Free math worksheets

- Decide whether distance rises, stays constant, or falls in each labeled section. - A downward graph segment means the distance from the starting point is decreasing.

1. In section \(a\), distance increases as time increases, so the slope is positive. In section \(b\), distance remains constant at \(5\) miles, so the slope is zero. In section \(c\), distance decreases as time increases, so the slope is negative. 2. A negative slope means the hiker's distance from the starting point is decreasing. The hiker is traveling back toward the starting point.

a) \(a\): positive; \(b\): zero; \(c\): negative b) The hiker is moving back toward the starting point.

- Read the axis labels before checking each statement. - A horizontal part of the graph means the elevation does not change. - Find the greatest vertical value to identify the highest point.

1. At hour \(0\), the graph shows \(400\,\text{ft}\), so statement 1 is true. 2. From hour \(3\) to hour \(4\), the graph is horizontal at \(1000\,\text{ft}\), so statement 2 is true. 3. The greatest elevation is \(1400\,\text{ft}\) at hour \(5\), so statement 3 is false.

1. **True:** The graph begins at \(400\,\text{ft}\). 2. **True:** The elevation remains \(1000\,\text{ft}\) from hour \(3\) to hour \(4\). 3. **False:** The highest point is \(1400\,\text{ft}\).

- Use the horizontal axis to locate the time and the vertical axis to read the height. - For question 2, start at \(150\,\text{cm}\) on the vertical axis and find the matching time on the graph.

1. At week \(4\), the graph has a height of \(75\,\text{cm}\). 2. The graph reaches \(150\,\text{cm}\) at week \(6\).

1. \(75\,\text{cm}\) 2. \(6\) weeks

- Think about how much horizontal space the water has at each height. - A steeper graph represents a faster increase in water height. - For equal volumes of water, compare the thickness of a layer in a wide section with one in a narrow section.

1. The cross-sectional area increases as the water rises. Because each equal amount of water spreads over a larger area, the water height rises quickly at first and then more slowly. 2. The graph increases, but its slope decreases over time. Therefore, the graph becomes less steep. 3. If the container narrows near the top, the same amount of water produces a larger rise in height. The graph becomes steeper again in that region.

a) The water height rises more slowly over time because the container becomes wider. b) The increasing graph becomes less steep. c) The graph becomes steeper again where the container narrows.

- Read horizontal graph heights as constant speeds over the corresponding distance intervals. - At \(s=1\), distinguish the open point from the filled point before stating Tim's speed. - To compare riders at one distance, compare the vertical positions of their graphs at the same x-value. - Equal speeds would appear as an intersection of the two graphs.

1. On \(0\le s<1\), Tim's graph is horizontal at \(15\,\text{mph}\). At \(s=1\), the open point at \(15\) and filled point at \(9\) show an instantaneous drop to \(9\,\text{mph}\). On \(1\le s\le3\), his graph is horizontal at \(9\,\text{mph}\). 2. At \(s=0.5\), Tim's graph is at \(15\,\text{mph}\), above Lisa's \(12\,\text{mph}\), so Tim is faster. 3. At \(s=2.5\), Tim's graph is at \(9\,\text{mph}\), below Lisa's \(12\,\text{mph}\), so Lisa is faster. 4. The two graphs never intersect, so there is no distance at which their speeds are equal.

a) Tim travels at \(15\,\text{mph}\) for \(0\le s<1\), then drops to \(9\,\text{mph}\) at \(s=1\) and stays there through \(s=3\). b) Tim at \(0.5\) mile; Lisa at \(2.5\) miles c) No; the graphs do not intersect.

- Read the time on the horizontal axis and temperature on the vertical axis. - The maximum is the highest point of the graph. - Find both intersections with the horizontal line \(y=32\). - Subtract the earlier temperature from the later temperature.

1. The graph gives \(23\,\text{°F}\) at \(6{:}00\) a.m., \(34\,\text{°F}\) at \(10{:}00\) a.m., \(45\,\text{°F}\) at \(2{:}00\) p.m., \(34\,\text{°F}\) at \(6{:}00\) p.m., and \(23\,\text{°F}\) at \(10{:}00\) p.m. 2. The highest point occurs at \(2{:}00\) p.m., with a maximum temperature of \(45\,\text{°F}\). 3. The graph crosses \(32\,\text{°F}\) at about \(9{:}32\) a.m. and \(6{:}28\) p.m. Thus, the temperature is above freezing between those times. 4. The increase is \(45-23=22\,\text{°F}\).

a) \(6{:}00\) a.m.: \(23\,\text{°F}\); \(10{:}00\) a.m.: \(34\,\text{°F}\); \(2{:}00\) p.m.: \(45\,\text{°F}\); \(6{:}00\) p.m.: \(34\,\text{°F}\); \(10{:}00\) p.m.: \(23\,\text{°F}\) b) \(2{:}00\) p.m.; \(45\,\text{°F}\) c) About \(9{:}32\) a.m. to \(6{:}28\) p.m. d) \(22\,\text{°F}\)

- Read each requested output directly from the graph. - Compare the eclipse section with the otherwise smooth daily pattern. - Percent decrease is \(\frac{\text{decrease}}{\text{starting value}}\cdot100\%\).

