Sketch Function Graphs

lessonbunny.com

Answer Key & Explanations

Sketch Function Graphs · Grade 8 · Worksheet 1

1. Sophia is filling a cylindrical tank with water at a constant rate. She then stops filling for 6 minutes to check the tank. After that, she continues filling at the same constant rate until the tank is full. Sketch a graph of the water height in the tank versus time. Answer: A graph with three distinct segments: a straight line with positive slope from the origin, a horizontal line segment, and another straight line with the same positive slope as the first segment, ending at a higher height. Solution: Identify the three phases of the scenario: filling, stopping, and filling again. During the first filling phase, water is added at a constant rate, so the height increases steadily over time. This is represented by a straight line with a positive slope starting from the origin (0,0).
   Full step-by-step solution

   Step 1: Identify the three phases of the scenario: filling, stopping, and filling again. Step 2: During the first filling phase, water is added at a constant rate, so the height increases steadily over time. This is represented by a straight line with a positive slope starting from the origin (0,0). Step 3: During the 6-minute stop, no water is added, so the height remains constant. This is represented by a horizontal line segment at the height reached at the end of the first filling phase. Step 4: During the second filling phase, water is again added at the same constant rate, so the height increases at the same rate as before. This is represented by another straight line with the same positive slope as the first segment, starting from the end of the horizontal segment and continuing until the tank is full. Step 5: The final graph has three connected segments: a rising line, a flat line, and another rising line with the same slope as the first. The answer is a sketch showing this pattern.

2. Olivia is watching her dog run in the park. The dog starts at Olivia's feet, runs away from her at a steady pace for 10 seconds, then stops to sniff a bush for 5 seconds. After sniffing, the dog runs back toward Olivia at a faster steady pace for 5 seconds until it reaches her. Sketch a qualitative graph of the dog's distance from Olivia over time. Describe the shape of the graph. Answer: The graph is composed of three line segments: first a line sloping upward (distance increasing), then a horizontal line (distance constant), then a line sloping downward (distance decreasing) that is steeper than the first segment. Solution: Identify the three phases of motion. Phase 1: Dog runs away for 10 seconds at a steady pace. Distance from Olivia increases at a constant rate, so the graph is a straight line sloping upward.
   Full step-by-step solution

   Step 1: Identify the three phases of motion. Phase 1: Dog runs away for 10 seconds at a steady pace. Distance from Olivia increases at a constant rate, so the graph is a straight line sloping upward. Phase 2: Dog stops for 5 seconds. Distance stays the same, so the graph is a horizontal line. Phase 3: Dog runs back to Olivia at a faster pace for 5 seconds. Distance decreases at a constant rate, so the graph is a straight line sloping downward. Since the dog runs back faster, the downward slope is steeper than the upward slope. The graph is a piecewise linear graph with three segments: rising, flat, then falling more steeply.

3. 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 65 square meters. What are the dimensions of Liam's garden? Answer: width = 5 m, length = 13 m Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \).
   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 rectangle = length × width Given area = 65 m², so: \[ (2w + 3) \times w = 65 \] --- **Step 3: Expand and rearrange** \[ 2w^2 + 3w = 65 \] Subtract 65 from both sides: \[ 2w^2 + 3w - 65 = 0 \] --- **Step 4: Solve the quadratic equation** We can use the quadratic formula: For \( a w^2 + b w + c = 0 \), \( a = 2 \), \( b = 3 \), \( c = -65 \). \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ w = \frac{-3 \pm \sqrt{3^2 - 4(2)(-65)}}{2 \times 2} \] \[ w = \frac{-3 \pm \sqrt{9 + 520}}{4} \] \[ w = \frac{-3 \pm \sqrt{529}}{4} \] \[ w = \frac{-3 \pm 23}{4} \] --- **Step 5: Evaluate the two possible solutions** First solution: \[ w = \frac{-3 + 23}{4} = \frac{20}{4} = 5 \] Second solution: \[ w = \frac{-3 - 23}{4} = \frac{-26}{4} = -6.5 \] Since width can't be negative, we take \( w = 5 \). --- **Step 6: Find the length** \[ l = 2w + 3 = 2(5) + 3 = 10 + 3 = 13 \] --- **Step 7: Check the area** Area = \( 13 \times 5 = 65 \) m², which matches the problem. --- **Final answer:** Width = 5 m, Length = 13 m

