Quarry School

Read domain and range as shadows of a graph

Explain it like I am five

Imagine shining a light straight down onto a graph. The shadow on the horizontal number line shows every input position the graph visits. That shadow is the domain. Now shine a light from the side. The shadow on the vertical number line shows every output height the graph reaches. That shadow is the range. You are collecting coordinates, not measuring the length of the curve. A curve can turn around, cross the same height twice, or have separate pieces. You still collect each covered input or output once, and leave a gap only when no included point reaches that coordinate.

−3−2−1123−1123456789domainrange(0, 0)
For the full square function the horizontal shadow continues both ways, while the vertical shadow begins at zero.
Reminder
  • Coordinates. The point (2, 4) has input coordinate 2 and output coordinate 4.
  • Solving for an input. y = −2x + 3 gives x = 3−y2, an input for any chosen real y.
Why it works. Every point on a function graph is an ordered pair (x, y). Its horizontal coordinate tells which input was used; its vertical coordinate tells the resulting output. Projecting every included point onto an axis therefore collects exactly the relevant set. A hollow point excludes that point alone. Another point with the same x or y may keep that coordinate in a shadow. Arrows indicate continuation beyond the viewing window, so the edge of a picture is not automatically an endpoint. The answer describes the function's graph, not the camera frame.
RuleDomain = the graph's horizontal shadow on the x-axis.
Range = its vertical shadow on the y-axis; include a coordinate if any included point reaches it.
The same idea, five ways
Say it

Say domain as horizontal shadow and range as vertical shadow.

Write it

An included graph point contributes its x-coordinate to the domain and its y-coordinate to the range.

In math
  • For the pictured original curve:
  • Horizontal extent: x ≥ 6
  • Domain: {x | x ≥ 6} = [6, ∞)
  • Vertical extent: y ≤ 2
  • Range: {y | y ≤ 2} = (−∞, 2]
  • For the textbook graph in the ladder:
  • Horizontal extent: −3 < x ≤ 1
  • Domain: {x | −3 < x ≤ 1} = (−3, 1]
  • Vertical extent: −4 ≤ y ≤ 0
  • Range: {y | −4 ≤ y ≤ 0} = [−4, 0]
Like

Two lights project the same curve onto two different walls.

See it
46810121416−224domainrangeincluded start(7, 1)(10, 0)
For this pictured curve, the input shadow starts at 6 and the output shadow ends at 2. The textbook bounds listed separately belong to the textbook graph in the ladder.
The same idea, other ways
As two shadows

Drop every graph point onto the x-axis to gather the domain. Push every graph point sideways onto the y-axis to gather the range. The two shadows can have different limits.

2468−11234domainrange
The square root graph starts at zero on both axes and continues upward and rightward.
As addresses

Every graph point carries an address (input, output). If the graph contains (2, 4), then 2 belongs to the domain and 4 belongs to the range. One address supplies membership on both axes.

With a turning graph

For y = x2 on −2 < x ≤ 2, the left point (−2, 4) is absent. The right point (2, 4) is present. The input −2 is excluded, but the output 4 is still reached.

As a camera frame

A window may show a line only between x = −3 and x = 3. If arrows continue past the frame, those marks are viewing limits, not domain endpoints. Ask what the graph does beyond the picture.

.1A bounded graph with endpoints

A bounded graph fits between finite input limits or output limits. Inclusion belongs to a coordinate whenever the graph actually reaches it. A turn in the curve can make its lowest output occur in the middle rather than at an endpoint.

  • Read all reached heights, including a high or low point inside the interval.
  • Use brackets for reached endpoint coordinates.
  • One missing point does not remove a coordinate reached somewhere else.
−22246lowest heightleft boundary locationright boundary location
This is the full square shape; keep only −2 < x ≤ 2. The stated conditions make (−2, 4) open and (2, 4) closed in the restricted graph.
Reminder
  • Coordinates. The point (2, 4) has input coordinate 2 and output coordinate 4.
  • Solving for an input. y = −2x + 3 gives x = 3−y2, an input for any chosen real y.
The same idea, five ways
Say it

Say inputs after negative two through two; outputs zero through four.

Write it

Removing (−2, 4) does not remove height 4 when (2, 4) still belongs.