1. The graph shows \(6\,\text{kW}\) at \(10{:}00\) a.m. and \(2\,\text{kW}\) at \(12{:}00\) p.m. 2. The eclipse effect begins at \(11{:}00\) a.m., reaches its greatest effect at \(12{:}00\) p.m., and ends at \(1{:}00\) p.m. 3. Output decreases from \(8\) to \(2\) kilowatts, a decrease of \(6\) kilowatts. The percent decrease is \(\frac{6}{8}=0.75=75\%\).

a) \(6\,\text{kW}\) at \(10{:}00\) a.m.; \(2\,\text{kW}\) at \(12{:}00\) p.m. b) Beginning: \(11{:}00\) a.m.; greatest effect: \(12{:}00\) p.m.; end: \(1{:}00\) p.m. c) \(75\%\)

- Identify the quantities and units on each axis. - For a specified time, move vertically to the graph and then horizontally to the temperature axis. - For a specified temperature, trace a horizontal level and check for more than one intersection. - The lowest point of the graph gives the daily minimum.

1. Read the graph at the requested times: <table> <tr><th>Time</th><td>9 a.m.</td><td>3 p.m.</td><td>11 p.m.</td></tr> <tr><th>Temperature</th><td>\(12\,^{\circ}\text{C}\)</td><td>\(18\,^{\circ}\text{C}\)</td><td>\(9\,^{\circ}\text{C}\)</td></tr> </table> 2. The horizontal line at \(15\,^{\circ}\text{C}\) meets the graph at \(t=11\) and \(t=19\), corresponding to \(11\) a.m. and \(7\) p.m. 3. The lowest point is \((3, 6)\), so the minimum temperature was \(6\,^{\circ}\text{C}\) at \(3\) a.m.

a) <table> <tr><th>Time</th><td>9 a.m.</td><td>3 p.m.</td><td>11 p.m.</td></tr> <tr><th>Temperature</th><td>\(12\,^{\circ}\text{C}\)</td><td>\(18\,^{\circ}\text{C}\)</td><td>\(9\,^{\circ}\text{C}\)</td></tr> </table> b) \(11\) a.m. and \(7\) p.m. c) \(3\) a.m.; \(6\,^{\circ}\text{C}\)

- Match straight, curved, and horizontal sections to the verbal descriptions. - A horizontal segment means the water level is constant. - Divide the total decrease by the elapsed time.

1. Graph A matches description 2: it rises steadily, remains constant during a rain pause, and then falls quickly. 2. Graph B matches description 1: the water level falls slowly from \(t=0\) to \(t=3\), then much faster from \(t=3\) to \(t=6\). 3. Graph C matches description 3: its increasing curve becomes steeper and then changes to a horizontal segment. 4. In Graph B, the level falls from \(9\) feet to \(0\) feet over \(3\) hours. The average decrease is \(\frac{9}{3}=3\) feet per hour.

a) \(A\to2\); \(B\to1\); \(C\to3\) b) \(3\,\text{ft/h}\) downward

- The initial temperature is the graph's value when \(t=0\). - The maximum temperature occurs at the highest point of the graph. - For part c), compare the graph with a horizontal line at \(40\,\text{°C}\).

1. At \(t=0\), the graph has a value of \(20\,\text{°C}\), so the initial temperature is \(20\,\text{°C}\). 2. The highest point on the graph is \((20, 45)\). Therefore, the maximum temperature is \(45\,\text{°C}\), reached after \(20\) minutes. 3. The graph is on or above \(40\,\text{°C}\) from \(t=10\) to \(t=40\). The duration is \(40-10=30\) minutes.

a) \(20\,\text{°C}\) b) After \(20\) minutes; \(45\,\text{°C}\) c) From \(10\) to \(40\) minutes; \(30\) minutes total

- Read the first point and the highest point on the graph. - A horizontal segment represents no change in elevation. - Elevation gain is summit elevation minus starting elevation. - For part d), locate the requested elevation on the descending segment.

1. The graph begins at \(600\,\text{ft}\) and reaches a maximum of \(3000\,\text{ft}\). 2. The horizontal segment at the summit runs from \(t=4\) to \(t=6\), so the rest lasts \(6-4=2\) hours. 3. The elevation gain during the climb is \(3000-600=2400\,\text{ft}\). 4. On the descending segment, the graph reaches \(2100\,\text{ft}\) at \(t=7\).

a) Starts at \(600\,\text{ft}\); maximum \(3000\,\text{ft}\) b) \(2\) hours c) \(2400\,\text{ft}\) d) After \(7\) hours

- Age \(0\) gives the initial estimated resale value shown by the graph. - The vertical-axis values are in thousands of dollars. - For part b), first identify half of the age-\(0\) graph value, then locate that output on the graph. - For part c), locate \(6\) on the vertical axis and find the first corresponding age.

1. At age \(0\), the graph shows \(24\) thousand dollars, so the initial estimated resale value is \(\$24{,}000\). At age \(4\), the graph shows about \(9.5\) thousand dollars, or about \(\$9500\). 2. Half of \(\$24{,}000\) is \(\$12{,}000\). The graph reaches \(12\) thousand dollars at age \(3\), so the estimated resale value is half its age-\(0\) value after \(3\) years. 3. The graph reaches \(6\) thousand dollars at age \(6\). Therefore, the car is worth \(\$6000\) or less after \(6\) years.

a) \(\$24{,}000\) at age \(0\); about \(\$9500\) after \(4\) years b) \(3\) years c) \(6\) years

- Peaks are local high points on the graph. - Use the axis scales when estimating values. - Separate what the graph shows from a possible explanation for why it happened.

