Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
Most BC free-response questions hand you one graph and ask about a different function. The graph supplied is often f′, and the questions are about f. Moving between the two is a fixed set of translations: the sign of f′ is the direction of f, the slope of f′ is the concavity of f, and the area under f′ is the change in f. Going the other way, a peak on f sits above a zero of f′, and an inflection of f sits above an extremum of f′. Learn the six translations in both directions and the graph questions stop being puzzles, because each one names which translation it wants.
Figure 17 — A function, its derivative and its second derivative stacked in one column. Read downward for signs and slopes, and read upward to find where a peak or an inflection must sit.
The three graphs share an axis, so a vertical line through the column shows three related facts at once. At a peak of f the derivative is zero and the second derivative is negative. At an inflection of f the second derivative is zero and the derivative has a peak or a valley.
| Supplied | Asked | What links them |
|---|---|---|
| graph of f | where f′ is positive | the segments where f rises |
| graph of f | where f′′ is positive | the segments where f bends up |
| graph of f′ | where f rises | the segments where f′ is above the axis |
| graph of f′ | where f is concave up | the segments where f′ climbs |
| graph of f′ | the change in f | the signed area under f′ |
| graph of f′′ | where f is concave up | the segments where f′′ is positive |
A graph of f′ fixes the shape of f and not its height. Two functions with the same derivative differ by a constant, so a question about a value of f needs either an initial condition or an accumulation from a known point. Reading a height off the graph of f′ is the single most common error on these questions.
Many parts ask two questions about one region. "Where is f increasing and concave down" needs the sign of f′ and the slope of f′ at the same time, and the answer is the overlap of the two sets. Keeping the two readings separate on paper prevents the answers from being blended into one.
A table of f′ values replaces the graph. Increasing entries mean f is concave up, sign changes mean extrema of f, and a difference of consecutive entries approximates a change in f. The translations survive the change of format.
Borrowed idea: Name both. Write two words at the top of your work. One names the graph you were given. The other names the function the question asks about. Every translation then follows from those two words. Without them the eye drifts to the picture and answers about the picture. So what does the first line of a good response look like? It says "given f′, asked about f" and nothing else.
| Question word | Translation to apply | Where to look |
|---|---|---|
| "increasing" | sign of f′ | above or below the axis |
| "concave" | slope of f′ | rising or falling |
| "local maximum" | sign change of f′ | axis crossing, high to low |
| "inflection" | extremum of f′ | peak or valley |
| "change in f" | area under f′ | between two inputs |
① Never read a height off the wrong graph. A value of f is never visible on a graph of f′.
② Use the axis, not the shape, for direction. The curve's own height is irrelevant to where f rises.
③ Mark the crossings first. Axes crossings and extrema of the supplied graph generate every answer.
Core Archetype (Simplest Form): The graph of f′ crosses the axis at x=1 from positive to negative and has a minimum at x=3. Describe f.
Standard Execution:
- Direction. f′>0 before x=1 and f′<0 after, so f rises then falls.
- Turning point. The crossing at x=1 makes a local maximum there.
- Bend. The minimum of f′ at x=3 is where the slope of f is smallest, so f has an inflection at x=3 and is concave up for x>3.
- Values. No height of f is available from this graph, only changes obtained by accumulating area.
Exam Variation (Context / Parameter Shift): The graph of f is supplied on [−6,7], and f is linear on [0,7] with a horizontal tangent at x=−2. Where is f′ increasing, and where is f′′ positive?
Solution: The derivative increases wherever f bends upward, which is the portion of the graph that curves up rather than down. The second derivative is positive on exactly those intervals, and it is zero on the linear piece where the graph has no bend at all.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
The graph of f′ crosses the axis at x=2, going from positive to negative. What happens on the graph of f at x=2?
MCQ 2
On an interval, f is decreasing and concave up. What is true of f′ there?
MCQ 3
The graph of f′′ is positive on (1,4). What does that say about f?
The graph of f′, the derivative of f, is defined on [−5,6] and consists of line segments. It is positive on (−5,1), crosses the axis at x=1, is negative on (1,4), and crosses back at x=4. The graph of f′ has a minimum at x=2 and a maximum at x=5.
(a) Find the open intervals on which f is increasing and those on which f is decreasing.
(b) Identify the x-coordinates of all local extrema of f, and classify each.
(c) Find the open intervals on which the graph of f is concave up and those on which it is concave down.
(d) A student claims f has a local minimum at x=2 because f′ has a minimum there. Explain the error.
| Feature on the supplied graph | Meaning for the other function | |
|---|---|---|
| f | rising segment | f′>0 there |
| f | upward bend | f′′>0 there |
| f′ | above the axis | f increasing |
| f′ | rising segment | f concave up |
| f′ | axis crossing with sign change | local extremum of f |
| f′ | peak or valley | inflection of f |
| f′′ | positive segment | f concave up |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
Most BC free-response questions hand you one graph and ask about a different function. The graph supplied is often f′, and the questions are about f. Moving between the two is a fixed set of translations: the sign of f′ is the direction of f, the slope of f′ is the concavity of f, and the area under f′ is the change in f. Going the other way, a peak on f sits above a zero of f′, and an inflection of f sits above an extremum of f′. Learn the six translations in both directions and the graph questions stop being puzzles, because each one names which translation it wants.
