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Interpret slope and intercept contextually

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5513128
The amount of water in a container is modeled by \(V(t)=40-3t\), where \(t\) is the number of minutes after draining begins and \(V(t)\) is the amount of water in gallons. What do the initial value and the rate of change mean in this situation? Include units.

Hints

- Think about what the model gives when the elapsed time is zero. - The coefficient multiplying time tells how the output changes for each additional minute. - Interpret the sign of the rate in terms of whether the amount is increasing or decreasing.

Solution

In \(V(t)=40-3t\), the constant term \(40\) is the value when \(t=0\), so the container starts with \(40\,\text{gal}\) of water. The coefficient of \(t\) is \(-3\), so the amount changes by \(-3\,\text{gal}\) each minute. In context, the container loses \(3\,\text{gal}\) of water per minute.

Answer

Initial value: \(40\,\text{gal}\), the amount of water when draining begins. Rate of change: \(-3\,\text{gal/min}\), meaning the amount decreases by \(3\,\text{gal}\) each minute.
5520078
A parking garage charges according to the table. <table><tr><th>Hours \(h\)</th><th>Cost \(C\)</th></tr><tr><td>\(0\)</td><td>\(\$6\)</td></tr><tr><td>\(1\)</td><td>\(\$8\)</td></tr><tr><td>\(2\)</td><td>\(\$10\)</td></tr></table> Find the rate of change and the initial value. Explain what each means in this situation.

Hints

- Compare costs for two rows whose hour values differ by one. - Look at the row where the number of hours is zero to identify the initial value.

Solution

1. The cost increases from \(\$6\) to \(\$8\) when time increases from \(0\) to \(1\) hour, so the rate of change is \(\$2\) per hour. 2. When \(h=0\), the cost is \(\$6\), so the initial value is \(\$6\). 3. The garage charges a \(\$6\) starting fee and then \(\$2\) for each hour parked.

Answer

Rate of change: \(\$2\) per hour, the hourly parking charge. Initial value: \(\$6\), the starting fee at \(0\) hours.
5513148
The graph models the cost \(C\), in dollars, of a ride that travels \(x\) miles. a) Use the graph to find the slope and the y-intercept. b) Interpret both values in the context of the ride.
Figure for problem 551314

Hints

- Use two clearly marked points to compare the change in cost with the change in miles. - The y-intercept is the point where the graph meets the y-axis. - Attach the graph's units to each value when you explain what it means.

Solution

1. Two marked points are \((0,4)\) and \((2,10)\). The slope is \(m=\frac{10-4}{2-0}=\frac{6}{2}=3\), and the line crosses the y-axis at \(4\), so the y-intercept is \(4\). 2. The slope means the cost increases by \(\$3\) for each mile traveled. The y-intercept means the ride has an initial charge of \(\$4\) when \(x=0\).

Answer

a) Slope: \(3\); y-intercept: \(4\). b) The ride costs \(\$3\) per mile plus an initial charge of \(\$4\).
5520088
Two equipment-rental plans charge according to Plan A: \(C = 3h\) Plan B: \(C = 5 + 3h\), where \(C\) is cost in dollars and \(h\) is the number of hours rented. Compare the slopes and y-intercepts of the two models and explain what each comparison means in context.

Hints

- Compare the coefficients of \(h\) first. - Then compare the values of \(C\) when \(h=0\). - Interpret the rate and starting value separately.

Solution

1. Both models have slope \(3\), so both plans charge \(\$3\) per rental hour. 2. Plan A has y-intercept \(0\), so it has no starting fee. 3. Plan B has y-intercept \(5\), so it has a \(\$5\) starting fee. 4. Because the hourly rates are equal, Plan B remains \(\$5\) more expensive for every rental duration.

Answer

Both slopes are \(3\), meaning \(\$3\) per hour. Plan A's y-intercept is \(0\), while Plan B's is \(5\), meaning Plan B has a \(\$5\) starting fee.
5520098
A candle is \(18\,\text{cm}\) tall when it is lit and burns down at a constant rate of \(1.5\,\text{cm}\) per hour. Write a linear equation for its height \(H\), in centimeters, after \(t\) hours. Identify the slope and y-intercept and explain what each means.

Hints

- Decide which quantity is the starting value and which quantity changes each hour. - Use a negative rate because the candle's height decreases. - Put the model in the form initial value plus rate times time.

Solution

1. The starting height is \(18\), so the y-intercept is \(18\). 2. The candle loses \(1.5\) centimeters each hour, so the slope is \(-1.5\). 3. The model is \(H = 18 - 1.5t\).

Answer

\(H = 18 - 1.5t\). The slope is \(-1.5\), meaning the candle loses \(1.5\,\text{cm}\) per hour. The y-intercept is \(18\), the starting height in centimeters.
5239388
Two monthly cell phone plans have costs, in dollars, modeled by the following expressions. The variable \(m\) is the number of gigabytes of data used. Plan 1: \(10+2m\) Plan 2: \(4m\) a) Describe the base fee and cost per gigabyte for each plan. b) Let \(E\) represent the amount saved by choosing Plan 1 instead of Plan 2. Write and simplify an expression for \(E\). c) Find \(E\) when \(m=3\) and when \(m=7\). Interpret each result.

Hints

- In a cost expression, what does a constant term represent? What does the coefficient of \(m\) represent? - Savings equal the cost not chosen minus the cost chosen. - Use parentheses when subtracting an entire cost expression. - What does a negative savings value mean?

