begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Given that $f(x)=3^x$

a) $g(x)=3^x+3=f(x)+3implies$ vertical translation 3 units upward

b) $g(x)=3^{x+3}=f(x+3)implies$ horizontal translation 3 units to the left

c) $g(x)=dfrac{1}{3}(3^x)=dfrac{1}{3}f(x)implies$ vertical compression by factor of $3$ ($y$-values decreased by $dfrac{1}{3}$

d) $g(x)=3^{frac{x}{3}}=fleft(dfrac{x}{3}right)implies$ horizontal stretching by a factor of $3$

a) vertical translation 3 units upward

b) horizontal translation 3 units to the left

c) vertical compression by a factor of 3

d) horizontal stretching by a factor of $3$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

a) Base function: $4^x$

(1) horizontal translation to the left $implies 4^{x+1}$

(2) vertical stretching by a factor of 3 $implies 3(4^{x+1})$

(3) reflecting in the $x$-axis $implies -3(4^{x+1})$

b) Base function: $left(dfrac{1}{2}right)^x$

(1) horizontal compression by a factor of $dfrac{1}{2}implies left(dfrac{1}{2}right)^{2x}$

(2) vertical stretching by a factor of $2implies 2left(dfrac{1}{2}right)^{2x}$

(3) vertical translation 3 units upward $implies 2left(dfrac{1}{2}right)^{2x}+3$

c) Base function: $(0.5)^x$

(1) horizontal translation 4 units to the right $implies 0.5^{x-4}$

(2) vertical stretching by a factor of $7implies 7(0.5^{x-4})$

(3) vertical translation 1 unit downwards $implies 7(0.5^{x-4})-1$

d) $h(x)=5^{3x-6}=5^{3(x-2)}$

Base function: $5^x$

(1) horizontal compression by a factor of $dfrac{1}{3}implies 5^{3x}$

(2) horizontal translation 2 units to the right $implies5^{3(x-2)}$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
a) & begin{tabular}[c]{@{}l@{}}Base function: $4^x$\ (1) horizontal translation to the left $implies 4^{x+1}$\ (2) vertical stretching by a factor of 3 $implies 3(4^{x+1})$\ (3) reflecting in the $x$-axis $implies -3(4^{x+1})$end{tabular} \ hline
b) & begin{tabular}[c]{@{}l@{}}Base function: $left(frac{1}{2}right)^x$\ (1) horizontal compression by a factor of $frac{1}{2}implies left(frac{1}{2}right)^{2x}$\ (2) vertical stretching by a factor of $2implies 2left(frac{1}{2}right)^{2x}$\ (3) vertical translation 3 units upward $implies 2left(frac{1}{2}right)^{2x}+3$end{tabular} \ hline
c) & begin{tabular}[c]{@{}l@{}}Base function: $(0.5)^x$\ (1) horizontal translation 4 units to the right $implies 0.5^{x-4}$\ (2) vertical stretching by a factor of $7implies 7(0.5^{x-4})$\ (3) vertical translation 1 unit downwards $implies 7(0.5^{x-4})-1$end{tabular} \ hline
d) & begin{tabular}[c]{@{}l@{}}$h(x)=5^{3x-6}=5^{3(x-2)}$\ Base function: $5^x$\ (1) horizontal compression by a factor of $dfrac{1}{3}implies 5^{3x}$\ (2) horizontal translation 2 units to the right $implies5^{3(x-2)}$end{tabular} \ hline
end{tabular}
end{table}

For a general exponential function $g(x)=acdot b^{k(x-d)}+c$

$y$-intercept: value of $y$ when $x=0$

asymptote: $y=c$

domain = ${ xinbold{R}}$

$$
{text{range = }}left{ {begin{array}{c}
{left{ {y in {mathbf{R}};|;y > c{text{ }}} right}{text{ if }}a > 0,,,} \
{left{ {y in {mathbf{R}};|;y < c{text{ }}} right}{text{ if }}a < 0}
end{array}} right.
$$

begin{table}[]
defarraystretch{2.2}%
begin{tabular}{|l|l|l|l|l|}
hline
Function & $y$-intercept & Asymptote & Domain & Range \ hline
$y=3^x+3$ & $4$ & $y=3$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>3}$ \ hline
$y=3^{x+3}$ & $27$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
$y=dfrac{1}{3}(3^x)$ & $dfrac{1}{3}$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
$y=3^{frac{x}{3}}$ & $1$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
$y=-3(4^{x+1})$ & $-12$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y3}$ \ hline
$y=7(0.5^{x-4})-1$ & 111 & $y=-1$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>-1}$ \ hline
$y=5^{3x-6}$ & $6.4times10^{-5}$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
end{tabular}
end{table}

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|l|l|}
hline
Function & $y$-intercept & Asymptote & Domain & Range \ hline
$y=3^x+3$ & $4$ & $y=3$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>3}$ \ hline
$y=3^{x+3}$ & $27$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
$y=frac{1}{3}(3^x)$ & $frac{1}{3}$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
$y=3^{frac{x}{3}}$ & $1$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
$y=-3(4^{x+1})$ & $-12$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y3}$ \ hline
$y=7(0.5^{x-4})-1$ & 111 & $y=-1$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>-1}$ \ hline
$y=5^{3x-6}$ & $6.4times10^{-5}$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>0}$ \ hline
end{tabular}
end{table}