In math
  • −2 < x ≤ 2
  • Domain: {x | −2 < x ≤ 2} = (−2, 2]
  • 0 ≤ y ≤ 4
  • Range: {y | 0 ≤ y ≤ 4} = [0, 4]
Like

A fence marks finite positions a path may reach.

See it
−22246lowest heightleft boundary locationright boundary location
This is the full square shape; keep only −2 < x ≤ 2. The stated conditions make (−2, 4) open and (2, 4) closed in the restricted graph.
Worked exampleAn open point need not remove its height

The graph y = x2 is restricted to −2 < x ≤ 2. The point (−2, 4) is excluded, (2, 4) is included, and all intervening graph points are present. Find domain and range.

−22246lowest heightleft boundary locationright boundary location
This is the full square shape; keep only −2 < x ≤ 2. The stated conditions make (−2, 4) open and (2, 4) closed in the restricted graph.
−22(−2, 2]
Domain: (−2, 2]. The endpoint symbols record which limits belong.
04[0, 4]
Range: [0, 4]. The endpoint symbols record which limits belong.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Write (−2, 2] for the domain.These are exactly the supplied input limits, with −2 excluded and 2 included.
  3. The lowest height is 0, reached at x = 0.A square is nonnegative, and zero is an allowed input.
  4. The highest height is 4, reached at x = 2.The included right endpoint produces 4 even though the left point at that height is excluded.
  5. All heights between 0 and 4 occur, so write [0, 4].For a height y in this interval, the allowed input x = y lies between 0 and 2 and gives x2 = y.
Answer
  • Domain: (−2, 2].
  • Range: [0, 4].
Check At y = 1, x = 1 and x = −1 both work. At y = 4, x = 2 works. The absence of (−2, 4) removes one point, not every occurrence of height 4.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The open point (−2, 4) removes height 4 from the restricted square graph.
The included point (2, 4) still reaches that height.
✓ Instead: Domain (−2, 2]; range [0, 4].
Tips and tricks
  • Check the whole curve before removing an output height.
.2An unbounded graph

Unbounded means the graph continues without a finite limit in the direction being discussed. Arrows are a promise of continuation. They do not turn infinity into a point. A nonhorizontal line reaches arbitrarily large positive and negative heights as you travel along it.

  • Infinity always takes a parenthesis.
  • A viewing window is not an input restriction.
  • To check a line's full range, solve for the input that produces an arbitrary output.
−4−224−6−4−224681012domainrange(0, 3)(1.5, 0)
The line y = −2x + 3 continues past this window and reaches every real output height.
Reminder
  • Coordinates. The point (2, 4) has input coordinate 2 and output coordinate 4.
  • Solving for an input. y = −2x + 3 gives x = 3−y2, an input for any chosen real y.
The same idea, five ways
Say it

Say the line continues without an input or output limit.

Write it

The stated continuation extends the line beyond the viewing frame.

In math
  • y = −2x + 3
  • x = 3−y2
  • Domain: {x | x is real} = (−∞, ∞)
  • Range: {y | y is real} = (−∞, ∞)
Like

A long road continues beyond the edge of a photograph.

See it
−4−224−6−4−224681012domainrange(0, 3)(1.5, 0)
The line y = −2x + 3 continues past this window and reaches every real output height.
Worked exampleA viewing window does not stop a line

The picture shows part of y = −2x + 3. Its caption states that the line continues in both directions, and no input restriction is supplied. Find domain and range. This asks which input positions and output heights belong to the entire line.

−4−224−6−4−224681012domainrange(0, 3)(1.5, 0)
The line y = −2x + 3 continues past this window and reaches every real output height.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Keep every real x.Multiplication and addition are defined for every real input, and the caption states that the line continues.
  3. To reach a chosen output y, solve y = −2x + 3.Finding an input for every proposed output checks the entire range.
  4. Subtract 3 and divide by −2: x = 3−y2.The number −2 multiplying x is nonzero, so dividing by it gives a real input for every real y.
  5. Write all real numbers for both sets.Neither input positions nor output heights stop at the window's edges.
Answer
  • Domain: (−∞, ∞).
  • Range: (−∞, ∞).
Check To reach y = 13, choose x = −5: −2(−5) + 3 = 13. This output can lie beyond a small visible graph window.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The displayed frame supplies the line's last input.
The caption states that the line continues, and the formula supplies no finite endpoint.
✓ Instead: For y = −2x + 3, both domain and range are (−∞, ∞).
Tips and tricks
  • Distinguish an arrow from an endpoint marker.
.3A graph of real data

A graph from measurements has labels and units as well as a curve. Read the years on one axis and the measured quantity on the other. If the curve falls between grid lines, say that your limits are approximate instead of treating a visual estimate as an exact number.