1. The three largest local peaks occur at about \(8{:}00\) a.m., \(1{:}00\) p.m., and \(8{:}00\) p.m. 2. At \(10{:}00\) a.m., the graph is at about \(0.65\,\text{GB/min}\). 3. The data rate decreases from about \(2.5\,\text{GB/min}\) at \(1{:}00\) p.m. to about \(0.55\,\text{GB/min}\) at \(3{:}00\) p.m. One possible explanation is that a temporary period of heavy activity ended. The graph alone cannot establish the cause. 4. The lowest displayed rate is about \(0.5\,\text{GB/min}\).

a) About \(8{:}00\) a.m., \(1{:}00\) p.m., and \(8{:}00\) p.m. b) About \(0.65\,\text{GB/min}\) c) The data rate decreases sharply; a temporary high-activity period may have ended. d) About \(0.5\,\text{GB/min}\)

- Decide whether each quantity increases, decreases, or repeats. - Look for constant change, curved change, periodic behavior, or growth that levels off.

1. Description a) matches Graph 1 because the graph repeatedly rises and falls in a periodic pattern. 2. Description b) matches Graph 2 because a constant amount used each day produces a linear decrease. 3. Description c) matches Graph 3 because the elephant's mass increases slowly at first, then more rapidly, and eventually levels off as it approaches adult size. 4. Description d) matches Graph 4 because radioactive decay decreases rapidly at first and then approaches zero more slowly.

a) Graph 1 b) Graph 2 c) Graph 3 d) Graph 4

- Move vertically from each time value to the graph, then horizontally to the distance axis. - A horizontal segment means the total distance is not changing. - Compare segment steepness to compare speeds.

1. The graph shows \(4\) miles after \(2\) hours and \(7\) miles after \(5\) hours. 2. Reading the graph at each whole hour gives: <table> <tr><th>Time (h)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><th>Distance (mi)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(6\)</td><td>\(7\)</td></tr> </table> 3. During the first \(3\) hours, the hiker travels at a constant rate of \(2\,\text{mph}\). From hour \(3\) to hour \(4\), the total distance does not change, so the hiker likely rests. During the final hour, the hiker travels at \(1\,\text{mph}\), which is slower than before.

a) After \(2\) hours: \(4\) miles; after \(5\) hours: \(7\) miles b) <table> <tr><th>Time (h)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><th>Distance (mi)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(6\)</td><td>\(7\)</td></tr> </table> c) The hiker travels steadily for \(3\) hours, likely rests for \(1\) hour, and then continues at a slower rate.

- Read the graph at each requested time. - The courier is stopped wherever the graph lies on the time axis. - The maximum speed is the graph's highest value.

1. Reading the graph gives \(10\,\text{mph}\) at \(5\) minutes, \(20\,\text{mph}\) at \(15\) minutes, \(0\,\text{mph}\) at \(30\) minutes, and \(25\,\text{mph}\) at \(50\) minutes. 2. The graph is at \(0\) from \(t=25\) to \(t=40\), so she is stopped for \(40-25=15\) minutes. 3. The greatest y-value is \(25\), so her maximum speed is \(25\,\text{mph}\).

a) \(10\,\text{mph}\), \(20\,\text{mph}\), \(0\,\text{mph}\), \(25\,\text{mph}\) b) \(15\) minutes c) \(25\,\text{mph}\)

- In a distance-time graph, slope represents speed. - A horizontal section means no distance is added. - Compare whether each graph becomes steeper or less steep over time.

1. Graph a is a straight line through the origin, so it has constant slope and represents constant speed. 2. Graph b has a horizontal section from \(t = 20\) to \(t = 30\), representing a stop. Its slope after the stop, \(0.3\,\text{mi/min}\), is greater than its slope before the stop, \(0.2\,\text{mi/min}\). 3. Graph c curves upward, so its slope increases over time. 4. Graph d curves downward, so its slope decreases over time. 5. Therefore, Leon matches a, Sarah matches b, Tim matches c, and Maya matches d.

Leon \(\rightarrow\) graph a; Sarah \(\rightarrow\) graph b; Tim \(\rightarrow\) graph c; Maya \(\rightarrow\) graph d

- Follow the marked measurements from left to right and describe each change in direction. - For part b), compare the heights of the marked points rather than the smooth curve between them. - For part c), identify the greatest and least measured outputs before finding their difference.

1. From midnight to \(3\) a.m., the water depth rises from \(3.0\,\text{m}\) to \(5.2\,\text{m}\). It then falls to \(3.0\,\text{m}\) at \(6\) a.m. and to \(0.8\,\text{m}\) at \(9\) a.m. By noon, it rises again to \(3.0\,\text{m}\). 2. The greatest measured depth is \(5.2\,\text{m}\), recorded at \(3\) a.m. and \(3\) p.m. 3. The least measured depth is \(0.8\,\text{m}\). The difference is \(5.2\,\text{m}-0.8\,\text{m}=4.4\,\text{m}\).

a) The depth rises from \(3.0\,\text{m}\) to \(5.2\,\text{m}\), falls to \(0.8\,\text{m}\), and then rises to \(3.0\,\text{m}\). b) \(3\) a.m. and \(3\) p.m.; \(5.2\,\text{m}\) c) \(4.4\,\text{m}\)

- Performances are likely associated with sustained high sound levels. - A stage-change break is likely associated with a lower, flatter section.

1. After the initial rise, the sound level remains near \(85\,\text{dB}\) from about minute \(5\) to minute \(40\), so this is when the opening band likely played. The headliner likely began around minute \(65\), when the sound level rises sharply to about \(105\,\text{dB}\). 2. The stage-change break likely occurred from about minute \(45\) to minute \(60\), when the sound level remains near \(60\,\text{dB}\).

1. Opening band: about minutes \(5\)–\(40\); headliner: beginning about minute \(65\) 2. Stage-change break: about minutes \(45\)–\(60\)

- Consider how people in a room affect carbon dioxide concentration. - Think about actions that can improve ventilation quickly. - Distinguish a plausible explanation from something the graph proves.

