Quarry School

Increasing, decreasing, and constant

Explain it like I am five

Picture a hiking trail seen from the side. You walk toward the right and watch its height. A stretch is increasing when every move right takes you higher, decreasing when every move right takes you lower, and constant when the height stays the same. The height can be below zero and still rise. You name each stretch by its inputs, the positions along the horizontal axis, rather than its heights. A turning point is where the graph switches between rising and falling. This course reports behavior using open intervals, which leave the boundary inputs out. That reporting convention describes stretches. It does not make an endpoint disappear from the function's domain.

−3−2−1123−1123456789(0, 0)turns at x = 0
Left to right, y = x2 drops until x = 0 and then climbs: decreasing on (−∞, 0), increasing on (0, ∞).
Reminder
  • Interval notation. (1, 3) means 1 < x < 3. [1, 3] includes the endpoints. Infinity always has a parenthesis.
  • Subtracting negatives. −1 − (−2) = −1 + 2 = 1: output −1 is an increase from −2.
  • Point coordinates. In (2, 5), 2 is the horizontal input and 5 is the vertical output.
Why it works. Choose any two inputs a < b within one stretch. The change in input, b − a, is positive. If f(b) > f(a), the change in output is positive, so the average rate of change is positive. If f(b) < f(a), that rate is negative. Equal outputs give rate 0. This explains the three rows of the table. The letters a and b name any two inputs, rather than fixed endpoints to solve for. We split a graph at turns, flat boundaries, and breaks because a single label must describe the entire stretch between them.
RuleOn an interval, whenever a < b: increasing means f(a) < f(b); decreasing means f(a) > f(b); constant means f(a) = f(b). This course reports intervals of behavior with open endpoints. These are comparisons between inputs, not a direction assigned to one point.
The same idea, five ways
Say it

Say: 'the output rises, falls, or stays the same as the input moves right.'

Write it

Write: 'f is increasing on (1, 3), decreasing on (3, 4), and constant on (4, 5).'

In math
  • 1 < x < 3
  • (1, 3)
  • {x : 1 < x < 3}
  • If 1 < a < b < 3, then f(a) < f(b).
  • Decreasing: a < b gives f(a) > f(b).
  • Constant: a < b gives f(a) = f(b).
Like

Walk east along an uphill, downhill, or level trail.

See it
13(1, 3)
After 'increasing on,' (1, 3) names all inputs strictly between the boundaries.
The same idea, other ways
As a picture

On y = x2, you move downhill toward 0 from the left, then uphill after 0. The vertex is the turning point of this U shaped graph, called a parabola. The heights are positive on both sides, but one side decreases and the other increases.

−222468(0, 0)
Read left to right to name direction, rather than checking whether the height is positive.
With two numbers

For f(x) = x2 on (0, ∞), let a = 1 and b = 3. Then f(1) = 1 < 9 = f(3). The letters are names for those inputs. On (−∞, 0), inputs −3 and −1 move right while outputs 9 and 1 move down.

input xoutput x²−39−111139
Compare inputs in increasing order within one side of the turn.
With recorded prices

The gas-price columns rise at each recorded year from 2005 through 2008, fall from 2008 to 2009, and rise at each recorded year afterward. These are sampled yearly data. They describe recorded changes and do not tell you the price at every time between the recorded years.

input yearoutput price ($ per gallon)20052.3120062.6220072.8420083.320092.4120102.8420113.5820123.68
Compare successive columns; the data do not establish unmeasured behavior between years.
Directions and rate signs

The behavior table links each direction to the sign of every average rate taken within that stretch. The drawn columns repeat all three comparisons.

input behavioroutput average rate signincreasing: climbspositivedecreasing: dropsnegativeconstant: levelzero
The three columns preserve the direction, name, and average-rate facts of the behavior table.
Graph moving left to rightNameSign of every average rate of change there
climbsincreasingpositive
dropsdecreasingnegative
stays levelconstantzero
.1Increasing: larger inputs give larger outputs

Think of walking up a ramp. You can begin below street level and still climb. Increasing describes the direction of the outputs as the inputs move right.

  • Any a < b within the interval gives f(a) < f(b).
  • Every average rate of change within that interval is positive.
  • Increasing does not mean positive outputs.
−4−2−4−3−2−11
The negative heights still rise as the input moves right.
Worked exampleIncreasing below zero

For f(x) = x on (−3, 0), name the output's direction.

