AP Precalculus · Free-response skill

Function concepts from a table

The move. From a table of an invented function, find an average rate of change, say where its table values fall and how the rates suggest it bends, and justify a claim about its zeros.

How to do it

  1. Start with the rates. Work out the change in output over each step of the table and write the rates in a row under it. Most parts use them.
  2. For an average rate of change, show the quotient f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a} with the table values in it, then give the number with units if there are any.
  3. Outputs that go up suggest increasing; outputs that go down suggest decreasing. Rates that grow suggest concave up; rates that shrink suggest concave down. A table shows only the listed inputs, so say which values or rates you compared.
  4. To justify a zero, name continuity (every polynomial is continuous), show the sign change with the two table values, and name the Intermediate Value Theorem.
  5. To count zeros, list every interval with a sign change, use end behavior for the parts of the graph beyond the table, and compare with the degree, which caps the number of real zeros.
  6. This kind of question sits in the calculator part of the free-response section, but every step here can be done by hand.

Practice prompt Try it

The function ff is a polynomial of degree 3 with a positive leading coefficient. Selected values of ff are shown in the table.

Selected values of the polynomial ff

xx0123456
f(x)f(x)484-2-4428

Use the table and the facts about ff to answer each part.

  1. Work out the average rate of change of ff from x=1x=1 to x=4x=4. Show the numbers you used.
  2. Name the longest stretch of the table on which the values of ff go down at every step. Then use the rates of change over [3,4][3,4], [4,5][4,5] and [5,6][5,6] to say whether the graph of ff looks concave up or concave down on 3≤x≤63 \le x \le 6, and why.
  3. Give a reason, based on the table, why ff must equal 0 for some xx between 2 and 3.
  4. Ana says ff has exactly three real zeros. Use the table, the end behavior of ff and its degree to back up her claim.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) The average rate of change from x=1x=1 to x=4x=4 is f(4)−f(1)4−1=−4−83=−123=−4\frac{f(4)-f(1)}{4-1}=\frac{-4-8}{3}=\frac{-12}{3}=-4.

  • Point 1 Shows the quotient with the table values before giving the answer -4.

(b) The values go down at every step from x=1x=1 to x=4x=4: 8, 4, -2, -4. On [3,4][3,4], [4,5][4,5] and [5,6][5,6] the rates of change are −2-2, 88 and 2424. These rates increase, so the graph of ff looks concave up on 3≤x≤63 \le x \le 6.

  • Point 2 Names the interval and lists the falling values as support.
  • Point 3 Computes the three rates and ties their growth to concave up.

(c) Since ff is a polynomial, it is continuous. f(2)=4f(2)=4 is positive and f(3)=−2f(3)=-2 is negative, and 0 lies between them, so by the Intermediate Value Theorem f(c)=0f(c)=0 for some cc between 2 and 3.

  • Point 4 States continuity and the reason for it.
  • Point 5 Gives both values, the sign change and the theorem by name.

(d) Because ff has odd degree and a positive leading coefficient, f(x)f(x) falls without bound as xx decreases, yet f(0)=4>0f(0)=4>0, so ff has a zero left of x=0x=0. The sign also changes between x=2x=2 and x=3x=3 (4 to -2) and between x=4x=4 and x=5x=5 (-4 to 4). That gives at least three real zeros. A degree 3 polynomial has at most three real zeros, so ff has exactly three.

  • Point 6 Finds all three intervals and uses end behavior for the one outside the table.
  • Point 7 Uses the degree as a cap to turn at least three into exactly three.