Question

I really need your help to answer this question for Tables 1 - 5. Thank you so much I appreciate it!!!

For each set of values, determine whether an exponential function is a good model. If so, find the function. If not, explain why.

Part C-Linear and Exponential Models (possible 15 points) You can transform an exponential function into a linear function by taking the logarithm of each side. Since linear models are easy to recognize, you can then determine whether an exponential function is a good model for a set of values. Write the general form of an exponential function. y = ab log y = log ab* log y = log a + x log b log y = (log b)x + log a Take the logarithm of each side. Product property and power property. Rewrite. If log b and log a are constants, then log y = (log b)x+ log a is a linear equation in slope-intercept form. To confirm that log b is a constant, check that the graph of log y = (log b)x+ log a is a line. Activity: Determine whether an exponential function is a good model for the values in the table. x y 0 3.0 2 5.1 4 8.6 6 14.5 8 24.5 10 41.4 Step Step 2 Enter the values into STAT lists L1 and L2. To enter the values for log y place cursor in the heading of L3 and press LOG 2ND 2 ENTER. To graph log y, set-up STAT PLOT feature and press 1. Then enter L3 next to YList. To Zoom results, Press ZOOM 9. 10 L1 L2 13 15.1 8.6 14.5 1 24.5 414 L1(1) = 0 L31 0.47712 0.70757 0.9345 1.1614 1.3892 1.617 Since the graph of log y = (log b)x + log a is linear, the slope log b is constant, and b also is constant. An exponential function therefore is a suitable model. Step Press STAT arrow over right O ENTER to find the exponential function; y = 3(1.3) For each set of values, determine whether an exponential function is a good model. If so, find the function. If not, explain why.

Answer 1

1.

y = 17.9x - 23.7 R2 = 0.9697 y = 5.5774e0.3952x R2 = 0.9492 54 102 145 9 10 10 Linear is better fit to the data

2

40.2 19.8 0NM 9.9) y = -9.01x + 24.51 R= 0.8663 y = 5.5774e0.3952x R2 = 0.9492 5.1 2.5 1 2 3 4 Expo is better fit to the data

3

1200 400 1200 500 1000 y = 143.45x + 369.68 R2 = 0.9831 1000 y = 5.5774e0.3952x R= 0.9492 wN 625 781.3 976.6 800 800 600 600 400 400 200 200 2 3 4 5 3 Expo is better fit to the data

4

y = 0.4286x + 1.0476 RP = 0.9643 1.5 •1. y = 5.577480.3952 • 1 R2 = 0.9492 0.5 in -10 O -10 N w Linear is better fit to the data

5

48 192 3 y = 27.9x + 38.1 R= 0.336 100 y = 5.5774e0.3952x R2 = 0.9492 50 Both the equation do not fit the data properly, the extremely points are miss both, hence none is better.