B Use the quadratic polynomial to estimate the outdoor tempe

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 72 degrees.

That is, solve the quadratic equation 72 = -0.3476t² + 10.948t - 6.0778

Solution

y = -0.3476t2 + 10.948t - 6.0778

estimate the time(s) of day when the outdoor temperature y was 72 degrees

So, we need to solve : 72 = -0.3476t^2 + 10.948t - 6.0778

-0.3476t^2 + 10.948t - 78.0778 =0

use quadratic root formula:

x = ( - b +/- sqrt(b^2 -4ac) )/2a

x = (- 10.948 + /- sqrt( ( 10.948^2+ 4*78.0778*0.3476 ) )/(-2*0.3476)

x = 10.91 hr , 20.58 hr