SOLUTION: I could really use some help on this quadratic, I'm lost! : QUADRATIC REGRESSION Data: On a particular spring day, the outdoor temperature was recorded at 8 times of the d

Question 1039862: I could really use some help on this quadratic, I'm lost! :
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 = -.2t^2 + 6.3t + 42.9 for 0 less than or equal to t less than or equal to 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 = 22 means 10 pm.
t = 18.25 hours means 6:15 pm
(a) Use the quadratic polynomial to estimate the outdoor temperature at 6:30 am, to the nearest tenth of a degree.

(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 = -0.2t^2 + 6.3t + 42.9 together with algebra to estimate the time(s) of day when the outdoor temperature y was 82 degrees.
That is, solve the quadratic equation 82 = -0.2t^2 + 6.3t + 42.9.
Show algebraic work in solving. Round the results to the nearest tenth. Write a concluding sentence to report the time(s) to the nearest quarter-hour, in the usual time notation. (Use more paper if needed)

I really appreciate any help that can be offered! Thank you so much!

Answer by josgarithmetic(39616) (Show Source): You can put this solution on YOUR website!

6:30 AM is .

You have two ways to find the maximum temperature and when it will be expected.
Either find the roots using general formula for quadratic equation and find the middle t
value and evaulate y for this t; or put the y=quadraticExpression into standard form to read the vertex directly.
YOU try either of those. Tell if you have trouble.

Maximum y is at t=15.75 hours.

--
Keep on working with it. You may have the right idea. Note that your given expression is for a parabolic function y=-0.2t^2+6.3t+42.9, and when you convert into standard form and graph it, this appears

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