−2−3−2−1−2−1
Direction and output sign are different facts.
  1. f(−2) = −2 and f(−1) = −1, so these outputs rise.−2 < −1, and −1 is greater than −2.
  2. The line rises throughout the interval: increasing on (−3, 0).For this formula any larger input is itself a larger output.
Answer
Increasing on (−3, 0), with negative outputs there.
Check −1−(−2)−1−(−2) = 1, a positive average rate.
.2Decreasing: larger inputs give smaller outputs

Walking down a ramp lowers your height as you move forward. On a decreasing stretch, the output becomes smaller as the input becomes larger, even when every output stays above zero.

  • Any a < b within the interval gives f(a) > f(b).
  • Every average rate of change within that interval is negative.
2246
The line falls to the right while the visible heights are nonnegative.
Worked examplePositive heights can decrease

Read the direction of f(x) = 6 − 2x on (0, 3).

224642
Positive outputs 4 and 2 go down from left to right.
  1. At x = 1 the height is 4; at x = 2 it is 2.6 − 2(1) = 4 and 6 − 2(2) = 2.
  2. Report decreasing on (0, 3).Every larger input subtracts a larger amount from 6.
Answer
Decreasing on (0, 3).
Check 2−42−1 = −2, a negative average rate.
.3Constant: the output stays level

Imagine a shelf at one fixed height. Walking along it changes your position but never its height. Constant means every input in the stretch produces the same output.

  • Any a < b within the interval gives f(a) = f(b).
  • Every average rate of change there is 0.
  • A constant stretch is neither strictly increasing nor strictly decreasing.
212345
Every input shown gives height 4.
Worked exampleFinding the level stretches on a water-tank graph

A graph shows the water depth h(t), in feet, in a tank, where the horizontal axis is labeled t (time in hours). The graph is made of straight segments with no breaks. It starts at a closed dot at (0, 2) and rises straight to (3, 5). It stays level at depth 5 from (3, 5) to (7, 5). It falls straight from (7, 5) to (9, 1). It stays level at depth 1 from (9, 1) to a closed dot at (12, 1), where the graph ends. Using open intervals of t, give where h is increasing, where h is decreasing, and where h is constant.

  1. Read the horizontal-axis label. The input is t (hours), so every interval is an interval of t-values, not depths.Intervals of behavior are always reported as intervals of the input variable.
  2. Mark the turning inputs and flat boundaries: t = 3 (rising becomes level), t = 7 (level becomes falling), and t = 9 (falling becomes level). There are no breaks.The behavior can only change at these inputs, so they split the graph into separate stretches.
  3. Mark the domain ends: closed dots at t = 0 and t = 12. Both boundaries are finite, and no interval may use infinity.A finite end dot gives a finite boundary. Only arrows or stated continuation justify infinity.
  4. On 0 to 3, moving right, the depth goes from 2 up to 5. Whenever a < b there, h(a) < h(b), so h is increasing.Increasing means larger inputs always give larger outputs.
  5. On 3 to 7, the depth stays at 5. For any a < b there, h(a) = h(b) = 5, so h is constant.Constant means every pair of inputs in the stretch gives equal outputs.
  6. On 7 to 9, the depth goes from 5 down to 1. Whenever a < b there, h(a) > h(b), so h is decreasing.Decreasing means larger inputs always give smaller outputs.
  7. On 9 to 12, the depth stays at 1. For any a < b there, h(a) = h(b) = 1, so h is constant again.This is a second level stretch. It is separate from the first because a decreasing stretch lies between them.
  8. Write each behavior with open endpoints: increasing (0, 3); decreasing (7, 9); constant (3, 7) and (9, 12), written (3, 7) ∪ (9, 12).This course reports intervals of behavior with open endpoints. The two constant stretches are joined piece by piece with 'and' or ∪, never merged into (3, 12).
Answer
Increasing on (0, 3). Decreasing on (7, 9). Constant on (3, 7) and (9, 12), that is, (3, 7) ∪ (9, 12).
Check Test inputs from each interval. Increasing: h(1) is about 3 and h(2) is about 4, so 3 < 4. Decreasing: h(7.5) = 4 and h(8.5) = 2, so 4 > 2. Constant: h(4) = h(6) = 5 and h(10) = h(11) = 1. The intervals (0, 3), (3, 7), (7, 9) and (9, 12) cover the domain from 0 to 12, leaving out only the boundary inputs, and they do not overlap. The combined interval (3, 12) would be wrong, because h(8) = 3 differs from h(4) = 5.

