Abby is baking brownies, so she turns the oven on. As soon as Abby turns the oven on, the oven heats up at a constant rate. Write an equation to represent this situation. Then use the equation to make a prediction.

Question

This graph represents the linear relationship between the time in minutes and the temperature of the oven in degrees Fahrenheit. For each minute, the temperature increases at a constant rate. What do the points (2, 100) and (6, 220) represent?
x
y
(2, 100)(6, 220)
0
2
4
6
8
10
12
14
16
0
20
40
60
80
100
120
140
160
180
200
220
240
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300
320
340
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Time (minutes)
Temperature (°F)
The point (2, 100) means that after
minutes, the oven’s temperature was
degrees Fahrenheit.

The point (6, 220) means that after
minutes, the oven’s temperature was
degrees Fahrenheit.

Answer 1

To represent the situation of the oven heating up at a constant rate, we can use a linear equation of the form:

\[ T = mt + b \]

where:

• \( T \) is the temperature of the oven in degrees Fahrenheit,
• \( t \) is the time in minutes,
• \( m \) is the rate of change (the increase in temperature per minute), and
• \( b \) is the starting temperature of the oven (when \( t = 0 \)).

From the points provided on the graph:

• The point (2, 100) means that after 2 minutes, the oven's temperature was 100 degrees Fahrenheit.
• The point (6, 220) means that after 6 minutes, the oven's temperature was 220 degrees Fahrenheit.

To find \( m \), the constant rate of change, we can use the two points (2, 100) and (6, 220):

1. Calculate the change in temperature: \[ \Delta T = 220 - 100 = 120 \text{ degrees} \]

2. Calculate the change in time: \[ \Delta t = 6 - 2 = 4 \text{ minutes} \]

3. Now, calculate the rate of change (slope \( m \)): \[ m = \frac{\Delta T}{\Delta t} = \frac{120}{4} = 30 \text{ degrees per minute} \]

Assuming the oven starts from room temperature (which we can estimate to be 0 degrees Fahrenheit for this equation), we have the following equation:

\[ T = 30t \]

Now we can verify the points and use the equation to make predictions.

Let’s check the points:

• For \( t = 2 \): \[ T = 30 \times 2 = 60 \text{ degrees} \text{ (which does not match the expected 100 degrees)} \]

This means that our starting temperature \( b \) was higher. We need to re-evaluate:

If we choose point (2, 100), we can plug in \( t = 2 \) and \( T = 100 \) into the generic equation \( T = mt + b \) to find \( b \):

\[ 100 = 30(2) + b \] \[ 100 = 60 + b \] \[ b = 100 - 60 = 40 \]

So the updated equation is: \[ T = 30t + 40 \]

To check again:

• For \( t = 2 \): \[ T = 30(2) + 40 = 60 + 40 = 100 \text{ degrees} \]

• For \( t = 6 \): \[ T = 30(6) + 40 = 180 + 40 = 220 \text{ degrees} \]

Both points match correctly!

That means:

• The point (2, 100) means that after 2 minutes, the oven’s temperature was 100 degrees Fahrenheit.
• The point (6, 220) means that after 6 minutes, the oven’s temperature was 220 degrees Fahrenheit.

Prediction:

Using the equation \( T = 30t + 40 \), we can predict the oven temperature at any given time \( t \). For example, after 10 minutes:

\[ T = 30(10) + 40 = 300 + 40 = 340 \text{ degrees Fahrenheit} \]

So, after 10 minutes, the oven would reach 340 degrees Fahrenheit.