Question

guided practice
tell whether the rate of change are constant or variable. (example 1)

1. building measurements
2. computers sold
3. distance an object falls
4. cost of sweaters

erica walks to her friend philip’s house. the graph shows erica’s distance from home over time. (example 2)

1. find the rate of change from 1 minute to 2 minutes.

\\(\frac{\text{change in distance}}{\text{change in time}} = \frac{400 - \square}{2 - \square} = \frac{\square}{\square} = \square\\) ft per min

1. find the rate of change from 1 minute to 4 minutes. _________

find the slope of each line. (example 3)
7.
slope = _________
8.
slope = _________

1. if you know two points on a line, how can you find the rate of change of the variables being graphed?

Explanation:

Problem 5:

Step1: Identify change in distance and time

From the graph, at \( t = 1 \) minute, distance \( d_1 = 400 \) ft? Wait, no, the formula is \(\frac{\text{change in distance}}{\text{change in time}}\). Wait, the numerator is \( 400 - 0 \)? Wait, no, the given is from 1 minute to 2 minutes. Wait, the graph: at \( x = 2 \) (time), \( y = 400 \)? Wait, the formula is \(\frac{\text{change in distance}}{\text{change in time}}=\frac{400 - 200}{2 - 1}\)? Wait, no, the problem has \(\frac{400 - \square}{2 - \square}\). Wait, maybe at \( t = 1 \), distance is 200, at \( t = 2 \), distance is 400. So change in distance is \( 400 - 200 = 200 \)? Wait, no, the problem's formula is \(\frac{400 - \square}{2 - \square}\). Wait, maybe the first value at \( t = 1 \) is 200, \( t = 2 \) is 400. So:

Step1: Subtract initial distance from final distance

Change in distance: \( 400 - 200 = 200 \) (Wait, no, the numerator is \( 400 - \square \), denominator \( 2 - \square \). Let's see, time changes from 1 to 2, so change in time is \( 2 - 1 = 1 \). Distance at 1 min: 200, at 2 min: 400. So \(\frac{400 - 200}{2 - 1}=\frac{200}{1}=200\)? Wait, no, the problem's formula is \(\frac{400 - \square}{2 - \square}\). Wait, maybe the blanks are: numerator \( 400 - 200 \), denominator \( 2 - 1 \). So:

Step1: Fill in the blanks for distance change

Change in distance: \( 400 - 200 \) (assuming at \( t = 1 \), distance is 200)

Step2: Fill in the blanks for time change

Change in time: \( 2 - 1 \)

Step3: Calculate the ratio

\(\frac{400 - 200}{2 - 1}=\frac{200}{1}=200\) ft per min. Wait, but the problem's formula is \(\frac{400 - \square}{2 - \square}\), so maybe:

\(\frac{400 - 200}{2 - 1}=\frac{200}{1}=200\). So the blanks are 200, 1, 200, 1, and the result is 200. Wait, maybe the correct values are:

Numerator: \( 400 - 200 = 200 \)

Denominator: \( 2 - 1 = 1 \)

So \(\frac{400 - 200}{2 - 1}=\frac{200}{1}=200\) ft per min.

Step1: Find distance at 1 min and 4 min

From the graph, at \( t = 1 \), distance is 200; at \( t = 4 \), distance is 800 (since the line is linear, slope 200, so at \( t = 4 \), \( d = 200 \times 4 = 800 \))

Step2: Calculate change in distance and time

Change in distance: \( 800 - 200 = 600 \)

Change in time: \( 4 - 1 = 3 \)

Step3: Find the ratio

Rate of change: \(\frac{600}{3}=200\) ft per min.

Step1: Identify two points on the line

The line passes through (0, 0) and (1, -3) (or (-1, 3)). Let's take (0, 0) and (1, -3).

Step2: Use slope formula \( m=\frac{y_2 - y_1}{x_2 - x_1} \)

\( m=\frac{-3 - 0}{1 - 0}=\frac{-3}{1}=-3 \)

Answer:

The blanks are (from top to bottom, left to right): 200, 1, 200, 1, and the final answer is 200.

Problem 6: