Sketch Function Graphs

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Sketch Function Graphs · Grade 8 · Worksheet 1

1. Liam fills a cylindrical tank with water at a constant rate for 5 minutes, then stops for 3 minutes, then drains the tank at a constant rate that is twice the filling rate until empty. Sketch a qualitative graph of the water height in the tank versus time. Answer: A graph with three segments: a straight line sloping upward for 5 minutes, a horizontal line for 3 minutes, then a straight line sloping downward twice as steeply as the first segment until it reaches the starting height. Solution: Identify the three phases: filling (0 to 5 min), stopped (5 to 8 min), draining (8 min until empty).
   Full step-by-step solution

   Step 1: Identify the three phases: filling (0 to 5 min), stopped (5 to 8 min), draining (8 min until empty). Step 2: During filling, the height increases at a constant rate, so the graph is a straight line with positive slope from (0,0) to (5, h_max). Step 3: During the stop, the height stays constant, so the graph is a horizontal line from (5, h_max) to (8, h_max). Step 4: During draining, the rate is twice the filling rate, so the height decreases twice as fast. The line from (8, h_max) to (t_end, 0) has a negative slope that is twice as steep (in absolute value) as the filling slope. Since the draining rate is double, the time to empty is half the filling time, so t_end = 8 + 2.5 = 10.5 minutes. The graph is a line from (8, h_max) to (10.5, 0). The final sketch shows three connected segments: upward slope, flat, then steeper downward slope.

2. Mason is filling a rectangular swimming pool with water. He turns on the hose and the water level rises steadily for 12 minutes until the pool is half full. He then stops to check the water temperature, so the water level remains constant for 7 minutes. After that, he turns the hose back on, but now the water level rises at a slower, steady rate for 22 minutes until the pool is completely full. Finally, Mason turns off the hose and immediately gets in the pool, causing the water level to drop suddenly by 2 inches as water splashes out. Sketch a qualitative graph showing the water level (in inches) over time (in minutes) for the entire process. Answer: A sketch with time on the x-axis and water level on the y-axis: an upward sloping line from (0,0) to (12, half), a horizontal line from (12, half) to (19, half), a shallower upward sloping line from (19, half) to (41, full), then a vertical drop at time 41. Solution: Identify the time intervals and what happens in each. From 0 to 12 minutes, water level rises steadily (positive slope). From 12 to 19 minutes, water level is constant (zero slope).
   Full step-by-step solution

   Step 1: Identify the time intervals and what happens in each. From 0 to 12 minutes, water level rises steadily (positive slope). From 12 to 19 minutes, water level is constant (zero slope). From 19 to 41 minutes, water level rises steadily but at a slower rate (positive but shallower slope). At 41 minutes, water level drops suddenly (vertical drop). Step 2: Draw the axes. Label the x-axis 'Time (minutes)' and the y-axis 'Water Level (inches)'. Step 3: Plot the first segment: a line from (0,0) to (12, some height, say H) with a positive slope. Step 4: Plot the second segment: a horizontal line from (12, H) to (19, H). Step 5: Plot the third segment: a line from (19, H) to (41, 2H) with a positive slope that is less steep than the first. Step 6: At time 41, draw a vertical line downward from (41, 2H) to (41, 2H - 2) to represent the sudden drop. The final sketch shows these features.

3. Sophia is on a hiking trip. She starts at the trailhead and hikes uphill for 30 minutes at a steady pace. She then rests for 15 minutes at a scenic overlook. After resting, she hikes downhill back to the trailhead, taking 45 minutes because the downhill path is longer but easier. Sketch a qualitative graph of Sophia's distance from the trailhead (vertical axis) versus time (horizontal axis). Describe the shape of your graph and explain what each segment represents. Answer: The graph is a line that rises steadily for 30 minutes, stays flat for 15 minutes, then falls steadily for 45 minutes back to the starting height. Solution: Identify the three phases of the hike: uphill (moving away), rest (stationary), downhill (returning). For the uphill phase (0 to 30 minutes), distance from the trailhead increases steadily, so the graph has a positive slope (rising line) from (0,0) to (30, d), where d is the maximum distance…
   Full step-by-step solution

   Step 1: Identify the three phases of the hike: uphill (moving away), rest (stationary), downhill (returning). Step 2: For the uphill phase (0 to 30 minutes), distance from the trailhead increases steadily, so the graph has a positive slope (rising line) from (0,0) to (30, d), where d is the maximum distance reached. Step 3: For the rest phase (30 to 45 minutes), distance remains constant, so the graph is a horizontal line (flat) at height d from time 30 to 45. Step 4: For the downhill phase (45 to 90 minutes), distance decreases steadily back to 0, so the graph has a negative slope (falling line) from (45, d) to (90, 0). Step 5: The final graph is a piecewise linear graph: a rising line, then a flat line, then a falling line. The downhill segment takes 45 minutes (longer time) compared to 30 minutes uphill, so the downhill slope is less steep. Answer: The graph rises for 30 minutes, stays flat for 15 minutes, then falls for 45 minutes back to the starting point.

