Answer
A) $$
\begin{aligned}
& H(t+15)=68+93(0.91)^{t+15} \\
& H(t)+15=83+93(0.91)^t
\end{aligned}
$$
B) See the figure below.
C) $H(t+15)$ is the function shifted to the left 15 units. It means that the cup of coffee was brought to the class 15 minutes earlier, while $H(t)+15$ is the function shifted up by 15 units or $15^{\circ} F$ and could mean that it was brought to a warmer classroom.
D) $68^{\circ} \mathrm{F}$, $83^{\circ} \mathrm{F}$.
Work Step by Step
A) If $H(t)=68+93(0.91)^t$
then
$$
\begin{aligned}
& H(t+15)=68+93(0.91)^{(t+15)}=68+93(0.91)^t(0.91)^{15} = 68+22.59976(0.91)^t\\
& H(t)+15=\left(68+93(.91)^t\right)+15=83+93(0.91)^t
\end{aligned}
$$
B) See the figure below.
C) $H(t+15)$ is the function shifted to the left 15 units. It means that the cup of coffee was brought to the class 15 minutes earlier, while $H(t)+15$ is the function shifted up by 15 units or $15^{\circ} F$ and could mean that it was brought to a warmer classroom.
D) As $t$ gets larger, both $\boldsymbol{H}(t+15)$ and $\boldsymbol{H}(t)$ approach a final temperature of $68^{\circ} \mathrm{F}$. On the other hand, $\boldsymbol{H}(t)+15$ approaches $68^{\circ} \mathrm{F}+15^{\circ} \mathrm{F}=83^{\circ} \mathrm{F}$
