College Algebra, Chapter 3, 3.5, Section 3.5, Problem 80

The equation $\displaystyle C(t) = \frac{1}{2} t^2 + 2$ represents the temperature on a certain afternoon where $t$ represents hours after 12 noon $(0 \leq t \leq 6)$ and $C$ is measured in $^{\circ} C$.

a.) How the equation $y = C(t)$ obtained from $y = t^2$?

$y = C(t)$ is obtained by shrinking the graph of $y = t^2$ vertically by a factor of 2, then the result is shifted 2 units upward.

b.) Suppose you want to measure the temperature in $^{\circ} F$ instead. What transformation would you have to apply to the function $y = C(t)$ to accomplish this? Recall that $\displaystyle F = \frac{9}{5} C + 32$. While the new functions $y = F(t)$ that results from this transformation.

The function $y = F(t)$ can be obtained by stretching the graph of $y =C(t)$ vertically by a factor of $\displaystyle \frac{9}{5}$, then the result is shifted 32 units upward.


$
\begin{equation}
\begin{aligned}

\text{Thus, if } F =& \frac{9}{5} C + 32, \text{ then}
&&
\\
\\
F =& \frac{9}{5} \left( \frac{t^2}{2} + 2\right) + 32
&& \text{Substitute } C(t) = \frac{t^2}{2} + 2
\\
\\
F =& \frac{9}{10} t^2 + \frac{18}{5} + 32
&& \text{Apply Distributive Property}
\\
\\
F =& \frac{9}{10} t^2 + \frac{178}{5}
&&

\end{aligned}
\end{equation}
$