4. Mere is watching her dog run in the park. The dog starts by sitting still for 2 minutes, then runs away from her at a constant speed for 4 minutes, stops to sniff a bush for 2 minutes, and then runs back to her at a faster constant speed for 2 minutes. Sketch a graph that shows the distance between Mere and her dog over time. Describe the shape of the graph for each segment of the dog's journey. Answer: The graph has 5 segments: (1) horizontal line at distance 0 for 2 minutes, (2) line sloping upward for 4 minutes, (3) horizontal line at a constant distance for 2 minutes, (4) line sloping downward for 2 minutes back to distance 0, (5) segment (4) is steeper than segment (2). Solution: Set up the axes. The x-axis is time (in minutes) from 0 to 10. The y-axis is distance (in meters) between Mere and her dog.
   Full step-by-step solution

   Step 1: Set up the axes. The x-axis is time (in minutes) from 0 to 10. The y-axis is distance (in meters) between Mere and her dog. Step 2: Segment 1 (0 to 2 minutes): The dog sits still next to Mere. Distance is 0 and remains constant. This is a horizontal line at y=0 from x=0 to x=2. Step 3: Segment 2 (2 to 6 minutes): The dog runs away at constant speed. Distance increases steadily. This is a straight line sloping upward from (2,0) to (6, d1) where d1 is some positive distance. The slope is positive and constant. Step 4: Segment 3 (6 to 8 minutes): The dog stops to sniff. Distance stays at d1 (constant). This is a horizontal line at y=d1 from x=6 to x=8. Step 5: Segment 4 (8 to 10 minutes): The dog runs back at a faster constant speed. Distance decreases steadily back to 0. This is a straight line sloping downward from (8, d1) to (10, 0). Since the dog runs faster, this line is steeper than the line in segment 2. Step 6: Final graph: The graph has a flat section at zero, then a sloped line up, then a flat section at the top, then a steeper sloped line down to zero. The answer is a qualitative sketch with the described five segments.

5. Noah is filling a swimming pool. The water level rises quickly at first, then slows down as the pool gets deeper, and finally stays constant when the pool is full. Sketch a graph of water depth vs. time. Answer: A qualitative graph showing a curve that increases steeply at first, then gradually flattens, and finally becomes horizontal. Solution: Identify the axes: horizontal axis is time, vertical axis is water depth. At the start (time = 0), depth is 0, so the graph begins at the origin.
   Full step-by-step solution

   Step 1: Identify the axes: horizontal axis is time, vertical axis is water depth. Step 2: At the start (time = 0), depth is 0, so the graph begins at the origin. Step 3: The water level rises quickly at first, so the graph has a steep positive slope initially. Step 4: As the pool gets deeper, the rate slows, so the slope becomes less steep (the curve bends toward horizontal). Step 5: When the pool is full, the depth stays constant, so the graph becomes a horizontal line (slope = 0). The final sketch is a curve that starts steep, gradually flattens, and ends with a horizontal segment.

6. Liam is designing a rectangular garden with a length that is 3 meters more than twice its width. If the area of the garden must be 54 square meters, what are the dimensions of the garden? Answer: width = 4.5 m, length = 12 m Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: \text{length} = 2w + 3 Area of a rectangle = length × width Given area = 54 square meters: (2w + 3) \times w = 54 2w^2 + 3w = 54 2w^2 + 3w - 54 = 0 w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here \(…
   Full step-by-step solution