1. Near \(9{:}30\) a.m. and \(11{:}15\) a.m., the concentration drops suddenly. The room was most likely ventilated by opening windows or doors. 2. A plausible explanation is that classes ended and students left the room. With fewer people exhaling carbon dioxide, the concentration decreases toward the outdoor level. The graph supports this explanation but does not prove the cause.

a) The room was most likely ventilated. b) Classes may have ended and the students may have left the room.

- Horizontal segments represent times when the van's distance from the depot is unchanged. - The most distant customer corresponds to the graph's greatest y-value. - Subtract the return-trip start time from its end time.

1. The horizontal segments last \(10\), \(15\), and \(10\) minutes. The total stop time is \(10+15+10=35\) minutes. 2. The greatest distance on the graph is \(8\) miles. 3. The return trip begins at \(t=75\) minutes and ends at \(t=90\) minutes, so it lasts \(15\) minutes.

a) \(35\) minutes b) \(8\) miles c) \(15\) minutes

- Check what each axis represents. - High points represent local or global maximum values; low points represent local minimum values. - The endpoint of this graph represents the end of the ride.

1. At a distance of \(600\,\text{ft}\), the graph shows a speed of \(70\,\text{mph}\). 2. The highest point of the graph is \((600, 70)\), so the greatest speed is \(70\,\text{mph}\). 3. The graph first reaches a local minimum at a distance of \(300\,\text{ft}\). The speed there is about \(3\,\text{mph}\). 4. The graph ends at \((2400, 0)\), so the track is \(2400\,\text{ft}\) long.

a) \(70\,\text{mph}\) b) \(70\,\text{mph}\) c) After \(300\,\text{ft}\), at about \(3\,\text{mph}\) d) \(2400\,\text{ft}\)

- Horizontal segments represent constant speed. - A speed of \(0\) means the ship is stopped. - Use the graph's corner points to identify each interval.

1. On the horizontal sections from \(0\) to \(8\) minutes and from \(12\) to \(20\) minutes, the ship travels at \(12\) knots. 2. The graph is at \(0\) knots from \(t=9\) to \(t=11\), so the ship is stopped for that interval. 3. At \(t=5\), the speed is \(12\) knots. At \(t=10\), the speed is \(0\) knots. 4. From \(t=11\) to \(t=12\), the ship speeds up from \(0\) to \(12\) knots. From \(t=12\) to \(t=13\), it maintains \(12\) knots.

a) \(12\) knots b) From \(9\) to \(11\) minutes c) \(12\) knots at \(5\) minutes; \(0\) knots at \(10\) minutes d) The ship speeds up to \(12\) knots and then maintains that speed.

- Look for the peaks of the graph and interpret what they represent. - Read the highest y-value and its corresponding time. - For the reverse relation, check whether one y-value can correspond to more than one x-value.

1. The two peaks indicate the busiest arrival times, about \(10{:}00\) a.m. and \(4{:}00\) p.m. 2. The greatest sound level is about \(90\,\text{dB}\), reached at \(10{:}00\) a.m. The second peak, at \(4{:}00\) p.m., is about \(80\,\text{dB}\). 3. No. A single sound level can occur at more than one time. Therefore, assigning a time to each sound level does not give exactly one output for every input.

a) About \(10{:}00\) a.m. and \(4{:}00\) p.m. b) About \(90\,\text{dB}\) at \(10{:}00\) a.m. c) No. The same sound level can correspond to more than one time.

- Decide whether each quantity changes at a constant rate, levels off, or repeats. - A periodic graph represents a repeating process.

1. Description 1 matches Graph A. Steady rainfall adds water at a constant volume rate, and a cylindrical barrel has constant cross-sectional area, so water height increases linearly. 2. Description 2 matches Graph B. A child grows rapidly at first, then more slowly, and height eventually levels off near adulthood. 3. Description 3 matches Graph C. A pendulum's height rises and falls in a repeating periodic pattern.

1) Graph A 2) Graph B 3) Graph C

- Decide whether each graph rises, falls, or eventually becomes constant. - For situation 1, look for a decreasing graph that becomes less steep over time. - For situation 2, look for slow growth, then faster growth, then leveling off; for situation 3, look for a rise followed by a horizontal segment.

1. Situation 1 matches Graph A. The graph decreases rapidly at first and then becomes less steep, so the rate of decrease becomes slower. 2. Situation 2 matches Graph B. Its S-shape represents slow initial growth, more rapid growth, and then leveling off near a limiting population. 3. Situation 3 matches Graph C. The speed first increases and then remains constant, represented by a rising segment followed by a horizontal segment.

1) Graph A 2) Graph B 3) Graph C

- Use the axis labels to identify the units. - The highest point is the greatest vertical value on the graph. - The hikers are traveling uphill wherever the graph rises from left to right.

1. The greatest elevation on the graph occurs at mile \(8\), where the elevation is \(1500\,\text{ft}\). 2. The least elevation is \(800\,\text{ft}\) at mile \(0\). 3. The graph rises from mile \(0\) to mile \(2\) and from mile \(5\) to mile \(8\).

a) At mile \(8\); \(1500\,\text{ft}\) b) At mile \(0\) c) From mile \(0\) to mile \(2\), and from mile \(5\) to mile \(8\)

- Read the graph at \(t=6\). - Compare the graph with a horizontal line at \(c=40\). - Because the problem says “greater than,” do not include the boundary times.

1. At \(t=6\), the graph has a value of \(30\,\text{mg/L}\). 2. The graph is at \(40\,\text{mg/L}\) at \(t=1\) and \(t=4\). Between those times, it is above \(40\,\text{mg/L}\). Because the inequality is strict, the concentration is greater than \(40\,\text{mg/L}\) for \(1<t<4\).

a) \(30\,\text{mg/L}\) b) \(1<t<4\) hours

- Check which quantity appears on each axis. - To find \(T(1)\), begin at \(1\) on the horizontal axis and move vertically to the graph. - Read both function values before subtracting. - For a specified temperature, look for every point where the graph has that y-value.

