QUESTION IMAGE
Question
2a) the table shows the distance penelope is from the park as she walks to soccer practice. assume the relationship between the two quantities is linear. \
find and interpret the rate of change. the rate of change is \underline{\qquad}, so penelope is \underline{\qquad} meters closer to the park every minute. find and interpret the initial value. the initial value is \underline{\qquad}, so penelope was initially \underline{\qquad} meters from the park. 2b) write the equation in the form $y = mx + b$.
2A) Rate of Change
Step1: Recall rate of change formula
The rate of change (slope) \( m \) for a linear relationship is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Let's take two points, say \( (x_1,y_1)=(5,1930) \) and \( (x_2,y_2)=(10,1380) \).
Step2: Calculate the rate of change
Substitute into the formula: \( m=\frac{1380 - 1930}{10 - 5}=\frac{- 550}{5}=- 110 \).
Interpretation: The negative sign indicates the distance from the park is decreasing. So Penelope is 110 meters closer to the park every minute.
2A) Initial Value
Step1: Use the linear equation \( y = mx + b \)
We know \( m=-110 \), and we can use a point, e.g., \( (5,1930) \). Substitute into \( y = mx + b \): \( 1930=-110(5)+b \).
Step2: Solve for \( b \)
Calculate \( -110(5)=-550 \), then \( 1930=-550 + b \). Add 550 to both sides: \( b = 1930 + 550 = 2480 \).
Interpretation: The initial value \( b \) is the distance from the park when time \( x = 0 \) (initial time), so Penelope was initially 2480 meters from the park.
2B) Equation
Step1: Identify \( m \) and \( b \)
From 2A, \( m=-110 \) and \( b = 2480 \).
Step2: Write the equation
Using \( y=mx + b \), substitute \( m \) and \( b \): \( y=-110x + 2480 \).
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s:
2A) The rate of change is \(-110\), so Penelope is \(110\) meters closer to the park every minute.
The initial value is \(2480\), so Penelope was initially \(2480\) meters from the park.
2B) The equation is \( y=-110x + 2480 \).