Aurora is planning to participate in an event at her school's field day that requires her to complete tasks at various stations in the fastest time possible. To prepare for the event, she is practicing and keeping track of her time to complete each station.

The x-coordinate is the station number, and the y-coordinate is the time in minutes since the start of the race that she completed the task.

(1, 3), (2, 6), (3, 12), (4, 24)

Part A: Is this data modeling a linear function or an exponential function? Explain your answer. (2 points)

Part B: Write a function to represent the data. Show your work. (4 points)

Part C: Determine the average rate of change between station 2 and station 4. Show your work. (4 points)

The rate of change of a linear function is constant, so the given data is modelling an exponential function since the y-values are doubling for each unit increase.

Part B

[tex]\boxed{\begin{minipage}{9 cm}\underline{General form of an Exponential Function}\\\\$f(x)=ab^x$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $b$ is the base (growth/decay factor) in decimal form.\\\end{minipage}}[/tex]

The y-values double per unit increase, so the base, b, is 2.

Substitute the value of b = 2 and point (1, 3) into the formula and solve for a:

[tex]\implies 3=a2^1[/tex]

[tex]\implies 3=2a[/tex]

[tex]\implies a=\dfrac{3}{2}=1.5[/tex]

Therefore, the function that represents the data is:

[tex]f(x)=1.5(2)^x[/tex]

Part C

[tex]\boxed{\begin{minipage}{6.3 cm}\underline{Average rate of change of function $f(x)$}\\\\$\dfrac{f(b)-f(a)}{b-a}$\\\\over the interval $a \leq x \leq b$\\\end{minipage}}[/tex]

Given interval:

2 ≤ x ≤ 4

Therefore, a = 2 and b = 4.

From the given points:

f(2) = 6
f(4) = 24

Therefore, the average rate of change is:

[tex]\implies \dfrac{f(4)-f(2)}{4-2}=\dfrac{24-6}{4-2}=\dfrac{18}{2}=9[/tex]