  • The oil graph's visible inputs run from about 1973 through 2008.
  • Its output unit is thousands of barrels of oil per day.
  • The visible range is approximately [180, 2010] in that unit.
  • A continuous plotted model fills a time span; a table listing separate yearly observations gives only those listed years.
Visible years: [1973, 2008]
Production: about [180, 2010]
Output unit: thousand barrels per day
Keep the input years separate from the output production units.
Reminder
  • Coordinates. The point (2, 4) has input coordinate 2 and output coordinate 4.
  • Solving for an input. y = −2x + 3 gives x = 3−y2, an input for any chosen real y.
The same idea, five ways
Say it

Say the visible years and production levels with their units.

Write it

The oil graph describes a shown time span and approximately read production levels.

In math
  • 1973 ≤ t ≤ 2008
  • Visible domain: [1973, 2008]
  • 180 ≤ b ≤ 2010, approximately
  • Visible range: approximately [180, 2010] thousand barrels per day
Like

Read a measuring instrument's units before interpreting its marks.

See it
Visible years: [1973, 2008]
Production: about [180, 2010]
Output unit: thousand barrels per day
Keep the input years separate from the output production units.
Worked exampleTextbook: a visible oil-production graph

For the portion of the textbook oil-production graph shown, the horizontal extent is about 1973 through 2008 and the vertical extent is about 180 through 2010 thousand barrels per day. State the visible domain and range with units.

Visible domain: [1973, 2008], in years.
Visible range: approximately [180, 2010], in thousand barrels per day.
These lines record the calculated results and their boundary choices.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Identify the input as year and the output as oil production.The horizontal axis records time, and the vertical axis records thousands of barrels per day.
  3. Write [1973, 2008] for the visible domain.These are the first and last years in the portion being described, with endpoints included in the displayed model.
  4. Write approximately [180, 2010] for the range.These are the estimated smallest and largest plotted production levels, not output years.
  5. Retain thousand barrels per day as the output unit.180 on this vertical axis represents about 180000 barrels per day, not 180 barrels per day.
  6. Limit the claim to the shown portion of the graph.The supplied picture does not establish the behavior outside its displayed time span.
Answer
  • Visible domain: [1973, 2008], in years.
  • Visible range: approximately [180, 2010], in thousand barrels per day.
Check Multiplying the vertical limits by 1000 gives approximately 180000 through 2010000 barrels per day. These are approximate graph readings rather than exact algebraic bounds.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The output 180 means 180 barrels per day.
The oil-production axis is measured in thousands of barrels per day.
✓ Instead: 180 on this axis represents about 180000 barrels per day.
Tips and tricks
  • Write the unit immediately beside every approximate limit.
Strategy: step by step
  1. 1. Identify the x-axis as the input axis and the y-axis as the output axis; read their units.
  2. 2. Sweep left to right and collect all input positions covered by included graph points.
  3. 3. Sweep bottom to top and collect all reached output heights.
  4. 4. Check boundary dots. A hollow point excludes only that point, so look for other points sharing its coordinate.
  5. 5. Follow arrows beyond the visible window, and preserve genuine gaps with unions.
  6. 6. Write each interval from smaller to larger and label estimates from a data graph as approximate.
Strategy
Read the graph's two shadows
1
Does a boundary point belong?
YesKeep its coordinates in the two shadows.
NoCheck whether another included point shares its x-coordinate or y-coordinate.
↓
2
Does an arrow continue beyond the window?
YesFollow the indicated continuation instead of using the frame as an endpoint.
NoRead the stated or drawn finite limits.
↓
3
Is a coordinate gap reached nowhere?
YesPreserve the gap with separate intervals.
NoInclude that coordinate in the shadow.
  1. 1. Identify the x-axis as the input axis and the y-axis as the output axis; read their units.
  2. 2. Sweep left to right and collect all input positions covered by included graph points.
  3. 3. Sweep bottom to top and collect all reached output heights.
  4. 4. Check boundary dots. A hollow point excludes only that point, so look for other points sharing its coordinate.
  5. 5. Follow arrows beyond the visible window, and preserve genuine gaps with unions.
  6. 6. Write each interval from smaller to larger and label estimates from a data graph as approximate.
Worked exampleRead a starting point and two directions

Read the pictured curve’s domain and range. It has an included start at (6, 2) and continues rightward and downward without bound. You are collecting its horizontal input positions and vertical output heights.

