Question

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

(1,900, 0)
left parenthesis 1,900 comma 0 right parenthesis

(1,500, 0)
left parenthesis 1,500 comma 0 right parenthesis

(0, 1,500)

Answer 1

To analyze the situation, we know the following:

• The school opened in 2018, starting with an enrollment (S) of 1,500 students. This means at time \( t = 0 \) (which corresponds to the year 2018), the number of students is 1,500.
• In 2022, which is 4 years after the school opened (i.e., \( t = 4 \)), there were 1,900 students enrolled.

To determine the y-intercept of this linear function of enrollment, we look for the point on the graph where \( t = 0 \). Since \( t = 0 \) corresponds to the year 2018 when the school started, the y-coordinate is the initial enrollment, which is 1,500.

Thus, the point that represents the y-intercept (the number of students enrolled when \( t = 0 \)) is:

\[ (0, 1,500) \]

Therefore, the correct choice for the y-intercept placement is:

(0, 1,500)