1. At \(t=1\), the graph has a value of \(38\), so \(T(1)=38\,\text{°F}\). 2. At \(t=3\), the graph has a value of \(30\), so \(T(3)=30\,\text{°F}\). 3. The graph shows \(T(2)=34\) and \(T(4)=34\). Therefore, \(T(2)-T(4)=34-34=0\,\text{°F}\). 4. The graph has a value of \(34\,\text{°F}\) at \(t=0\), \(t=2\), and \(t=4\) hours.

a) \(38\,\text{°F}\) b) \(30\,\text{°F}\) c) \(0\,\text{°F}\) d) At \(t=0\), \(t=2\), and \(t=4\) hours

- Account for the fact that the vertical axis is measured in thousands of dollars. - A function value is the y-coordinate corresponding to the given x-coordinate. - To find a change, subtract the earlier function value from the later one. - Break-even points are x-intercepts.

1. At \(x=3\), the graph has a y-value of \(4\). Because profit is measured in thousands of dollars, the profit is \(\$4000\). 2. At \(x=4\), the graph has a y-value of \(3\), so \(p(4)=3\), representing \(\$3000\). 3. The graph shows \(p(1)=0\) and \(p(2)=3\). Thus, \(p(2)-p(1)=3-0=3\), so profit increases by \(\$3000\). 4. The graph crosses the x-axis at \(x=1\) and \(x=5\). These are the break-even production levels.

a) \(\$4000\) b) \(p(4)=3\), or \(\$3000\) c) An increase of \(\$3000\) d) \(x=1\) and \(x=5\)

- The horizontal axis shows time, and the vertical axis shows distance from home. - A horizontal segment means time passes while the distance stays the same. - The return trip is the part where the distance from home decreases. - For the total distance, include both the trip away from home and the trip back.

1. At \(10\) minutes, the graph shows a distance of \(3\) miles. 2. The graph is horizontal from \(10\) to \(15\) minutes, so the stop lasts \(15 - 10 = 5\) minutes. 3. On the return trip, the distance decreases from \(3\) miles at \(15\) minutes to \(0\) miles at \(25\) minutes. The graph reaches \(1.5\) miles halfway through that interval, at \(20\) minutes. 4. Lucas rides \(3\) miles to his friend’s house and \(3\) miles home, for a total of \(3 + 3 = 6\) miles.

1) \(3\) miles 2) \(5\) minutes 3) \(20\) minutes 4) \(6\) miles

- The horizontal axis shows the hour of the day, and the vertical axis shows temperature. - To find when the temperature was \(32\,{}^\circ\text{F}\), find where the graph crosses that horizontal level. - The lowest point gives the minimum temperature. - The temperature rises wherever the graph moves upward from left to right.

1. At hour \(6\), the graph shows \(28\,{}^\circ\text{F}\). At hour \(14\), it shows \(43\,{}^\circ\text{F}\). 2. The graph reaches \(32\,{}^\circ\text{F}\) at hour \(4\) and hour \(11\), which are \(4{:}00\) a.m. and \(11{:}00\) a.m. 3. The lowest point is \(26\,{}^\circ\text{F}\) at \(8{:}00\) a.m. 4. The graph rises continuously from hour \(8\) to hour \(17\), or from \(8{:}00\) a.m. to \(5{:}00\) p.m.

1. \(28\,{}^\circ\text{F}\) at \(6{:}00\) a.m.; \(43\,{}^\circ\text{F}\) at \(2{:}00\) p.m. 2. \(4{:}00\) a.m. and \(11{:}00\) a.m. 3. \(26\,{}^\circ\text{F}\) at \(8{:}00\) a.m. 4. From \(8{:}00\) a.m. to \(5{:}00\) p.m.

- To find \(f(x)\), locate the input on the x-axis and read the corresponding y-coordinate. - Zeros occur where the graph meets the x-axis. - The domain is the horizontal extent of the graph, and the range is its vertical extent. - Positive outputs occur where the graph lies above the x-axis.

1. Read the y-coordinate of the graph at each input: \(f(-3)=0\), \(f(-1)=2\), and \(f(3)=-2\). 2. The graph meets the x-axis at \(x=-3\), \(x=1\), and \(x=5\). 3. The graph extends from \(x=-4\) through \(x=5\), so the domain is \([-4, 5]\). Its lowest output is \(-3\) and its highest output is \(2\), so the range is \([-3, 2]\). 4. The graph is above the x-axis between the zeros \(-3\) and \(1\), so \(f(x)>0\) for \(-3<x<1\).

1) \(f(-3)=0\), \(f(-1)=2\), and \(f(3)=-2\) 2) \(x\in\{-3, 1, 5\}\) 3) Domain: \([-4, 5]\); range: \([-3, 2]\) 4) \(-3<x<1\)

- Translate each event time and distance into a coordinate pair. - A rest keeps distance unchanged, so that part of the graph must be horizontal. - Returning home means the distance from home decreases to \(0\).

1. Devon starts at home, so the graph begins at \((0,0)\). 2. After \(30\) minutes, Devon is \(2\) miles from home, giving the point \((30,2)\). The first segment is increasing. 3. Devon rests for \(15\) minutes, so the distance stays \(2\) miles from minute \(30\) through minute \(45\). This gives the point \((45,2)\) and a constant segment from \((30,2)\) to \((45,2)\). 4. Devon reaches home at \(60\) minutes, giving the endpoint \((60,0)\). The final segment is decreasing.

