a) Input the times under L1 in your graphing calculator and the corresponding temperatures under L2. The types of regressions we can use to write a model are linear, quadratic, cubic, quartic, natural log, exponential, power, logistic, and sinusoidal.

The graph is in the shape of half of a U so it is likely an exponential model. Use the ExpReg feature on your calculator to find the model and the value of R2. This gives \(\displaystyle{a}≈{71}{\quad\text{and}\quad}{b}≈{0.8658}\) for \(\displaystyle{y}={a}⋅{b}^{{x}}\) which gives a model of \(\displaystyle{y}={71}{\left({0.8658}\right)}^{{x}}\). Since \(\displaystyle{R}^{{2}}≈{0.9965}\) which is very close to 1, then this model is a very good fit for the data.

b) For an exponential function, aa is the initial value and bb is the growth or decay factor. Since \(\displaystyle{a}≈{71}\), then the initial temperature is about \(71^{\circ}C\). Since \(b \approx 0.8658\) and \(1-0.8658=0.1342=13.42\%\), then the temperature is decreasing by about 13.42% each minute the temperature is decreasing by about 13.42% each minute.

c) After \(x=3\) minutes, the temperature is about \(\displaystyle{y}={71}{\left({0.8658}\right)}^{{3}}≈{46.1}^{\circ}{C}\).

d) After \(x=25\) minutes, the temperature is about \(\displaystyle{y}={71}{\left({0.8658}\right)}^{{25}}≈{1.9}^{\circ}{C}\).

e) Since \(x=3\) lies between two points that were used to create the model, I have high confidence in the estimate of \(46.1^{\circ}C\). Since \(x=25\) lies outside of the points used to create the model but is close to the largest point with \(x=20\), I am moderately confident in the estimate of \(1.9^{\circ}C\).