46810121416−224domainrangeincluded start(7, 1)(10, 0)
Read the start (6, 2), then follow the curve to the right and downward beyond the frame.
  1. Locate the start at input 6 and output 2.The first coordinate is horizontal input, and the second is vertical output.
  2. Project the entire curve onto the input axis: it begins at 6 and continues through every larger input.The curve continues rightward and has no gap in its horizontal shadow.
  3. Include input 6 and write [6, ∞).The starting point belongs, so the domain begins with a bracket.
  4. Project onto the output axis: the highest height is 2 and the curve continues through every smaller height.The curve moves downward without a break or a lowest height.
  5. Write (−∞, 2] for the range.Height 2 belongs, smaller heights continue without limit, and limits are written in increasing order.
Answer
  • Domain: [6, ∞).
  • Range: (−∞, 2].
Check The included start contributes 6 to the domain and 2 to the range. The visible points (7, 1) and (10, 0) confirm the rightward and downward direction; the stated continuation extends both shadows beyond the frame.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Read a starting point and two directions

Read the pictured curve’s domain and range. It has an included start at (6, 2) and continues rightward and downward without bound. You are collecting its horizontal input positions and vertical output heights.

46810121416−224domainrangeincluded start(7, 1)(10, 0)
Read the start (6, 2), then follow the curve to the right and downward beyond the frame.
  1. Locate the start at input 6 and output 2.The first coordinate is horizontal input, and the second is vertical output.
  2. Project the entire curve onto the input axis: it begins at 6 and continues through every larger input.The curve continues rightward and has no gap in its horizontal shadow.
  3. Include input 6 and write [6, ∞).The starting point belongs, so the domain begins with a bracket.
  4. Project onto the output axis: the highest height is 2 and the curve continues through every smaller height.The curve moves downward without a break or a lowest height.
  5. Write (−∞, 2] for the range.Height 2 belongs, smaller heights continue without limit, and limits are written in increasing order.
Answer
  • Domain: [6, ∞).
  • Range: (−∞, 2].
Check The included start contributes 6 to the domain and 2 to the range. The visible points (7, 1) and (10, 0) confirm the rightward and downward direction; the stated continuation extends both shadows beyond the frame.
Rung 2Textbook: read the full horizontal and vertical extents

The textbook graph has a hollow left endpoint at x = −3 and an included right endpoint at x = 1. It covers every x between them, reaches y = −4 and y = 0, and covers every height between them. Find domain and range. This asks which input positions and output heights belong, including the endpoint decisions.

−31(−3, 1]
The horizontal shadow excludes −3 and includes 1; this matches the textbook graph’s endpoint dots.
−40[−4, 0]
Range: [−4, 0]. The endpoint symbols record which limits belong.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Read the horizontal extent from −3 through 1.The x-coordinate is the input, so the horizontal shadow gives the domain.
  3. Exclude −3, include 1, and write (−3, 1].The left endpoint is hollow and no other point has x = −3. The right endpoint is included. Input endpoints and output endpoints are separate decisions.
  4. Read the vertical extent from −4 through 0.The y-coordinate is the output, and the smaller height must be written first.
  5. Include both reached heights and write [−4, 0].Every height between them is reached, including the endpoints.
Answer
  • Domain: (−3, 1].
  • Range: [−4, 0].
Check Input −3 is absent and input 1 is present, matching the horizontal shadow. The graph reaches heights −4 and 0, so both range endpoints stay included. An output height of −2 belongs, while 1 does not; an input of 0 belongs, while −4 does not.
Rung 3An open point need not remove its height

The graph y = x2 is restricted to −2 < x ≤ 2. The point (−2, 4) is excluded, (2, 4) is included, and all intervening graph points are present. Find domain and range.

