
Carol H. answered 08/01/14
Tutor
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Masters' Degree in Mathematics with 35 years of Teaching Experience
linear regression: y = -17.77x + 139.59
Using this equation, let y = 0 which would be the number of feet at ground level: 0 = -17.77x + 139.59. Thus solving for x, he would hit the ground in 7.86 sec.
Quadratic regression: y = -16.29x2 + 51.45x + 78.52
This equation isn't feasible because his jump would take the shape of a parabola. If he was being shot out of a cannon, this equation would make more sense.
Exponential regression: y = 154.98(.81)x
This equation isn't logical either since with a decreasing exponential equation, technically he would never hit the ground.
Do you know how to enter the data into a graphing calculator? If not, let me know.

Carol H.
I am using a TI-84 Plus Graphing Calculator. First you need to enter times and heights. STAT-EDIT-enter the times under list 1 (L1) and the heights under list 2 (L2). Hit 2nd Quit. Then, STAT-CALC-LINREG(ax + b)-ENTER. That gives you the equation for
a straight line.
Enter STAT-CALC-QUADREG-ENTER. Then, you have the equation for a quadratic function (parabola). Follow the same steps for other regressions.
To graph your equations, hit Y=, VARS, Statistics, Arrow over to EQ, Enter. Then hit graph.
Hope that helps.
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08/02/14
Nikky C.
08/02/14