# QUADRATIC REGRESSION Data: On a particular summer day, the outdoor temperature was recorded at 8 times of the day, and the following table was compiled. A scatterplot was produced and the parabola of best fit was determined.

Data: On a particular summer day, the outdoor temperature was recorded at 8 times of the day, and the following table was compiled. A scatterplot was produced and the parabola of best fit was determined.

 t = Time of day (hour) y = Outdoor Temperature (degrees F.) 7 52 9 67 11 73 13 76 14 78 17 79 20 76 23 61

y = -0.3476t2 + 10.948t – 6.0778  where t = Time of day (hour) and y = Temperature (in degrees)

REMARKS: The times are the hours since midnight. For instance, 7 means 7 am, and 13 means 1 pm.

(a) 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.

(b) Use the quadratic polynomial to estimate the outdoor temperature at 7:30 am, to the nearest tenth of a degree. (work optional)

(c) Use the quadratic polynomial y = -0.3476t2 + 10.948t – 6.0778  together with algebra to estimate the time(s) of day when the outdoor temperature  y was 75 degrees.

That is, solve the quadratic equation 75 = -0.3476t2 + 10.948t – 6.0778  .

Show algebraic work in solving. State your results clearly; report the time(s) to the nearest quarter hour.