begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|l|}
hline
$x$ & $f(x)$ & First Differences & Ratio of First Differences \ hline
$-1.5$ & 3.5 & & \ hline
$-1$ & 0.5 & $0.5-3.5=-3$ & \ hline
$-0.5$ & $-8.5$ & $-8.5-0.5=-9$ & $-9div -3=3$ \ hline
$0$ & $-35.5$ & $-35.5-(-8.5)=-27$ & $-27div -9=3$ \ hline
$1$ & $-116.5$ & $-116.5-(-35.5)=-81$ & $-81div -27=3$ \ hline
end{tabular}
end{table}

$f(x)=-frac{1}{2}left(3^{2(x+2)}right)+5$

(1) Reflect in the $x$-axis $implies y=-3^x$

(2) Horizontally compress by a factor of $dfrac{1}{2}implies y=-3^{2x}$

(3) Vertically compress by a factor of 2 $dfrac{1}{2}implies y=-dfrac{1}{2}left(3^{2x}right)$

(4) Horizontally translate 2 units to the left $y=-dfrac{1}{2}left(3^{2(x+2)}right)$

(5) Vertically translate 5 units upward $y=-dfrac{1}{2}left(3^{2(x+2)}right)+5$

The asymptote is $y=5$

b) If the ratios of the first differences is constant, it is exponential.

c) Perform the following transformations:

(1) Reflect in the $x$-axis

(2) Horizontally compress by a factor of $dfrac{1}{2}$

(3) Vertically compress by a factor of 2

(4) Horizontally translate 2 units to the left

(5) Vertically translate 5 units upward

$=dfrac{1}{-5^3}$

$$
=-dfrac{1}{125}
$$

$x^{-a}=dfrac{1}{x^a}$

$(-x)^a = x^a$ if $a$ is even

$(-x)^a=-x^a$ if $a$ is odd

$$
=3^2=9
$$

b) $9$

$(x^a)^b=x^{ab}$

$x^acdot x^b=x^{a+b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$(x^a)^b=x^{ab}$

$x^acdot x^b=x^{a+b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$x^{-a}=dfrac{1}{x^a}$

$(x^a)^b=x^{ab}$

$x^acdot x^b=x^{a+b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$(x^a)^b=x^{ab}$

$x^acdot x^b=x^{a+b}$

$dfrac{x^a}{x^b}=x^{a-b}$

$$
x^{-a}=dfrac{1}{x^a}
$$

b) $dfrac{1}{3125a^3b}$

c) $2x$

d) $dfrac{1}{4xy^4}$

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$

$bold{a);;}$ The decay rate is $3.6%$ so $r=-0.036$ and the initial value is 100. If $I$ is the intensity and $n$ is the number of gels added, then the model would be

$I=100[1+(-0.036)]^nimplies boxed{I=100(0.964)^n}$

$bold{b);;}$ We shall find the value of $I$ when $n=3$

$I=100(0.964)^3=boxed{89.6%}$

$bold{c);;}$ It is exponential because the independent variable is in the exponent. It is an example of decay because the function decreases as $n$ increases.

b) $89.6%$

c) The independent variable is in the exponent and the function is decreasing as $n$ increases.

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$

$bold{a);;}$ The initial population $(at 1990)$ is $P_0=2times 10^6$ and the growth rate is $r=4%=0.04$. Using the exponential model, the equation is

$$
begin{equation*} P=P_0(1+r)^nend{equation*}
$$

$$
begin{equation*}P=(2times 10^6)(1.04)^nend{equation*}
$$

$4times 10^6=2times10^6(1.04)^n$

$2=1.04^n$

By graphical approach, we get $n=17.67approx 18$

Therefore, the population will have doubled 18 years after 1990 which is year 2008.

b) 18 years after 1990 or in the year 2008

From the given graph, it is sharply decreasing as $x$ values increases. This indicates that the base function was reflected in the $x$-axis, that is $f(x)to -f(x)$. From this information, we can eliminate options (a) and (b) because both are reflected in the $y$-axis. We can also observe that it has asymptote at $y=-2$. Option (c) would have an asymptote at $y=0$ so the only possible answer is option (d).