Vertices: \((0,0)\), \((30,2)\), \((45,2)\), \((60,0)\). The graph is increasing from \(0\) to \(30\) minutes, constant from \(30\) to \(45\) minutes, and decreasing from \(45\) to \(60\) minutes.

- Consider what a speed of \(0\) means. - Identify the quantity on the horizontal axis. - Think about what an odometer does while a car is stopped.

1. A speed of \(0\,\text{mph}\) at mile \(5\), with positive speed immediately before and after, represents an idealized stop at that location, such as a red light or stop sign. 2. The horizontal axis shows distance, not time. During a stop, the car's distance does not change, so the entire waiting period occurs at the single distance \(s=5\). The graph can show that the speed there is \(0\), but it cannot show how long the stop lasts.

a) The car made an idealized stop at mile \(5\), perhaps at a traffic signal or stop sign. b) Distance remains fixed during the stop, and the graph has no time axis, so the waiting time is not represented.

- Read the two places where the blue line meets the dashed horizontal boundaries. - Follow the line from left to right and describe its position relative to the strip in each region. - An open strip excludes points exactly on either boundary. - Use the overall left-to-right direction of the line to describe whether the function is increasing or decreasing.

1. The line crosses the lower dashed boundary \(y=-3\) at \(x=-4\) and the upper dashed boundary \(y=2\) at \(x=6\). 2. Because the line is increasing, it lies below the strip for \(x<-4\), inside the strip for \(-4<x<6\), and above the strip for \(x>6\). 3. The strip is open, so the crossing points themselves are excluded. 4. The graph rises from left to right, so it must meet the lower boundary before the upper boundary.

a) Below for \(x<-4\); inside for \(-4<x<6\); above for \(x>6\) b) Crossings at \(x=-4\) and \(x=6\); inside interval: \(-4<x<6\) c) \(g\) is increasing; it crosses the lower boundary first and the upper boundary second.

- Think about what a corner in Graph A says about the cooling rate. - For a temperature at a given time, move vertically from the time to Graph B. - For a given temperature, move horizontally to Graph B and then down to the time axis.

1. A liquid's temperature changes continuously, and its cooling rate does not suddenly jump at each measurement time. The corners in Graph A represent abrupt rate changes, while Graph B represents a smooth physical process. 2. At \(t=12\) minutes, Graph B is at about \(121\,\text{°F}\). A reasonable graph-reading range is approximately \(119\)–\(122\,\text{°F}\). 3. The horizontal line \(T=104\) intersects Graph B at about \(18.4\) minutes, so an estimate of \(18\) to \(19\) minutes is reasonable.

a) Graph B is smoother and better represents continuous cooling without abrupt changes in rate. b) About \(121\,\text{°F}\) c) About \(18\) to \(19\) minutes

- Constant speed produces a straight line. - A horizontal segment means no distance is added. - A graph that becomes steeper represents increasing speed. - The first graph to reach the target distance represents the earliest arrival.

1. Lara rides at a constant speed, so her graph is the straight line \(a\). Max rides, stops, and then continues faster, so his graph is the piecewise-linear graph \(b\) with a horizontal segment. Sophie speeds up over time, so her graph becomes progressively steeper and is graph \(c\). 2. Graph \(b\) reaches \(12\) miles at \(70\) minutes, graph \(a\) at \(80\) minutes, and graph \(c\) at \(90\) minutes. Max arrives first in \(70\) minutes. 3. Max's horizontal segment runs from minute \(20\) to minute \(50\), so he is stopped for \(30\) minutes. 4. Graph \(a\) reaches \(6\) miles at \(40\) minutes.

a) Lara: \(a\); Max: \(b\); Sophie: \(c\) b) 1. Max, \(70\) minutes 2. \(30\) minutes 3. \(6\) miles

- Water height rises faster in a narrower cross section. - A greater slope means a faster increase in height. - The corner in each graph marks the boundary between the two cylinders.

1. In a narrow cylinder, the same added volume produces a faster rise in water height, so the graph is steeper. 2. Graph \(h\) begins steeply and then becomes less steep. Therefore, graph \(h\) represents Container A, which is narrow below and wide above. 3. Graph \(g\) begins less steeply and then becomes steeper. Therefore, graph \(g\) represents Container B, which is wide below and narrow above. 4. The change in diameter occurs at each graph's corner. For Container A, the corner is at a height of \(4\,\text{cm}\). For Container B, it is at \(2\,\text{cm}\).

a) Container A: graph \(h\); Container B: graph \(g\) b) Container A: \(4\,\text{cm}\); Container B: \(2\,\text{cm}\)

- Read the elevation at the starting and ending distances. - Find the highest point on each graph. - Total elevation gain includes only uphill changes, not downhill changes.

1. Trail A starts at \(400\,\text{ft}\) and ends at \(600\,\text{ft}\), for a net change of \(600 - 400 = 200\,\text{ft}\). Trail B starts at \(200\,\text{ft}\) and ends at \(900\,\text{ft}\), for a net change of \(900 - 200 = 700\,\text{ft}\). 2. Trail A reaches a maximum elevation of \(800\,\text{ft}\) at mile \(9\). Trail B reaches a maximum elevation of \(900\,\text{ft}\) at mile \(15\). 3. Trail A has one overall uphill section from \(300\,\text{ft}\) to \(800\,\text{ft}\), a gain of \(500\,\text{ft}\). 4. Trail B gains \(700 - 200 = 500\,\text{ft}\), then later gains \(900 - 500 = 400\,\text{ft}\). Its total elevation gain is \(500 + 400 = 900\,\text{ft}\). 5. Trail B has more total elevation gain.

a) Trail A: start \(400\,\text{ft}\), end \(600\,\text{ft}\), net change \(200\,\text{ft}\); Trail B: start \(200\,\text{ft}\), end \(900\,\text{ft}\), net change \(700\,\text{ft}\) b) Trail A: \(800\,\text{ft}\) at mile \(9\); Trail B: \(900\,\text{ft}\) at mile \(15\) c) Trail B; Trail A gains \(500\,\text{ft}\), while Trail B gains \(900\,\text{ft}\).

