9.QUADRATIC REGRESSION

Data: On a particular spring day, the outdoor temperature was recorded at 8 times of the day. The parabola of best fit was determined using the data.

Quadratic Polynomial of Best Fit: y = 0.10t²+ 2.9t + 44.6 for 0 t 24

where t = time of day (in hours) and y = temperature (in degrees)

REMARKS:The times are the hours since midnight.

For instance, t = 6 means 6 am. t = 20 means 8 pm. t = 18.25 hours means 6:15 pm

(a) Use the quadratic polynomial to estimate the outdoor temperature at 6:15 am, to the nearest tenth of a degree. (work optional)

(b) Using algebraic techniques we have learned, find themaximum temperaturepredicted by the quadratic modeland find thetime when it occurred. Report the time to the nearest quarter hour (i.e., __:00 or __:15 or __:30 or __:45). (For instance, a time of 18.25 hours is reported as 6:15 pm.) Report the maximum temperature to the nearest tenth of a degree.Show algebraic work.

(c) Use the quadratic polynomial y = 0.10t² + 2.9t + 44.6 together with algebra to estimate the time(s) of day when the outdoor temperature y was exactly 60 degrees. That is, solve the quadratic equation 60 = 0.10t² + 2.9t + 44.6. Show algebraic work in solving. Round the results to the nearest tenth. Writea concluding sentence to report the time(s) to the nearest quarter-hour, in the usual time notation.