Work to write

  1. Input variable: t
  2. Increasing: (0, 3)
  3. Decreasing: (7, 9)
  4. Constant: (3, 7) ∪ (9, 12)

Increasing on (0, 3). Decreasing on (7, 9). Constant on (3, 7) and (9, 12), that is, (3, 7) ∪ (9, 12).

Strategy: step by step
  1. Begin at the left of the whole given graph. Read the horizontal-axis label: the input might be x or t.
  2. Mark turning inputs, flat boundaries, and breaks.
  3. Mark actual domain ends. A finite end dot gives a finite boundary; arrows or stated continuation can justify infinity.
  4. Move right within each separate stretch. Name it increasing, decreasing, or constant.
  5. Write open intervals of inputs. Keep pieces on opposite sides of a break separate.
  6. Give each requested behavior on its own line. Say 'and' between stretches; ∪ joins the listed stretches piece by piece.
Strategy
Read, split, name each stretch
1
Does the graph break at an excluded input or a vertical asymptote?
YesSplit the interval at that input. A vertical asymptote is a vertical line the curve approaches while outputs grow in size.
NoContinue within the connected piece.
↓
2
Does the graph stop at a finite end dot, with no stated continuation?
YesUse that input as the finite boundary. A viewing-window edge alone does not prove the graph stops.
NoUse infinity only when arrows or the stated domain justify continuation.
↓
3
Does the graph turn or become flat?
YesSplit before labeling the stretches.
NoKeep the current stretch together.
↓
4
Does the height rise as you move right?
YesIncreasing.
NoFalling is decreasing; unchanged height is constant.
  1. Find the input axis and actual domain ends.
  2. Split at turns, flat boundaries, and gaps.
  3. Read left to right on each separate piece.
  4. Write open intervals of inputs.
Worked exampleReading increasing and decreasing intervals from a downward parabola

The graph shows y = f(x), where f(x) = -(x - 2)2 + 3. The horizontal axis is labeled x. The graph is a parabola that opens downward. Its highest point (turning point) is at (2, 3). It also passes through (0, -1), (1, 2), (3, 2) and (4, -1). Arrows on both ends show that the graph continues forever to the left and to the right, so the domain is all real numbers. State the open interval(s) on which f is increasing, the open interval(s) on which it is decreasing, and any interval on which it is constant.

−2246−6−4−224(2, 3)(0, -1)(4, -1)
Graph of f(x) = -(x - 2)2 + 3, a downward parabola with its highest point at (2, 3), passing through (0, -1) and (4, -1), with arrows on both ends.
  1. Start at the far left of the graph and read the horizontal-axis label. It is x, so every interval is written in terms of x-values.Intervals of behavior describe inputs, so we must know which variable is the input.
  2. Mark the turning input x = 2, where the graph changes from rising to falling at (2, 3). There are no flat stretches and no breaks.Behavior can only change at turning inputs, flat boundaries or breaks, so these split the domain into stretches.
  3. Mark the domain ends. Both ends carry arrows, so the graph continues without end in both directions. The left boundary is -∞ and the right boundary is ∞.Arrows show the graph continues, which justifies using infinity instead of a finite endpoint.
  4. Move right through the stretch from -∞ to 2. Here the graph rises. For example, at 0 < 1 we have f(0) = -1 < f(1) = 2. This stretch is increasing.Increasing means that whenever a < b on the interval, f(a) < f(b).
  5. Move right through the stretch from 2 to ∞. Here the graph falls. For example, at 3 < 4 we have f(3) = 2 > f(4) = -1. This stretch is decreasing.Decreasing means that whenever a < b on the interval, f(a) > f(b).
  6. Look for a stretch where the output stays the same. There is none, because the curve never runs level over an interval.Constant requires f(a) = f(b) for all a < b in an interval. A single highest point is not an interval.
  7. Write each behavior on its own line, using open intervals of x: increasing on (-∞, 2), decreasing on (2, ∞), constant on no interval.This course reports intervals with open endpoints. The turning input x = 2 belongs to neither stretch.
Answer
Increasing on (-∞, 2). Decreasing on (2, ∞). Constant on no interval.
Check Compare inputs on each side of x = 2 using the formula f(x) = -(x - 2)2 + 3. On the left, f(-1) = -(-3)2 + 3 = -6 and f(0) = -1. Since -1 < 0 and -6 < -1, the function is increasing. On the right, f(4) = -1 and f(5) = -(3)2 + 3 = -6. Since 4 < 5 and -1 > -6, the function is decreasing. Both results agree with the answer.

