Ans: (a) (i) $(36,3)$ (ii) $\dfrac{-1}{144}(x-32)^2+8$ (iii) $2^{\frac{-1}{144}(x-32)^2+3}+5$ (b) (i) $36^\circ\text{C}$ (ii) $v=2^{\frac{-1}{144}(t-32)^2+3}+5$
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- By the method of completing the square, we have
$\begin{array}{rcl}
f(x) & = & \dfrac{1}{2} x-\dfrac{1}{144}x^2 – 6 \\
& = & \dfrac{-1}{144}(x^2 – 72x) – 6 \\
& = & \dfrac{-1}{144}(x^2 – 72x + 36^2 36^2) – 6 \\
& = & \dfrac{-1}{144}(x-36)^2 + 9 – 6 \\
& = & \dfrac{-1}{144}(x-36)^2 + 3
\end{array}$Therefore, the vertex of the graph $f(x)$ is $(36,3)$.
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$\begin{array}{rcl}
g(x) & = & f(x + 4) + 5 \\
& = & \dfrac{-1}{144}[(x+4) – 36]^2 + 3 + 5 \\
& = & \dfrac{-1}{144}(x-32)^2 + 8
\end{array}$ -
$\begin{array}{rcl}
h(x) & = & 2^{f(x+4)} + 5 \\
& = & 2^{\frac{-1}{144}[(x+4) – 36]^2 + 3} +5 \\
& = & 2^{\frac{-1}{144}(x-32)^2 + 3} + 5
\end{array}$
- By the method of completing the square, we have
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- Sub. $u=8$ into $u=2^{f(s)}$, we have
$\begin{array}{rcl}
8 & = & 2^{f(s)} \\
2^3 & = & 2^{f(s)} \\
3 & = & f(s) \\
3 & = & \dfrac{-1}{144}(s-36)^2 + 3 \\
(s-36)^2 & = & 0 \\
s & = & 36
\end{array}$Therefore, the required temperature is $36^\circ\text{C}$.
- By observing Table 1(a) and Table 1(b), we have $s=t+4$ and $u=v-5$. Hence, we have
$\begin{array}{rcl}
u & = & 2^{f(s)} \\
v-5 & = & 2^{f(t+4)} \\
v & = & 2^{\frac{-1}{144}[(t+4)-36]^2 + 3} +5 \\
& = & 2^{\frac{-1}{144}(t-32)^2 + 3} + 5
\end{array}$
- Sub. $u=8$ into $u=2^{f(s)}$, we have
