A university opened in 2015 with an initial enrollment of 2,300 students. The number of students attending, N, grows as a linear function of time, t, where t is the number of years since the university’s opening. In 2023, there were 2,900 students enrolled. Sketch a graph of the situation and label the point of the y-intercept.
First, we know that the number of students attending, N, is a linear function of time, so it can be represented by an equation of the form N = mt + b, where m is the slope and b is the y-intercept.
Given that the initial enrollment in 2015 was 2,300 students, we know that the y-intercept is (0, 2300).
We are also given that in 2023, there were 2,900 students enrolled. Since 2023 is 8 years after the university's opening in 2015, we can plug in t = 8 and N = 2900 into the equation N = mt + b to solve for the slope, m.
2900 = 8m + 2300
600 = 8m
m = 75
So, the linear function representing the number of students attending the university is N = 75t + 2300.
Now we can sketch the graph of this function, with t on the x-axis and N on the y-axis. The y-intercept is (0, 2300) and the slope is 75. The graph will be a diagonal line starting at (0, 2300) and increasing by 75 for each additional year. In 2023, the point will be (8, 2900).
The graph will start at (0, 2300) and increase in a straight line towards the point (8, 2900).