   Let's solve the problem step by step. --- **Step 1: Define the variables** Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: \[ \text{length} = 2w + 3 \] --- **Step 2: Write the area equation** Area of a rectangle = length × width Given area = 54 square meters: \[ (2w + 3) \times w = 54 \] --- **Step 3: Expand and rearrange** \[ 2w^2 + 3w = 54 \] \[ 2w^2 + 3w - 54 = 0 \] --- **Step 4: Solve the quadratic equation** We can use the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here \( a = 2 \), \( b = 3 \), \( c = -54 \). First, compute the discriminant: \[ b^2 - 4ac = 3^2 - 4(2)(-54) = 9 + 432 = 441 \] \[ \sqrt{441} = 21 \] Now: \[ w = \frac{-3 \pm 21}{2 \times 2} = \frac{-3 \pm 21}{4} \] --- **Step 5: Two possible solutions for \( w \)** First: \[ w = \frac{-3 + 21}{4} = \frac{18}{4} = 4.5 \] Second: \[ w = \frac{-3 - 21}{4} = \frac{-24}{4} = -6 \] --- **Step 6: Interpret the solutions** Width cannot be negative, so \( w = 4.5 \) meters. --- **Step 7: Find the length** \[ \text{length} = 2w + 3 = 2(4.5) + 3 = 9 + 3 = 12 \] --- **Step 8: Verify** Area = length × width = \( 12 \times 4.5 = 54 \) — correct. --- **Final answer:** Width = 4.5 m, Length = 12 m

7. A rectangular garden is drawn on a coordinate plane with corners at (2, 1), (8, 1), (8, 5), and (2, 5). A diagonal line is drawn from the bottom-left corner (2, 1) to the top-right corner (8, 5). What is the length of this diagonal? Round your answer to the nearest tenth. Answer: 7.2 Solution: Identify the coordinates of the two endpoints of the diagonal. The problem says the diagonal goes from the bottom-left corner (2, 1) to the top-right corner (8, 5).
   Full step-by-step solution

   Step 1: Identify the coordinates of the two endpoints of the diagonal. The problem says the diagonal goes from the bottom-left corner (2, 1) to the top-right corner (8, 5). So we have: Point A = (2, 1) Point B = (8, 5) Step 2: Recall the distance formula. The distance between two points (x1, y1) and (x2, y2) is: Distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) Step 3: Substitute the coordinates into the formula. x2 - x1 = 8 - 2 = 6 y2 - y1 = 5 - 1 = 4 So: Distance = sqrt( (6)^2 + (4)^2 ) Step 4: Calculate the squares. 6^2 = 36 4^2 = 16 Step 5: Add the squares. 36 + 16 = 52 Step 6: Take the square root. Distance = sqrt(52) Step 7: Simplify sqrt(52). We can factor 52 as 4 * 13, so: sqrt(52) = sqrt(4 * 13) = sqrt(4) * sqrt(13) = 2 * sqrt(13) Step 8: Approximate sqrt(13). We know 3.60555 is a close approximation for sqrt(13). So: 2 * 3.60555 = 7.2111 Step 9: Round to the nearest tenth. 7.2111 rounded to the nearest tenth is 7.2 Final Answer: 7.2

8. Noah starts at his house, walks away from it at a steady pace for 6 minutes, stops to tie his shoe for 1 minute, then walks back home at a faster steady pace. Sketch a qualitative graph of Noah's distance from home versus time. Answer: A graph with three distinct segments: a straight line with positive slope from (0,0) to (6, d1), a horizontal line from (6, d1) to (7, d1), and a steeper straight line with negative slope from (7, d1) back to (t, 0) where t < 14. Solution: Identify the three phases of the story. Phase 1: Noah walks away from home for 6 minutes at a steady pace. On a distance-time graph, this is a straight line with a positive slope (distance increases).
   Full step-by-step solution

   Step 1: Identify the three phases of the story. Phase 1: Noah walks away from home for 6 minutes at a steady pace. On a distance-time graph, this is a straight line with a positive slope (distance increases). Since the pace is steady, the slope is constant. Phase 2: Noah stops to tie his shoe for 1 minute. During this time, his distance from home does not change, so the graph shows a horizontal line segment from t=6 to t=7. Phase 3: Noah walks back home at a faster steady pace. The distance decreases, so the line has a negative slope. Since he walks faster, the slope is steeper (more negative) than the slope in Phase 1. The line goes from (7, d1) back to the time axis (distance = 0) at some time t. Because he walks faster, the return trip takes less than 6 minutes, so t is less than 13. The final graph has three connected segments: a rising line, a horizontal line, and a steeper falling line.