Question

A new high school opened in 2018 with an initial enrollment of 1,500 students. The number of students attending, S , grows as a linear function of time, t, where t is the number of years since the school’s opening. In 2022, there were 1,900 students enrolled. Sketch a graph of the situation. Where would you place the y-intercept?(1 point) Responses (0, 1,900) left parenthesis 0 comma 1,900 right parenthesis (0, 1,500) left parenthesis 0 comma 1,500 right parenthesis (1,500, 0) left parenthesis 1,500 comma 0 right parenthesis (1,900, 0)

Answer 1

To model the enrollment as a linear function of time, let's denote the number of students \( S \) as a function of the number of years since the school opened, \( t \).

From the information given:
- In 2018 (when \( t = 0 \)), the enrollment was \( S(0) = 1,500 \).
- In 2022 (when \( t = 4 \)), the enrollment was \( S(4) = 1,900 \).

First, we can determine the slope \( m \) of the linear function using the two points: \( (0, 1,500) \) and \( (4, 1,900) \).

1. Calculate the slope \( m \):
\[
m = \frac{S(4) - S(0)}{4 - 0} = \frac{1,900 - 1,500}{4 - 0} = \frac{400}{4} = 100
\]

2. The linear function can be expressed in slope-intercept form \( S(t) = mt + b \).
Here, \( m = 100 \) and \( b \) is the y-intercept, which we found from \( S(0) = 1,500 \). Thus, \( b = 1,500 \).

The function is:
\[
S(t) = 100t + 1,500
\]

**Graphing the Function**:

- The y-intercept will be at the point \( (0, 1,500) \).
- The point for \( t = 4 \) years (2022) shows \( S(4) = 1,900 \), which is another point on the line.

Plotting those two points and connecting them will yield a straight line that represents the growth in school enrollment over time.

**Answer**:
The y-intercept is located at \( (0, 1,500) \).