 Recognizing Linear and Exponential Data

The problem: For each table of data, find either a linear or exponential function that models the data.

Part a).
 x 0 1 2 3 y -3 -1.5 0 1.5

Notice that as each x increases by 1, each corresponding y increases by 1.5 (a constant amount). Thus, the data can be modeled by a linear function, f(x) = mx + b. Recall that m is the slope of the line and b is the y-intercept. There are two ways to obtain the equation of this line.

1). The equation of the line can be determined symbolically as follows:
Since each y increases by 1.5 when each x increases by 1, m= 1.5.
Also, from the data point (0, - 3) the y-intercept can be identified as b = - 3. Thus, the equation is y = 1.5x - 3

2). The equation of the line can be determined using the calculator linear regression capability as follows:
Press STAT then select EDIT. Enter the x values in and the y values in . Check the menu to be sure that there are no functions listed and Plot1 is on (highlighted). Press ZOOM and select 6 to graph points as a scatterplot in the standard window [-10,10,1] by [-10,10,1]. Note that the data points appear to lie in a straight line and the y-differences calculated in L3 are equal.
Whenever the y-differences are all equal, a linear model will fit the data exactly.     Next press STAT, ARROW RIGHT to CALC, the select 4:LinReg(ax+b) by arrowing down to 4 and pressing ENTER . To copy the equation obtained into Y1, we scroll down to Store RegEQ: and press VARS followed by the RIGHT ARROW. Then press ENTER two times. Next arrow down to Calculate and press ENTER.    The results show that the equation form is y=ax+b (same as y=mx+b), where the slope a = 1.5 and the y-intercept b = -3.
Thus, the result is y = 1.5x - 3. To verify this visually, graph the function, y = 1.5x - 3, and the scatterplot simultaneously.  Therefore, the linear function which models the data is f(x) = 1.5x - 3.