Linear Change Practice

Complete this assessment to review what you’ve learned. It will not count toward your grade.
2 of 52 of 5 Items

Question
Use the table to answer the question.

Let x
represent the time in minutes and let y
represent the distance in feet.

Time (Minutes), x


1 2 3 4
Distance (Feet), y


75 150 225 300
Use the table to create a formula for the given information.

(1 point)

1 answer

To create a formula based on the provided table, we first observe the relationship between \( x \) (time in minutes) and \( y \) (distance in feet).

Given the data:

  • When \( x = 1 \), \( y = 75 \)
  • When \( x = 2 \), \( y = 150 \)
  • When \( x = 3 \), \( y = 225 \)
  • When \( x = 4 \), \( y = 300 \)

We can see that the distance \( y \) increases by 75 feet for each additional minute. This indicates a linear relationship, where the rate of change (slope) is constant.

To find the slope \( m \):

\[ m = \frac{\Delta y}{\Delta x} = \frac{150 - 75}{2 - 1} = \frac{75}{1} = 75 \]

This shows that for every 1 minute (for each increment of \( x \)), the distance \( y \) increases by 75 feet.

Next, we can use the slope-intercept form of a linear equation, which is given by:

\[ y = mx + b \]

Where:

  • \( m = 75 \) (the slope)
  • \( b \) is the y-intercept (the value of \( y \) when \( x = 0 \)).

To find \( b \), we can look at one of the given points. When \( x = 1 \) and \( y = 75 \):

\[ 75 = 75(1) + b \] \[ 75 = 75 + b \] \[ b = 75 - 75 = 0 \]

Now we can plug in the values of \( m \) and \( b \) into the equation:

\[ y = 75x + 0 \]

Thus, the formula relating distance to time in this case is:

\[ y = 75x \]