Work to write

  1. Turning input: x = 2 (highest point at (2, 3)); arrows on both ends, so the domain is (-∞, ∞)
  2. Increasing on (-∞, 2)
  3. Decreasing on (2, ∞)
  4. Constant: on no interval

Increasing on (-∞, 2). Decreasing on (2, ∞). Constant on no interval.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Reading one rising line: where is it increasing, decreasing, or constant?

The graph shows y = g(x), where g(x) = 12x + 1. The horizontal axis is labeled x. The graph is a straight line with arrows at both ends, so it continues forever to the left and to the right. It passes through (-4, -1), (0, 1) and (2, 2). Give the open intervals on which g is increasing, decreasing, and constant.

−6−4−2246−4−224(-4, -1)(0, 1)(2, 2)
The line y = g(x) = 12x + 1 rises from left to right through (-4, -1), (0, 1) and (2, 2), with arrows at both ends.
  1. Start at the far left of the graph and read the horizontal-axis label: the input is x.Intervals of behavior are written in terms of the input, so we must know which variable the intervals describe.
  2. Look for turning inputs, flat stretches, and breaks. There are none. The graph is one unbroken straight line with slope 12.A non-vertical line with nonzero slope never changes direction and never levels off, so there are no boundaries inside it.
  3. Check the domain ends. Both ends have arrows, so the domain runs from -∞ to ∞.Arrows show that the graph continues without end, which justifies using infinity. A finite end dot would have given a finite boundary instead.
  4. Move right along the single stretch. Compare the inputs -4 < 2: g(-4) = -1 and g(2) = 2, and -1 < 2. The outputs rise as x moves right, so the stretch is increasing.Increasing means that whenever a < b, f(a) < f(b). A positive slope makes this true for every pair of inputs on the line.
  5. Write the behavior as an open interval of inputs: increasing on (-∞, ∞).This course reports intervals with open endpoints, and ∞ is never included in any case.
  6. Report each requested behavior on its own line. Decreasing: none. Constant: none.No pair of inputs a < b gives g(a) > g(b) or g(a) = g(b), so neither behavior occurs anywhere.
Answer
Increasing on (-∞, ∞). Decreasing: none. Constant: none.
Check Test another pair of inputs, -4 < 0: g(-4) = 12(-4) + 1 = -1 and g(0) = 1. Since -1 < 1, the output rises again. The slope 12 is positive, which confirms that the line rises everywhere and has no flat or falling part.

Work to write

  1. Input variable: x; no turning points, flat stretches, or breaks
  2. Domain: arrows at both ends, so (-∞, ∞)
  3. Increasing on (-∞, ∞)
  4. Decreasing: none
  5. Constant: none

Increasing on (-∞, ∞). Decreasing: none. Constant: none.

Rung 2A parabola that opens upward with its turn at x = -3

The graph shows y = h(x), where h(x) = 12(x + 3)2 - 2. The horizontal axis is labeled x. The graph is a parabola that opens upward, with arrows on both ends, so it continues forever to the left and to the right. Its lowest point (turning point) is at (-3, -2). It also passes through (-7, 6), (-5, 0), (-4, -1.5), (-2, -1.5), (-1, 0) and (1, 6). Give the open intervals where h is increasing, decreasing, and constant.

−8−6−4−22−4−2246(-3, -2)(-5, 0)(-1, 0)(-7, 6)(1, 6)
The graph of y = h(x) = 12(x + 3)2 - 2 opens upward, with its lowest point at (-3, -2). It passes through (-7, 6), (-5, 0), (-1, 0) and (1, 6), and it continues forever in both directions.
  1. Read the horizontal-axis label. It is x, so every interval will be a set of x-values.Intervals of behavior describe inputs. They do not describe outputs.
  2. Mark the turning input: the lowest point is (-3, -2), so mark x = -3. There are no flat pieces and no breaks.The graph can change from falling to rising only at a turning input. Flat boundaries and breaks would also split stretches, but this graph has none.
  3. Check the domain ends. Both ends have arrows, so the inputs run from -∞ to ∞.Arrows show that the graph continues without end. That justifies using -∞ and ∞ as boundaries.
  4. Move right on the stretch left of x = -3. Take -5 < -4: h(-5) = 0 and h(-4) = -1.5, so h(-5) > h(-4). Also h(-7) = 6 > h(-5) = 0. Name this stretch decreasing.Decreasing means that whenever a < b, f(a) > f(b). The outputs fall as the inputs grow.
  5. Move right on the stretch right of x = -3. Take -2 < -1: h(-2) = -1.5 and h(-1) = 0, so h(-2) < h(-1). Also h(-1) = 0 < h(1) = 6. Name this stretch increasing.Increasing means that whenever a < b, f(a) < f(b). The outputs rise as the inputs grow.
  6. Write the open intervals: decreasing on (-∞, -3) and increasing on (-3, ∞). No stretch is constant.This course reports intervals with open endpoints. The turning input x = -3 is not placed in either interval, because a single point has no direction by itself. No two different inputs give equal outputs along a stretch, so there is no constant interval.
Answer
  • Increasing: (-3, ∞)
  • Decreasing: (-∞, -3)
  • Constant: none
Check Compare two inputs on each side of the turn. On the left, -7 < -5 and h(-7) = 12(16) - 2 = 6 > h(-5) = 12(4) - 2 = 0, so h is decreasing. On the right, -1 < 1 and h(-1) = 0 < h(1) = 12(16) - 2 = 6, so h is increasing. The symmetric points (-4, -1.5) and (-2, -1.5) lie on opposite sides of x = -3, which agrees with the turn being at x = -3.

Work to write

  1. Horizontal axis is x; turning input x = -3; arrows on both ends
  2. Decreasing on (-∞, -3)
  3. Increasing on (-3, ∞)
  4. Constant: none

Increasing: (-3, ∞)
Decreasing: (-∞, -3)
Constant: none

Rung 3A parabola that stops at two solid end dots

The graph shows y = p(t), where p(t) = (t - 1)2 - 4, drawn only for -2 ≤ t ≤ 3. The horizontal axis is labeled t. The graph is a parabola that opens upward. It starts at a solid (filled) dot at (-2, 5) and ends at a solid (filled) dot at (3, 0). There are no arrows. Its lowest point (turning point) is at (1, -4). It also passes through (-1, 0), (0, -3) and (2, -3). State the open intervals on which p is increasing, decreasing, and constant.

−4−224−6−4−2246(-2, 5)(1, -4)(3, 0)(-1, 0)(0, -3)(2, -3)
The graph of p(t) = (t - 1)2 - 4 for -2 ≤ t ≤ 3. It has solid end dots at (-2, 5) and (3, 0) and its lowest point at (1, -4).
  1. Start at the left end of the graph and read the horizontal axis. It is labeled t, so every interval is an interval of t-values.Intervals of behavior describe inputs, and the input variable here is t, not x.
  2. Mark the turning input t = 1, where the lowest point (1, -4) is. There are no flat pieces and no breaks.Behavior can change only at turning inputs, at the ends of flat pieces, or at breaks.
  3. Mark the domain ends t = -2 and t = 3. Both are solid end dots and there are no arrows.A finite end dot gives a finite boundary. Without arrows the graph does not continue, so -∞ and ∞ are not used.
  4. Move right from t = -2 to t = 1. The heights go 5, 0, -3, -4, so this stretch is decreasing.For a < b in this stretch, p(a) > p(b). For example, p(-1) = 0 > p(0) = -3.
  5. Move right from t = 1 to t = 3. The heights go -4, -3, 0, so this stretch is increasing.For a < b in this stretch, p(a) < p(b). For example, p(2) = -3 < p(3) = 0.
  6. Write each stretch as an open interval of t: decreasing on (-2, 1) and increasing on (1, 3). No stretch is constant.This course reports intervals of behavior with open endpoints, even when the end dots are solid. No two inputs in any stretch give equal outputs, so p is never constant.
Answer
  • Increasing: (1, 3)
  • Decreasing: (-2, 1)
  • Constant: none
Check Test two inputs inside each interval. On (-2, 1): p(-1.5) = 6.25 - 4 = 2.25 and p(0.5) = 0.25 - 4 = -3.75. Since 2.25 > -3.75, p is decreasing there. On (1, 3): p(1.5) = 0.25 - 4 = -3.75 and p(2.5) = 2.25 - 4 = -1.75. Since -3.75 < -1.75, p is increasing there. Both intervals stop at the finite ends -2 and 3, with no ∞.