- Identify what each axis represents. - Opening and closing correspond to when the graph first rises above 0 and later returns to 0. - For a specified number of visitors, imagine drawing a horizontal line through that y-value. - Average rate of change is change in visitors divided by change in time.

1. Before \(8{:}00\) a.m., the graph is at \(0\) visitors. It begins increasing at \(8{:}00\) a.m., so the pool opens then. The graph returns to \(0\) visitors at \(8{:}00\) p.m., so the pool closes then. 2. The highest point occurs at about \(3{:}15\) p.m. The maximum is about \(600\) visitors. 3. A horizontal line at \(400\) visitors intersects the graph at about noon and shortly after \(6{:}00\) p.m. 4. At \(10{:}00\) a.m., the graph shows about \(200\) visitors, and at noon it shows about \(400\) visitors. The estimated average increase is \(\frac{400-200}{12-10}=100\) visitors per hour.

a) Opens at \(8{:}00\) a.m.; closes at \(8{:}00\) p.m. b) About \(3{:}15\) p.m.; about \(600\) visitors c) About noon and shortly after \(6{:}00\) p.m. d) About \(100\) visitors per hour

- Profit occurs when revenue is above expenses. - The greatest loss occurs where the vertical gap is largest with expenses on top. - Equal amounts occur where the two curves intersect.

1. The company earns a profit when the revenue graph lies above the expense graph. The graphs cross between months \(6\) and \(7\), so revenue is greater than expenses during months \(7\) through \(12\). 2. The loss is greatest where the expense graph is farthest above the revenue graph. That occurs in month \(1\). 3. The graphs intersect at about \(x=6.5\), between months \(6\) and \(7\).

a) Months \(7\) through \(12\) b) Month \(1\) c) About \(x=6.5\), between months \(6\) and \(7\)

- A horizontal section means the distance from the owner is unchanged. - Distance from the owner is not the same as total distance traveled. - Add the absolute change in distance over each moving interval.

1. At \(t = 20\,\text{s}\), the distance is \(120\,\text{ft}\). 2. Horizontal sections show constant distance. These occur from \(20\) to \(50\) seconds and from \(70\) to \(90\) seconds. 3. The greatest distance is \(120\,\text{ft}\), reached from \(20\) to \(50\) seconds. 4. From \(0\) to \(20\) seconds, the dog runs \(120\,\text{ft}\) away. From \(50\) to \(70\) seconds, it runs \(120 - 30 = 90\,\text{ft}\) toward the owner. From \(90\) to \(120\) seconds, it runs another \(30\,\text{ft}\) toward the owner. 5. The total distance is \(120 + 90 + 30 = 240\,\text{ft}\).

a) \(120\,\text{ft}\) b) From \(20\) to \(50\) seconds and from \(70\) to \(90\) seconds c) From \(20\) to \(50\) seconds, at a distance of \(120\,\text{ft}\) d) \(240\,\text{ft}\)

- Maximum and minimum values occur at the graph's highest and lowest points. - For part b), compare the graph with a horizontal line at \(6\). - Compare the vertical changes over two-hour intervals. - For part d), subtract the later water level from the earlier one.

1. The maximum water level is \(15\,\text{ft}\) at \(3{:}00\) a.m. The minimum is \(3\,\text{ft}\) at \(9{:}00\) a.m. 2. The graph is at or above \(6\,\text{ft}\) from midnight to \(7{:}00\) a.m. and again from \(11{:}00\) a.m. to noon. 3. Comparing two-hour increases, the greatest rises occur from midnight to \(2{:}00\) a.m. and from \(10{:}00\) a.m. to noon. 4. At \(4{:}00\) a.m., the water level is about \(14.2\,\text{ft}\); at \(6{:}00\) a.m., it is \(9\,\text{ft}\). The level drops by about \(14.2-9=5.2\,\text{ft}\).

a) Highest: \(15\,\text{ft}\) at \(3{:}00\) a.m.; lowest: \(3\,\text{ft}\) at \(9{:}00\) a.m. b) Midnight–\(7{:}00\) a.m. and \(11{:}00\) a.m.–noon c) Midnight–\(2{:}00\) a.m. and \(10{:}00\) a.m.–noon d) About \(5.2\,\text{ft}\)

- Use the axis labels to identify time and height. - Follow grid lines between an axis value and the graph. - For day \(7\), use the straight segment between the measurements on days \(6\) and \(8\). - Growth is the difference between the final height and the initial height.