−22246lowest heightleft boundary locationright boundary location
This is the full square shape; keep only −2 < x ≤ 2. The stated conditions make (−2, 4) open and (2, 4) closed in the restricted graph.
−22(−2, 2]
Domain: (−2, 2]. The endpoint symbols record which limits belong.
04[0, 4]
Range: [0, 4]. The endpoint symbols record which limits belong.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Write (−2, 2] for the domain.These are exactly the supplied input limits, with −2 excluded and 2 included.
  3. The lowest height is 0, reached at x = 0.A square is nonnegative, and zero is an allowed input.
  4. The highest height is 4, reached at x = 2.The included right endpoint produces 4 even though the left point at that height is excluded.
  5. All heights between 0 and 4 occur, so write [0, 4].For a height y in this interval, the allowed input x = y lies between 0 and 2 and gives x2 = y.
Answer
  • Domain: (−2, 2].
  • Range: [0, 4].
Check At y = 1, x = 1 and x = −1 both work. At y = 4, x = 2 works. The absence of (−2, 4) removes one point, not every occurrence of height 4.
Rung 4A viewing window does not stop a line

The picture shows part of y = −2x + 3. Its caption states that the line continues in both directions, and no input restriction is supplied. Find domain and range. This asks which input positions and output heights belong to the entire line.

−4−224−6−4−224681012domainrange(0, 3)(1.5, 0)
The line y = −2x + 3 continues past this window and reaches every real output height.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Keep every real x.Multiplication and addition are defined for every real input, and the caption states that the line continues.
  3. To reach a chosen output y, solve y = −2x + 3.Finding an input for every proposed output checks the entire range.
  4. Subtract 3 and divide by −2: x = 3−y2.The number −2 multiplying x is nonzero, so dividing by it gives a real input for every real y.
  5. Write all real numbers for both sets.Neither input positions nor output heights stop at the window's edges.
Answer
  • Domain: (−∞, ∞).
  • Range: (−∞, ∞).
Check To reach y = 13, choose x = −5: −2(−5) + 3 = 13. This output can lie beyond a small visible graph window.
Rung 5Textbook: a visible oil-production graph

For the portion of the textbook oil-production graph shown, the horizontal extent is about 1973 through 2008 and the vertical extent is about 180 through 2010 thousand barrels per day. State the visible domain and range with units.

Visible domain: [1973, 2008], in years.
Visible range: approximately [180, 2010], in thousand barrels per day.
These lines record the calculated results and their boundary choices.
  1. We need every allowed input and every output the rule actually reaches.State what is being found before choosing the calculation.
  2. Identify the input as year and the output as oil production.The horizontal axis records time, and the vertical axis records thousands of barrels per day.
  3. Write [1973, 2008] for the visible domain.These are the first and last years in the portion being described, with endpoints included in the displayed model.
  4. Write approximately [180, 2010] for the range.These are the estimated smallest and largest plotted production levels, not output years.
  5. Retain thousand barrels per day as the output unit.180 on this vertical axis represents about 180000 barrels per day, not 180 barrels per day.
  6. Limit the claim to the shown portion of the graph.The supplied picture does not establish the behavior outside its displayed time span.
Answer
  • Visible domain: [1973, 2008], in years.
  • Visible range: approximately [180, 2010], in thousand barrels per day.
Check Multiplying the vertical limits by 1000 gives approximately 180000 through 2010000 barrels per day. These are approximate graph readings rather than exact algebraic bounds.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: An open point at (−2, 4) means 4 is not in the range.
The included point (2, 4) also supplies height 4 in the restricted square graph.
✓ Instead: For −2 < x ≤ 2 and y = x2, the range is [0, 4].
✗ Not this: The line picture ends at x = 4, so the domain ends at 4.
The stated continuation extends the line beyond the displayed frame. A window border is not an assigned endpoint.
✓ Instead: For y = −2x + 3 with no restriction, the domain is (−∞, ∞).
Tips and tricks
  • Label the axes input and output before reading their shadows.
  • Read the bottom height first when writing the range.
  • Look for another included point at a hollow point's height before excluding that height.
Trap. Deleting an output height because one point at that height is hollow. Look across the whole graph. A different included point can still put that height in the range.