Question

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)

Answer 1

Part A: you can check to see if this model is a linear or exponential function by checking the gradients of the given (x,y) values. If the gradients are the same for all the (x,y) values then the model would be linear but if they are not then the model would be exponential. Gradient is another name for the slope in this case. So to calculate gradient we do (y2-y1)/(x2-x1):

Gradient of (1,3) and (2,6) is (6-3)/(2-1) = 3
Gradient of (2,6) and (3,12) is (12-6)/(3-2) = 6
Gradient of (3,12) and (4,24) is 12
Since the gradients of these points are not a constant value then this would explain that this is an exponential function.
Part B: y = 1.5*2^x
So general formula of exponential is y=a*b^x. We need to find the values of a and b:
We can plug in the (1,3) coordinates : 3 = a*b^1 which is 3 =a*b
Now divide both sides by b - 3/b = (a*b) / b = 3/b = a
To find the value of b we plug in the 2nd coordinate (2,6) and we get 6 = a*b^2.
Now we have 3/b=a and 6=a*b^2. We substitute for a and we get 6=(3/b)*b^2
We simply to get 6=3b which gives b as 2.
To find a we just reuse (1,3) in the equation 3= a*2^1 which gives us a=3/2 or 1.5
So the function that represents the data is y=1.5*2^x
Part C: To find the average rate of change (the gradient or slope of the two coordinates) we use the formula we used in part A. (y2-y1)/(x2-x1)
(2,6) and (4,24) so (24-6)/(4-2) = 18/2 = 9 minutes per station.

Answer 2

Part A: This data is modeling an exponential function. The time it takes to complete each station is doubling as the station number increases by 1, which is characteristic of an exponential function.

Part B: To represent the data with an exponential function, we can use the form y = ab^x, where a is the initial value and b is the base. Using the given data, we can find the values of a and b.

Using the point (1, 3):
3 = ab^1
a = 3/b

Using the point (2, 6):
6 = ab^2
6 = (3/b)b^2
6 = 3b
b = 2

Now that we have the value of b, we can find the value of a using the point (1, 3):
3 = a(2)^1
a = 3/2

So, the function to represent the data is y = (3/2)(2)^x.

Part C: The average rate of change between station 2 and station 4 can be found by calculating the change in y-coordinates divided by the change in x-coordinates.

Station 2: (2, 6)
Station 4: (4, 24)

Average rate of change = (24 - 6) / (4 - 2) = 18 / 2 = 9

Therefore, the average rate of change between station 2 and station 4 is 9.