Figure 17 — A function, its derivative and its second derivative stacked in one column. Read downward for signs and slopes, and read upward to find where a peak or an inflection must sit.
The three graphs share an axis, so a vertical line through the column shows three related facts at once. At a peak of f the derivative is zero and the second derivative is negative. At an inflection of f the second derivative is zero and the derivative has a peak or a valley.
| Supplied | Asked | What links them |
|---|---|---|
| graph of f | where f′ is positive | the segments where f rises |
| graph of f | where f′′ is positive | the segments where f bends up |
| graph of f′ | where f rises | the segments where f′ is above the axis |
| graph of f′ | where f is concave up | the segments where f′ climbs |
| graph of f′ | the change in f | the signed area under f′ |
| graph of f′′ | where f is concave up | the segments where f′′ is positive |
A graph of f′ fixes the shape of f and not its height. Two functions with the same derivative differ by a constant, so a question about a value of f needs either an initial condition or an accumulation from a known point. Reading a height off the graph of f′ is the single most common error on these questions.
Many parts ask two questions about one region. "Where is f increasing and concave down" needs the sign of f′ and the slope of f′ at the same time, and the answer is the overlap of the two sets. Keeping the two readings separate on paper prevents the answers from being blended into one.
A table of f′ values replaces the graph. Increasing entries mean f is concave up, sign changes mean extrema of f, and a difference of consecutive entries approximates a change in f. The translations survive the change of format.
Borrowed idea: Name both. Write two words at the top of your work. One names the graph you were given. The other names the function the question asks about. Every translation then follows from those two words. Without them the eye drifts to the picture and answers about the picture. So what does the first line of a good response look like? It says "given f′, asked about f" and nothing else.
| Question word | Translation to apply | Where to look |
|---|---|---|
| "increasing" | sign of f′ | above or below the axis |
| "concave" | slope of f′ | rising or falling |
| "local maximum" | sign change of f′ | axis crossing, high to low |
| "inflection" | extremum of f′ | peak or valley |
| "change in f" | area under f′ | between two inputs |
① Never read a height off the wrong graph. A value of f is never visible on a graph of f′.
② Use the axis, not the shape, for direction. The curve's own height is irrelevant to where f rises.
③ Mark the crossings first. Axes crossings and extrema of the supplied graph generate every answer.
Core Archetype (Simplest Form): The graph of f′ crosses the axis at x=1 from positive to negative and has a minimum at x=3. Describe f.
Standard Execution:
- Direction. f′>0 before x=1 and f′<0 after, so f rises then falls.
- Turning point. The crossing at x=1 makes a local maximum there.
- Bend. The minimum of f′ at x=3 is where the slope of f is smallest, so f has an inflection at x=3 and is concave up for x>3.
- Values. No height of f is available from this graph, only changes obtained by accumulating area.
Exam Variation (Context / Parameter Shift): The graph of f is supplied on [−6,7], and f is linear on [0,7] with a horizontal tangent at x=−2. Where is f′ increasing, and where is f′′ positive?
Solution: The derivative increases wherever f bends upward, which is the portion of the graph that curves up rather than down. The second derivative is positive on exactly those intervals, and it is zero on the linear piece where the graph has no bend at all.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
The graph of f′ crosses the axis at x=2, going from positive to negative. What happens on the graph of f at x=2?
MCQ 2
On an interval, f is decreasing and concave up. What is true of f′ there?
MCQ 3
The graph of f′′ is positive on (1,4). What does that say about f?
The graph of f′, the derivative of f, is defined on [−5,6] and consists of line segments. It is positive on (−5,1), crosses the axis at x=1, is negative on (1,4), and crosses back at x=4. The graph of f′ has a minimum at x=2 and a maximum at x=5.
(a) Find the open intervals on which f is increasing and those on which f is decreasing.
(b) Identify the x-coordinates of all local extrema of f, and classify each.
(c) Find the open intervals on which the graph of f is concave up and those on which it is concave down.
(d) A student claims f has a local minimum at x=2 because f′ has a minimum there. Explain the error.
| Feature on the supplied graph | Meaning for the other function | |
|---|---|---|
| f | rising segment | f′>0 there |
| f | upward bend | f′′>0 there |
| f′ | above the axis | f increasing |
| f′ | rising segment | f concave up |
| f′ | axis crossing with sign change | local extremum of f |
| f′ | peak or valley | inflection of f |
| f′′ | positive segment | f concave up |
Strengthen this topic with AP-style practice, or continue to next topic.