Solution

1. Plan 1 has a base fee of \(\$10\) and costs \(\$2\) per gigabyte. Plan 2 has no base fee and costs \(\$4\) per gigabyte. 2. Savings from choosing Plan 1 are Plan 2 cost minus Plan 1 cost: \(E=4m-(10+2m)=4m-10-2m=2m-10\). 3. When \(m=3\), \(E=2\cdot3-10=-4\), so Plan 1 costs \(\$4\) more. When \(m=7\), \(E=2\cdot7-10=4\), so Plan 1 saves \(\$4\).

Answer

a) Plan 1: \(\$10\) base fee and \(\$2\) per gigabyte. Plan 2: no base fee and \(\$4\) per gigabyte. b) \(E=2m-10\) c) For \(m=3\), \(E=-4\), so Plan 1 costs \(\$4\) more. For \(m=7\), \(E=4\), so Plan 1 saves \(\$4\).
5513138
A delivery service models its total charge by \(C(m)=3m+7\), where \(m\) is the number of miles traveled and \(C(m)\) is the cost in dollars. a) Jordan says that \(3\) is the starting charge and \(7\) is the charge per mile. Casey says that \(7\) is the starting charge and \(3\) is the charge per mile. Who is correct? Explain. b) A second service charges the same amount per mile but has a starting charge that is \(\$2\) higher. Write a linear model for the second service.

Hints

- Decide which part of a linear model remains when the input is zero. - Decide which part tells how much the output changes when the input increases by one unit. - In part b), identify which feature of the original model stays the same and which feature changes.

Solution

1. Casey is correct. In \(C(m)=3m+7\), the coefficient \(3\) is the rate of change, so it represents \(\$3\) per mile. The constant term \(7\) is \(C(0)\), so it represents the \(\$7\) starting charge. 2. The per-mile rate stays \(3\). The new starting charge is \(7+2=9\), so the second service can be modeled by \(S(m)=3m+9\).

Answer

a) Casey is correct: \(\$7\) is the starting charge and \(\$3\) per mile is the rate of change. b) \(S(m)=3m+9\)
5520108
The graph shows the amount of water in two tanks over time. A student says, “Tank B will eventually catch up to Tank A because both amounts are decreasing.” Evaluate the claim using the slopes and y-intercepts of the two lines.
Figure for problem 552010

Hints

- Compare the lines' starting values at time zero. - Compare their vertical changes over the same time interval. - Think about what equal slopes imply about the gap between two lines.

Solution

1. Tank A starts at \(14\) gallons and Tank B starts at \(10\) gallons, so their y-intercepts differ by \(4\) gallons. 2. Both lines have slope \(-2\), so both tanks lose water at \(2\) gallons per minute. 3. Since the rates are equal, the \(4\)-gallon difference stays constant. The lines are parallel and do not meet. 4. The student's claim is false.

Answer

The claim is false. Both tanks decrease at \(2\,\text{gallons per minute}\), but Tank A starts \(4\) gallons higher, so the difference stays \(4\) gallons.
5520118
The graph shows a simplified linear model for air temperature \(T\), in degrees Fahrenheit, at elevation \(e\), measured in thousands of feet. Only elevations from \(2\) to \(8\) thousand feet are shown. Use the marked points when needed. a) Find the slope and interpret it in context. b) Write a linear model relating \(T\) and \(e\). c) Find the y-intercept and explain what it means. Why is that intercept an extrapolated value rather than a value displayed on the graph?
Figure for problem 552011

Hints

- Use the two marked points to find change in temperature divided by change in elevation. - After finding the slope, substitute one graphed point into \(T = me + b\) to determine the intercept. - Compare the input value represented by the y-intercept with the elevations actually shown.

Solution

1. Using \((2, 62)\) and \((4, 54)\), the slope is \(\frac{54-62}{4-2} = -4\). The model predicts a decrease of \(4\) degrees Fahrenheit for each increase of \(1000\) feet in elevation. 2. Write \(T = -4e + b\). Substitute \((2, 62)\): \(62 = -4(2) + b\), so \(b = 70\). Thus \(T = 70 - 4e\). 3. The y-intercept is \(70\), representing the model's predicted temperature at \(e=0\), or sea level. It is extrapolated because the displayed graph begins at \(e=2\), not \(e=0\).

Answer

a) Slope: \(-4\) degrees Fahrenheit per \(1000\) feet; the predicted temperature decreases by \(4\) degrees Fahrenheit for each additional \(1000\) feet of elevation. b) \(T = 70 - 4e\) c) The y-intercept is \(70\) degrees Fahrenheit, the model's predicted sea-level temperature. It is extrapolated because \(e=0\) is outside the displayed range.
5520128
The graph shows a device's battery percentage after it is unplugged. a) Determine the y-intercept and slope and interpret both in context. b) Write a linear model for battery percentage \(B\) after \(t\) hours. c) According to the model, after how many hours will the battery be at \(50\%\)?
Figure for problem 552012

Hints

- Read the battery level where the graph meets the vertical axis. - Use two clear points to find change in percentage divided by change in time. - Use the starting value and rate to write the model before answering part c).

Solution

1. The graph starts at \((0, 90)\), so the y-intercept is \(90\): the battery begins at \(90\%\). 2. From \((0, 90)\) to \((5, 50)\), the change is \(-40\) percentage points over \(5\) hours, so the slope is \(-8\) percentage points per hour. 3. The model is \(B = 90 - 8t\). 4. Set \(B=50\): \(50 = 90 - 8t\). Then \(8t = 40\), so \(t=5\).

Answer

a) y-intercept: \(90\%\), the starting battery level; slope: \(-8\) percentage points per hour. b) \(B = 90 - 8t\) c) \(5\) hours

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