The temperatures of Charleston, South Carolina on May 3rd are recorded in the table below. Determine the equation that models this data.

Ok, the first step to see the behavior of this data is doing a graph of it,

So we can see that it looks like an reversed V. (at the bottom of the answer you can see the graph)

Then, we can model this with a piecewise function like:

For the first part we can take the two pairs:

if 11 am is our t = 0h

(0, 85) and (6, 93)

then the linear equation

y = a*t + b

a = (93 - 85)/(6 -0) = 8/6.

and b = 85.

then the linear equation is:

y = (8/6)*t + 85. for 0h ≤ t ≤ 6h

The second part of this equation can be modeled with the points

(6, 93) and (11, 84)

then we have:

y = s*t + c

s = (84 - 93)/(11 - 6) = -7/5

and we can find the value of c as:

93 = (-7/5)*6 + c

93 + (7/5)*6 = 101.4

then this second part is:

y = (-7/5)*t + 101.4 for 6  ≤ t ≤ 11h

then, in conclusion we have:

         y = (8/6)*t + 85. for 0h ≤ t ≤ 6h

f(x) =

         y = (-7/5)*t + 101.4 for 6  ≤ t ≤ 11h