Answer
(a) $y=1.95x+34.2$ for $[0,8]$, $y=0.48x+45.96$ for $(8,13]$
(b) $f(x)=\begin{cases} 1.95x+34.2 \ \ \ \ \ \ \ 0\le x\le 8 \\ 0.48x+45.96\ \ \ \ \ 8\lt x\le 13 \end{cases}$
Work Step by Step
(a)
Step 1. Use the first two points $(0,34.2)$ and $(8,49.8)$, assume the equation as $y=mx+b$, we have $m=\frac{49.8-34.2}{8-0}=1.95$ and $b=34.2$, thus $y=1.95x+34.2$ for $[0,8]$
Step 2. Use the last two points $(8,49.8)$ and $(13,52.2)$, assume the equation as $y=m'x+b'$, we have $m'=\frac{52.2-49.8}{13-8}=0.48$ and $49.8=0.48(8)+b'$ or $b'=45.96$, thus $y=0.48x+45.96$ for $(8,13]$
(b) Combine the above results, we have $f(x)=\begin{cases} 1.95x+34.2 \ \ \ \ \ \ \ 0\le x\le 8 \\ 0.48x+45.96\ \ \ \ \ 8\lt x\le 13 \end{cases}$
