Consider the data points (–4, –14), (–2, –7), (0, –3), (2, –2), and (6, –6). Which type of function would best model the data set?
A. exponential
B. linear
C. quadratic
Answer (2)
The best function model for the given data points is a quadratic function because the changes in y-values are not constant but display a pattern indicative of quadratic behavior. The analysis of the first and second differences supports this conclusion. Therefore, the answer is C. Quadratic.
;
To determine which type of function best models the given data points ( − 4 , − 14 ) , ( − 2 , − 7 ) , ( 0 , − 3 ) , ( 2 , − 2 ) , ( 6 , − 6 ) , we can analyze the pattern they form when plotted on a graph. Let's consider the characteristic patterns of exponential, linear, and quadratic functions:
Linear Function : A linear function has a constant rate of change and forms a straight line, which can be described by the equation y = m x + b , where m is the slope and b is the y-intercept.
Quadratic Function : A quadratic function forms a parabolic curve, described by the equation y = a x 2 + b x + c , where a , b , and c are constants.
Exponential Function : An exponential function shows rapid growth or decay and is represented by the equation y = a b x , where a is the initial value and b is the base of the exponential.
To identify the best fitting model, let's compute differences in the y-values as the x-values increase:
The change from ( − 4 , − 14 ) to ( − 2 , − 7 ) is − 7 − ( − 14 ) = 7 .
The change from ( − 2 , − 7 ) to ( 0 , − 3 ) is − 3 − ( − 7 ) = 4 .
The change from ( 0 , − 3 ) to ( 2 , − 2 ) is − 2 − ( − 3 ) = 1 .
The change from ( 2 , − 2 ) to ( 6 , − 6 ) is − 6 − ( − 2 ) = − 4 .
The differences are not constant, which suggests the data is not linear. Now, calculate the second differences:
The second difference from 7 to 4 is 4 − 7 = − 3 .
The second difference from 4 to 1 is 1 − 4 = − 3 .
The second difference from 1 to − 4 is − 4 − 1 = − 5 .
The second differences are not constant either, which suggests the data is not quadratic.
Given that neither the first nor second differences are constant, the data does not neatly fit a linear or quadratic model. However, without rapid exponential growth or decay, the data seems closest to a linear pattern in simple analysis. Thus, a linear model is most practical without complex models, though exact fit calculations could determine a precise model.
The chosen option is linear.