1. At day \(4\), the graph shows \(10\,\text{mm}\). 2. Day \(7\) is halfway between day \(6\), at \(30\,\text{mm}\), and day \(8\), at \(50\,\text{mm}\). The straight segment gives \(30 + \frac{50 - 30}{2} = 40\,\text{mm}\). 3. The graph reaches \(60\,\text{mm}\) at day \(10\). 4. The plant’s height changes from \(10\,\text{mm}\) on day \(4\) to \(65\,\text{mm}\) on day \(12\). The growth is \(65 - 10 = 55\,\text{mm}\). 5. The values read from the graph are: <table> <thead> <tr><th>Day</th><th>\(0\)</th><th>\(2\)</th><th>\(4\)</th><th>\(6\)</th><th>\(8\)</th><th>\(10\)</th><th>\(12\)</th><th>\(14\)</th></tr> </thead> <tbody> <tr><td>Height (mm)</td><td>\(0\)</td><td>\(0\)</td><td>\(10\)</td><td>\(30\)</td><td>\(50\)</td><td>\(60\)</td><td>\(65\)</td><td>\(70\)</td></tr> </tbody> </table>

1. \(10\,\text{mm}\) 2. \(40\,\text{mm}\) 3. \(10\) days 4. \(55\,\text{mm}\) 5. <table> <thead> <tr><th>Day</th><th>\(0\)</th><th>\(2\)</th><th>\(4\)</th><th>\(6\)</th><th>\(8\)</th><th>\(10\)</th><th>\(12\)</th><th>\(14\)</th></tr> </thead> <tbody> <tr><td>Height (mm)</td><td>\(0\)</td><td>\(0\)</td><td>\(10\)</td><td>\(30\)</td><td>\(50\)</td><td>\(60\)</td><td>\(65\)</td><td>\(70\)</td></tr> </tbody> </table>

- Intersection points occur where the two functions have the same output. - Compare which graph is higher between successive intersection points. - Use the intersection x-values as boundaries when stating the interval.

1. The graphs intersect at \((0, 0)\) and \((4, 2)\). 2. Between the two intersection points, the graph of \(f\) lies above the graph of \(g\). Therefore, \(f(x)>g(x)\) for \(0<x<4\). 3. At \(x=4\), the functions are equal. For \(4<x<5\), the graph of \(g\) lies above the graph of \(f\).

1. The intersection points are \((0, 0)\) and \((4, 2)\). 2. \(0<x<4\)

- Restrict your attention to the portion of the graph whose x-values are in the stated interval. - Compare all local high and low points and both endpoints of the interval. - The greatest function value is the highest y-value; the least function value is the lowest y-value.

1. On \([-3, 0]\), the least value is \(-4\) at \(x=-2\). The greatest value is the endpoint value \(f(0)=0\). 2. On \([0, 3]\), the greatest value is \(4\) at \(x=2\). The least value is the endpoint value \(f(0)=0\). 3. On \([-4, 4]\), the greatest value is \(4\), reached at \(x=-4\) and \(x=2\). The least value is \(-4\), reached at \(x=-2\) and \(x=4\).

1) Greatest: \(0\); least: \(-4\) 2) Greatest: \(4\); least: \(0\) 3) Greatest: \(4\); least: \(-4\)

- Determine the two possible values of \(f(x)\) whose squares equal \(4\). - Find where the graph has either of those y-values. - Check which listed interval contains one of the corresponding x-values.

1. The equation \((f(x))^2=4\) means \(f(x)=2\) or \(f(x)=-2\). 2. The graph intersects \(y=2\) at approximately \(x=-3.46\) and \(x=3.46\). 3. The graph never reaches \(y=-2\), because its minimum value is \(-1\). 4. Of the listed intervals, only \([-4, -3]\) contains one of the solutions.

1) \([-4, -3]\)

- Convert the stated starting and ending temperatures into graph endpoints. - “Always decreases” rules out any rising or horizontal interval. - Compare what “quickly” and “more slowly” imply about the magnitude of the graph's slope over time.

1. The graph must start at \((0,180)\) and end at \((30,100)\). 2. Because the temperature always decreases, the graph must be decreasing throughout the interval from \(0\) to \(30\) minutes. 3. “Drops quickly at first” means the graph should have a steep negative slope near the start. 4. “Then more slowly” means the negative slope should become less steep in magnitude as time passes, so the graph gradually flattens while still decreasing.

A valid graph starts at \((0,180)\), ends at \((30,100)\), and decreases for the entire interval. It is steeply decreasing near \(t=0\) and gradually becomes less steep while remaining decreasing toward \(t=30\).

- Compare the story and the proposed graph one time interval at a time rather than judging the graph only by its final point. - A rest at a fixed distance must appear as a horizontal segment over the same time interval. - A steady return must be one straight decreasing segment beginning when the return starts and ending when the runner reaches the starting line. - After locating the mismatches, write the coordinates forced by the stated times and distances.

1. From \(0\) to \(2\) minutes, the proposed graph rises from \((0,0)\) to \((2,400)\), so that interval matches the story. 2. From \(2\) to \(3\) minutes, the story requires a horizontal segment at \(400\,\text{m}\), but the proposed graph rises from \(400\) to \(500\). The correct point at \(t=3\) is \((3,400)\). 3. From \(3\) to \(4\) minutes, the story says the runner should already be moving back, but the proposed graph is horizontal at \(500\,\text{m}\). The correct graph should be decreasing during this interval. 4. From \(4\) to \(5\) minutes, the proposed graph returns from \(500\,\text{m}\) to \(0\) in one minute. The story instead requires one steady return segment from \((3,400)\) to \((5,0)\), whose rate of change is \(\frac{0-400}{5-3}=-200\,\text{m/min}\). 5. The correct key vertices are \((0,0)\), \((2,400)\), \((3,400)\), and \((5,0)\).

The proposed graph is correct only from \(0\) to \(2\) minutes. From \(2\) to \(3\), it should be constant at \(400\,\text{m}\), not increasing. From \(3\) to \(5\), it should decrease steadily from \((3,400)\) to \((5,0)\); the proposed rest from \(3\) to \(4\) and one-minute return from \(4\) to \(5\) are both incorrect. Correct vertices: \((0,0)\), \((2,400)\), \((3,400)\), \((5,0)\).