4. Liam is designing a rectangular garden with a length that is 3 meters more than twice its width. The area of the garden needs to be 35 square meters. What are the dimensions of Liam's garden? Answer: width = 3.5 m, length = 10 m Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \). Area of a rectangle = length × width.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define variables** Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \). --- **Step 2: Write the area equation** Area of a rectangle = length × width. Given area = 35 square meters: \[ l \times w = 35 \] Substitute \( l = 2w + 3 \): \[ (2w + 3) \times w = 35 \] --- **Step 3: Expand and rearrange** \[ 2w^2 + 3w = 35 \] \[ 2w^2 + 3w - 35 = 0 \] --- **Step 4: Solve the quadratic equation** Use the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here \( a = 2 \), \( b = 3 \), \( c = -35 \). First, discriminant: \[ b^2 - 4ac = 3^2 - 4(2)(-35) = 9 + 280 = 289 \] \[ \sqrt{289} = 17 \] So: \[ w = \frac{-3 \pm 17}{2 \times 2} = \frac{-3 \pm 17}{4} \] --- **Step 5: Two possible solutions** Case 1: \( w = \frac{-3 + 17}{4} = \frac{14}{4} = 3.5 \) Case 2: \( w = \frac{-3 - 17}{4} = \frac{-20}{4} = -5 \) (not valid, width can't be negative) So \( w = 3.5 \) meters. --- **Step 6: Find length** \[ l = 2w + 3 = 2(3.5) + 3 = 7 + 3 = 10 \] --- **Step 7: Final answer** Width = 3.5 m, Length = 10 m.

5. A right triangle is drawn on a coordinate plane with vertices at (0,0), (12,0), and (12,5). A rectangle is inscribed inside this triangle such that one side lies along the x-axis from (0,0) to (x,0), and the opposite vertices touch the hypotenuse of the triangle. What is the area of the largest possible rectangle that can be inscribed in this triangle under these conditions? Answer: 30 Solution: Find the equation of the hypotenuse. The hypotenuse connects (0,0) and (12,5). The slope is (5-0)/(12-0) = 5/12.
   Full step-by-step solution

   Step 1: Find the equation of the hypotenuse. The hypotenuse connects (0,0) and (12,5). The slope is (5-0)/(12-0) = 5/12. The equation is y = (5/12)x. Step 2: Let the rectangle have width x along the x-axis. The top-right corner touches the hypotenuse at (x, y), where y = (5/12)x. Step 3: The rectangle's dimensions are width = x and height = y = (5/12)x. Step 4: The area function is A(x) = width × height = x × (5/12)x = (5/12)x². Step 5: The maximum occurs at the maximum x-value, which is when the rectangle spans the full base of the triangle. The triangle's base is from (0,0) to (12,0), so maximum x = 12. Step 6: Calculate the maximum area: A(12) = (5/12) × 12² = (5/12) × 144 = 5 × 12 = 30. The answer is 30.

6. Mason is observing a hot air balloon during a 12-minute flight. The balloon rises steadily for the first 7 minutes, then stays at a constant altitude for 2 minutes, and finally descends quickly for the remaining 3 minutes. Sketch a qualitative graph of altitude (in meters) versus time (in minutes). Describe the shape of your graph and explain what each section represents. Answer: A graph with three distinct segments: a straight line sloping upward for 0 to 7 minutes (rising), a horizontal line from 7 to 9 minutes (constant), and a steeper straight line sloping downward from 9 to 12 minutes (descending). Solution: Draw axes. Label the x-axis 'Time (minutes)' from 0 to 12, and the y-axis 'Altitude (meters)' from 0 to some maximum (e.g., 700 m). Step 2: For the first 7 minutes, the balloon rises steadily.
   Full step-by-step solution

   Step 1: Draw axes. Label the x-axis 'Time (minutes)' from 0 to 12, and the y-axis 'Altitude (meters)' from 0 to some maximum (e.g., 700 m). Step 2: For the first 7 minutes, the balloon rises steadily. This means the altitude increases at a constant rate. Draw a straight line with a positive slope from (0,0) to (7, 700). Step 3: For the next 2 minutes (from 7 to 9 minutes), the altitude stays constant. Draw a horizontal line from (7, 700) to (9, 700). Step 4: For the final 3 minutes (from 9 to 12 minutes), the balloon descends quickly. This means the altitude decreases at a faster rate than it rose. Draw a straight line with a negative slope that is steeper than the rising segment, from (9, 700) to (12, 0). The final graph shows three distinct linear segments: rising, constant, and descending. The answer is a graph with these three segments.