ones come in conjugate pairs such as 2+i2+i and 2−i2-i.

At a real zero of odd multiplicity the graph crosses the xx-axis; at an even multiplicity it touches and turns back. A polynomial is continuous, so if it changes sign between two inputs, the Intermediate Value Theorem says it has a zero between them.

Calculator On the exam

On the exam, a graphing calculator is required in multiple-choice Part B and free-response Part A, set to radian mode. It is not allowed in multiple-choice Part A or free-response Part B, so practice rates of change, zeros and asymptotes by hand too.

Trap A positive average does not mean always rising

A positive average rate of change on [a,b][a,b] only says f(b)>f(a)f(b)>f(a). The function may still fall for part of the interval. To claim it is increasing on the whole interval, you need more than the two endpoints.

Worked example Easy

Let f(x)=x2−3x+1f(x)=x^2-3x+1. What is the average rate of change of ff on the interval 1≤x≤51 \le x \le 5?

  1. 1212
  2. −3-3
  3. 33Answer
  4. 13\frac{1}{3}

How to solve it

  1. Find the outputs: f(1)=1−3+1=−1f(1)=1-3+1=-1 and f(5)=25−15+1=11f(5)=25-15+1=11.
  2. Divide the change in output by the change in input: 11−(−1)5−1=124\frac{11-(-1)}{5-1}=\frac{12}{4}.
  3. The average rate of change is 3.

Why each choice is right or wrong

  • A. Incorrect. The value 12 is the change in output, f(5)−f(1)f(5)-f(1), before dividing by the change in input, 4.
  • B. Incorrect. The value -3 comes from subtracting in the wrong order on top, f(1)−f(5)f(1)-f(5), while keeping 5−15-1 on the bottom.
  • C. Correct. f(1)=−1f(1)=-1 and f(5)=11f(5)=11, so the rate is 11−(−1)5−1=124=3\frac{11-(-1)}{5-1}=\frac{12}{4}=3.
  • D. Incorrect. The value 13\frac{1}{3} flips the fraction, dividing the input change 4 by the output change 12.

Explain Explain it

Task verbs: calculate, justify

To justify a claim about a zero, name the fact you use and show the numbers that make it apply. For example: the polynomial is continuous, it is negative at one input and positive at another, so it equals 0 somewhere between them.

Worked example Medium

Let p(x)=(x−3)2(x+1)(x−6)p(x)=(x-3)^2(x+1)(x-6). At which input does the graph of pp touch the xx-axis without crossing it?

  1. x=−1x=-1
  2. x=3x=3Answer
  3. x=6x=6
  4. x=−3x=-3

How to solve it

  1. List the zeros from the factors: 3 (multiplicity 2), -1 (multiplicity 1) and 6 (multiplicity 1).
  2. An even multiplicity means the squared factor stays non-negative, so the sign of pp does not switch there.
  3. The graph touches without crossing at x=3x=3.

Why each choice is right or wrong

  • A. Incorrect. The factor x+1x+1 appears once, an odd multiplicity, so the graph crosses the axis at x=−1x=-1.
  • B. Correct. The factor x−3x-3 is squared, an even multiplicity, so the sign of pp does not change at x=3x=3 and the graph only touches the axis.
  • C. Incorrect. The factor x−6x-6 appears once, an odd multiplicity, so the graph crosses the axis at x=6x=6.
  • D. Incorrect. The input -3 is not a zero at all, since p(−3)=36⋅(−2)⋅(−9)=648p(-3)=36\cdot(-2)\cdot(-9)=648, which is not 0.

Justify Your turn

A polynomial qq has q(1)=−3q(1)=-3 and q(4)=2q(4)=2. Justify why qq must have a zero somewhere between x=1x=1 and x=4x=4.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says that qq is continuous because it is a polynomial
  • Uses the sign change from -3 to 2 to conclude q(c)=0q(c)=0 for some cc between 1 and 4

Model answer

Because qq is a polynomial, it is continuous on [1,4][1,4]. Its value goes from q(1)=−3q(1)=-3, below 0, to q(4)=2q(4)=2, above 0. Since 0 is between -3 and 2, the Intermediate Value Theorem says q(c)=0q(c)=0 for some cc with 1<c<41 < c < 4.

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