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).

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.


TI 84 Screen Image

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.

TI 84 Screen

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.

TI 84 Screen

Therefore, the linear function which models the data is f(x) = 1.5x - 3.





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