(a)

\begin{align} \text{When } & x = 0, \\ y & = 4^{(0)} \\ & = 1 \\ \implies & y\text{-intercept is } 1 \end{align}

(b)

\begin{align} \text{When } & x = 0, \\ y & = \left( 1 \over 3 \right)^{(0)} \\ & = 1 \\ \implies & y \text{-intercept is } 1 \end{align}

(c)

\begin{align} \text{When } & x = 0, \\ y & = e^{(0)} \\ & = 1 \\ \implies & y \text{-intercept is } 1 \end{align}

(d)

\begin{align} \text{When } & x = 0, \\ y & = \pi^{(0)} \\ & = 1 \\ \implies & y \text{-intercept is } 1 \end{align}

(a)

\begin{align} \text{When } & x = 0, \\ y & = 6(e^0) \\ & = 6(1) \\ & = 6 \\ \implies & y \text{-intercept is } 6 \end{align}

(b)

\begin{align} \text{When } & x = 0, \\ y & = 2\left(1 \over e\right)^0 \\ & = 2(1) \\ & = 2 \\ \implies & y \text{-intercept is } 2 \end{align}

(c)

\begin{align} \text{When } & x = 0, \\ y & = 2\left(1 \over 3\right)^0 \\ & = 2(1) \\ & = 2 \\ \implies & y \text{-intercept is } 2 \end{align}

(d)

\begin{align} \text{When } & x = 0, \\ y & = 3(5.1)^0 \\ & = 3(1) \\ & = 3 \\ \\ \implies & y \text{-intercept is } 3 \end{align}

(i) The amount of money that James invested into the savings account is the amount he invested at the start

\begin{align} \text{When } & x = 0, \\ y & = 50 \phantom{.} 000(1.04)^{0} \\ & = 50 \phantom{.} 000(1) \\ & = \$50 \phantom{.} 000 \end{align}

(ii)

\begin{align} \text{When } & x = 5, \\ y & = 50 \phantom{.} 000(1.04)^{5} \\ & = 60 \phantom{.} 832.645\phantom{.} 12 \\ & \approx \$60 \phantom{.} 833 \end{align}

(i)

\begin{align} \text{When } & x = 0, \\ y & = 40(1.4)^0 \\ & = 40(1) \\ & = 40 \end{align}

(ii)

\begin{align} \text{When } & x = 6, \\ y & = 40(1.4)^{6} \\ & = 301.181 \phantom{.} 44 \\ & \approx 301 \end{align}

(iii) Since x in hours, we are looking for the number of bacteria after 24 hours (1 day)

\begin{align} \text{When } & x = 24, \\ y & = 40(1.4)^{24} \\ & = 128 \phantom{.} 567.988 \phantom{.} 017 \\ & \approx 128 \phantom{.} 568 \end{align}

(i)

\begin{align} \text{When } & x = 0, \\ T & = 90(0.98)^0 \\ & = 90^\circ \text{C} \end{align}

(ii)

\begin{align} \text{When } & x = 10, \\ T & = 90(0.98)^{10} \\ & = 73.536 \\ & \approx 73.5^\circ \text{C} \end{align}

(iii) Since x is in minutes, we are looking for the temperature of the liquid after 60 minutes (1 hour)

\begin{align} \text{When } & x = 60, \\ T & = 90(0.98)^{60} \\ & = 26.779 \\ & \approx 26.8^\circ \text{C} \end{align}

The sketch for parts (i) and (ii):

(i)

\begin{align} \text{When } & x = 0, \\ y & = 3(2^0) \\ & = 3(1) \\ & = 3 \\ \implies & y \text{-intercept is } 3 \end{align}

(ii)

\begin{align} y & = 6 - x \\ \\ \text{When } & x = 0, \\ y & = 6 - (0) \\ & = 6 \\ \implies & y \text{-intercept is } 6 \\ \\ \text{When } & y = 0, \\ 0 & = 6 - x \\ x & = 6 \\ \implies & x \text{-intercept is } 6 \end{align}

(iii)

\begin{align} \underbrace{3(2^x)}_{\text{Curve }} & = \underbrace{6 - x}_{\text{Line}} \end{align} $$ \therefore \text{From the graph, only 1 solution} $$

(i)

\begin{align} \text{When } & t = 0, \\ P & = 600(2 + e^{-0.2(0)}) \\ & = 600(2 + 1) \\ & = 1800 \end{align}

(ii) Since t is in months, we are looking for the number of fishes in the lake after 12 months (1 year)

\begin{align} \text{When } & t = 12, \\ P & = 600(2 + e^{-0.2(12)}) \\ & = 1 \phantom{.} 254.430 \phantom{.} 77 \\ & \approx 1254 \end{align}

(iii)

\begin{align} P & = 600(2 + e^{-0.2t}) \\ & = 600\left( 2 + {1 \over e^{0.2t}} \right) \end{align} \begin{align} \text{As } t \rightarrow \infty &, \phantom{0} {1 \over e^{0.2t}} \rightarrow 0 \\ \\ \therefore P & = 600(2 + 0) \\ & = 1200 \end{align}

(i) Starting from the end of 2011 to the end of 2015, 4 years has elapsed

\begin{align} \text{When } & x = 4, \\ y & = 45 \phantom{.} 000 (1.02)^4 \\ & = 48 \phantom{.} 709.447 \\ & \approx 48 \phantom{.} 709 \end{align}