Work to write

  1. Input variable: t
  2. Turning input: t = 1; domain ends: t = -2 and t = 3 (solid dots, no arrows)
  3. Increasing on (1, 3)
  4. Decreasing on (-2, 1)
  5. Constant: none

Increasing: (1, 3)
Decreasing: (-2, 1)
Constant: none

Rung 4Rung 4: a break splits the branches

Read the behavior of f(x) = 1x. Why are there two decreasing intervals?

−4−224−4−224
Each branch falls separately; 0 is an excluded input.
  1. Exclude input 0.10 has no value.
  2. On the left, outputs at −2 and −1 fall from −12 to −1; on the right, outputs at 1 and 2 fall from 1 to 12.Each branch decreases as its inputs move right.
  3. Keep (−∞, 0) and (0, ∞) separate.No connected behavior interval can cross the missing input 0.
Answer
  • Decreasing on (−∞, 0) and on (0, ∞), each branch separately.
  • Increasing on no interval.
  • Constant on no interval.
Check Across the break, −1 < 1 but f(−1) = −1 < 1 = f(1), so the function is not decreasing across the whole union.
Rung 5Rung 5: a level stretch between sloping pieces

The connected straight pieces continue at both ends. They fall to (−3, −1), rise to (0, 4), stay at height 4 until x = 2, then rise again. Report all behaviors.

−4−224−22468(−3, −1)flat beginsflat ends
Dots sample connected straight pieces; the sample height at −1 is rounded and both ends continue.
  1. Mark −3, 0, and 2.Those inputs bound changes in direction or flatness.
  2. Read down, up, level, up as you move right.These are the directions of the given pieces.
  3. List the flat interval separately.Equal outputs there fail the strictly increasing comparison.
Answer
  • Decreasing on (−∞, −3).
  • Increasing on (−3, 0) and (2, ∞).
  • Constant on (0, 2).
Check Height 4 repeats across the middle stretch, so its average rate is 0.
Rung 6Rung 6: ends, a turn, and a level piece

This complete graph consists of straight segments joining corners (−4, −1), (−2, 3), (1, −2), (3, −2), and (5, 4). It stops at included endpoints. Report all behaviors.

−4−224−224filled endturnflat startsflat endsfilled end
Dots sample the connected straight segments; intermediate heights on the falling segment are rounded and the filled endpoints are actual ends.
  1. Mark endpoints −4 and 5 and inner boundaries −2, 1, and 3.Those inputs split the given segments.
  2. Read up, down, level, up across the successive segments.The segments connect the stated corners with no additional turns.
  3. Keep the two rising intervals separate and finite.The falling and level intervals lie between them and the graph stops.
Answer
  • Increasing on (−4, −2) and (3, 5).
  • Decreasing on (−2, 1).
  • Constant on (1, 3).
  • Domain: [−4, 5].
Check Each answer describes one segment direction. The endpoint inputs remain in the domain.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Increasing means positive outputs.
For f(x) = x on (−3, 0), every output is negative but larger inputs give larger outputs.
✓ Instead: Read whether height rises as the input moves right.
✗ Not this: (1, 3) always names a point.
After 'increasing on' it names all inputs between 1 and 3; after 'passes through' it names input 1 and output 3.
✓ Instead: Read the surrounding words.
✗ Not this: Because 1x decreases on both branches, it decreases across their whole union.
−1 < 1 but f(−1) = −1 < 1 = f(1), which fails the decreasing comparison.
✓ Instead: Report the two decreasing branches separately.
✗ Not this: Using (0, ∞) for x means its domain excludes 0.
0 = 0 is defined. Domain [0, ∞) and course behavior interval (0, ∞) answer different questions.
✓ Instead: Keep the domain answer separate from the open behavior-interval answer.
Tips and tricks
  • Read right, then name the height's direction.
  • Put inputs on interval boundaries and keep outputs for heights.
  • Use 'and' between behavior pieces. With ∪, mean the listed intervals piece by piece, especially across a gap.
  • Use the stated domain, endpoint marks, and arrows to distinguish an actual end from a viewing-window edge.
Trap. Writing the intervals in y-values, for example 'increasing from −2 to 5' because the heights run from −2 to 5. Intervals of increase and decrease are always inputs (x-values or t-values), read along the horizontal axis.