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 the maximum temperature predicted by the quadratic model and find the time 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. help conclude sentence to report the time(s) to the nearest quarter-hour, in the usual time notation.