To model the situation, we can write a linear equation that relates the time (x) in minutes to the temperature (y) in degrees Fahrenheit. Since the temperature increases at a constant rate, we can express the relationship as:
\[ y = mx + b \]
Where:
- \( y \) is the temperature in degrees Fahrenheit.
- \( x \) is the time in minutes.
- \( m \) is the rate of temperature increase per minute.
- \( b \) is the initial temperature of the oven when \( x = 0 \).
From the two points given:
- (4, 120): at 4 minutes, the temperature is 120°F.
- (14, 340): at 14 minutes, the temperature is 340°F.
Step 1: Find the rate of temperature increase (m)
We can find \( m \) using the formula for slope:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Using the points (4, 120) and (14, 340):
- \( y_1 = 120 \), \( x_1 = 4 \)
- \( y_2 = 340 \), \( x_2 = 14 \)
\[ m = \frac{340 - 120}{14 - 4} = \frac{220}{10} = 22 \]
So, the oven temperature increases at a rate of 22°F per minute.
Step 2: Find the initial temperature (b)
To find \( b \), we can use one of the points in the equation. We'll use the point (4, 120):
\[ y = mx + b \] \[ 120 = 22(4) + b \] \[ 120 = 88 + b \] \[ b = 120 - 88 = 32 \]
The equation for the temperature of the oven:
Thus, our linear equation is:
\[ y = 22x + 32 \]
Interpret the given points
Now we interpret the points:
-
The point (4, 120) means that after 4 minutes, the oven’s temperature was 120 degrees Fahrenheit.
-
The point (14, 340) means that after 14 minutes, the oven’s temperature was 340 degrees Fahrenheit.
Feel free to ask if you have any further questions!