(ii) Starting from the end of 2011 to the end of 2021, 10 years has elapsed

\begin{align} \text{When } & x = 10, \\ y & = 45 \phantom{.} 000 (1.02)^{10} \\ & = 54 \phantom{.} 854.748 \phantom{.} 899 \\ & \approx 54 \phantom{.} 854 \phantom{00000} [\text{Rounded down}] \end{align}

(iii)

\begin{align} \text{When } & x = 0, \\ y & = 45 \phantom{.} 000(1.02)^0 \\ & = 45 \phantom{.} 000(1) \\ & = 45 \phantom{.} 000 \\ \implies & y \text{-intercept is } 45 \phantom{.} 000 \end{align}
(Since x represents the number of years, start from x = 0)

(i)

\begin{align} \text{When } & t =0, \\ C & = 4.86e^{-0.047(0)} \\ & = 4.86(0) \\ & = 4.86 \text{ μg/ml} \end{align}

(ii)

\begin{align} \text{When } & t = 10, \\ C & = 4.86e^{-0.047(10)} \\ & = 3.037 \phantom{.} 511 \\ & \approx 3.04 \text{ μg/ml} \end{align}

(iii) Since t is in hours, we are looking for the concentration of Fenitoine in the blood after 24 hours (1 day)

\begin{align} \text{When } & t = 10, \\ C & = 4.86e^{-0.047(24)} \\ & = 1.573 \phantom{.} 084 \\ & \approx 1.57 \text{ μg/ml} \end{align}

(iv)

\begin{align} C & = 4.86e^{-0.047t} \\ & = 4.86\left(1 \over e^{0.047t}\right) \end{align} \begin{align} \text{As } t \rightarrow \infty &, \phantom{0} {1 \over e^{0.047t}} \rightarrow 0 \\ \\ \therefore C & = 4.86(0) \\ & = 0 \text{ μg/ml} \end{align}

(v)

\begin{align} C & = 4.86e^{-0.047t} \\ & = 4.86(e^{-0.047})^t \\ & = 4.86\left(1 \over e^{0.047}\right)^t \end{align} \begin{align} \text{When } & t = 0, \\ C & = 4.86 \text{ (From part i)} \\ \implies & \text{Vertical intercept is } 4.86 \end{align}
(Since t is the number of hours after administering of the drug, t cannot be a negative value. Thus start from t = 0)

(i)

\begin{align} \text{When } & t = 0, \\ m & = 405\left(2 \over 3\right)^0 \\ & = 405(1) \\ & = 405 \text{ g} \end{align}

(ii)

\begin{align} \text{When } & t = 3, \\ m & = 405\left(2 \over 3\right)^3 \\ & = 120 \text{ g} \end{align}

(iii)

\begin{align} \text{Amount of substance decayed in 3 years } & = \text{Initial mass } - \text{Final mass } \\ & = 405 - 120 \\ & = 285 \text{ g} \end{align}

(iv)

\begin{align} \text{When } & m = 80, \\ 80 & = 405\left(2 \over 3\right)^t \\ {80 \over 405} & = \left(2 \over 3\right)^t \\ {16 \over 81} & = \left(2 \over 3\right)^t \\ {2^4 \over 3^4} & = \left(2 \over 3\right)^t \\ \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \phantom{00000000} \left(2 \over 3\right)^4 & = \left(2 \over 3\right)^t \\ \\ \therefore t & = 4 \\ \\ \therefore \text{No. of years} & = 4 \end{align}

(i) The initial value of the car is $60,000

\begin{align} V & = Ae^{-kt} \\ \\ \text{When } t = 0 & \text{ and } V = 60 \phantom{.} 000, \\ 60 \phantom{.} 000 & = Ae^{-k(0)} \\ 60 \phantom{.} 000 & = Ae^0 \\ 60 \phantom{.} 000 & = A(1) \\ \\ A & = 60 \phantom{.} 000 \\ \\ \therefore V & = 60 \phantom{.} 000e^{-kt} \\ \\ \text{When } t = 4 & \text{ and } V = 39 \phantom{.} 366, \\ 39 \phantom{.} 366 & = 60 \phantom{.} 000e^{-k(4)} \\ 39 \phantom{.} 366 & = 60 \phantom{.} 000e^{-4k} \\ {39 \phantom{.} 366 \over 60 \phantom{.} 000} & = e^{-4k} \\ {6561 \over 10 \phantom{.} 000} & = e^{-4k} \\ {9^4 \over 10^4} & = (e^{-k})^4 \\ \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \phantom{00000000} \left(9 \over 10\right)^4 & = (e^{-k})^4 \\ \\ \therefore e^{-k} & = {9 \over 10} \\ & = 0.9 \\ \\ \text{Given that } V & = 60 \phantom{.} 000e^{-kt} \\ & = 60 \phantom{.} 000(e^{-k})^t \\ & = 60 \phantom{.} 000(0.9)^t \text{ (Shown)} \end{align}

(ii) For V = 60 \phantom{.} 000(0.9)^t, as t increases, V decreases

\begin{align} \text{When } & t = 0, \\ V & = 60 \phantom{.} 000(0.9)^0 \\ & = 60 \phantom{.} 000(1) \\ & = 60 \phantom{.} 000 \\ \implies & \text{Vertical intercept is } 60 \phantom{.} 000 \end{align}
(Since time, t, cannot be negative, the graph starts from t = 0)