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

The base function is $h(x)=left(dfrac{1}{2}right)^x$

a) Horizontal compression by a factor of $dfrac{1}{|k|} = frac 1 2$ and then reflecting in the $x$-axis and reflection about $y$ axis.

b) Horizontal translation 3 units to the right, reflecting in the $y$-axis and then vertical stretching by a factor of 5.

c) $g(x)=-4left(dfrac{1}{2}right)^{3x+9}-6=-4left(dfrac{1}{2}right)^{3(x+3)}-6$

Horizontal translation 3 units to the left, horizontal compression by a factor of $dfrac{1}{3}$, vertical stretching by a factor of 4, reflecting in the $x$-axis, and vertical translation 6 units downward.

a) Horizontal compression by a factor of $dfrac{1}{|k|}$ and then reflecting in the $x$-axis and reflection about $y$ axis.

b) Horizontal translation 3 units to the right, reflecting in the $y$-axis and then vertical stretching by a factor of 5.

c) Horizontal translation 3 units to the left, horizontal compression by a factor of $dfrac{1}{3}$, vertical stretching by a factor of 4, reflecting in the $x$-axis, and vertical translation 6 units downward.

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}
For a general exponential function $g(x)=acdot b^{k(x-d)}+c$\\
$y$-intercept: value of $y$ when $x=0$ \\
asymptote: $y=c$\\
domain = ${ xinbold{R}}$\\
${text{range = }}left{ {begin{array}{*{20}{c}}
{left{ {y in {mathbf{R}};|;y > c{text{ }}} right}{text{ if }}a > 0,,,} \
{left{ {y in {mathbf{R}};|;y < c{text{ }}} right}{text{ if }}a < 0}
end{array}} right.$

In the following, the base function is $f(x)=4^x$

a) reflecting the function in the $y$-axis, vertical stretching by a factor of $0.5$ and vertical translation 2 units upward.

$g(x)=0.5(4^{-x})+2$

$y$-intercept: $y=0.5(4^{0})+2=0.5(1)+2=2.5$

asymptote: $y=2$

domain: ${ xinbold{R}}$

range: ${ yinbold{R};|;y>2}$

b) $h(x)=-f(0.25x+1)-1=-f(0.25(x+4))-1$

Horizontal stretching by a factor of $dfrac{1}{0.25}=4$, horizontal translation 4 units to the left, reflecting in the $x$-axis, and vertical translation 1 unit downward.

$h(x)=-(4^{0.25x+1})-1$

$y$-intercept: $y=-4^{0+1}-1=-5$

asymptote: $y=-1$

domain = ${ xinbold{R}}$

range = ${ yinbold{R};|;y<-1}$

c) $g(x)=-2f(2x-6)=-2f(2(x-3))$

Horizontal compression by a factor of $dfrac{1}{2}$, horizontal translation 3 units to the right, vertical stretching by a factor of 2, reflection in the $x$-axis.

$g(x)=-2(4^{2x-6})$

$y$-axis: $y=-2(4^{2(0)-6})=-2(4^{-6})=dfrac{-2}{4^6}=-dfrac{1}{2048}$

asymptote: $y=0$

domain: ${ xinbold{R}}$

range: ${ yinbold{R};|;y<0}$

d) $h(x)=f(-0.5x+1)=f(-0.5(x-1))$

Horizontal stretching by a factor of $dfrac{1}{0.5}=2$, reflecting in the $y$-axis, horizontal translation 1 unit to the right.

$h(x)=4^{-0.5x+1}$

$y$-intercept: $y=4^{-0.5(0)+1}=4$

asymptote: $y=0$

domain: ${ xinbold{R};}$

range: ${ yinbold{R};|;y>0}$

begin{table}[]
defarraystretch{1.7}%
begin{tabular}{|l|l|l|l|l|}
hline
Function & $y$-intercept & Asymptote & Domain & Range \ hline
a) $g(x)=0.5(4^{-x})+2$ & $2.5$ & $y=2$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y>2}$ \ hline
b) $h(x)=-4^{0.25x+1}-1$ & $-5$ & $y=-1$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y<-1}$ \ hline
c) $g(x)=-2(4^{2x-6})$ & $-frac{1}{2048}$ & $y=0$ & ${ xinbold{R}}$ & ${ yinbold{R};|;y0}$ \ hline
end{tabular}
end{table}

$f(x)=6^x$

$g(x)=3^{2x}=9^x$

Both functions have the same $y$-intercept $(0,1)$, asymptote $y=0$, domain ${xinbold{R}}$ and range ${ yinbold{R};|;y>0}$

Since $g(x)$ has greater base, it increases faster than $f(x)$.

Both have the same $y$-intercept, asymptote, domain and range, but $g(x)$ increases faster than $f(x)$.

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

For a general exponential function $g(x)=acdot b^{k(x-d)}+c$\\
$y$-intercept: value of $y$ when $x=0$ \\
asymptote: $y=c$\\
domain = ${ xinbold{R}}$\\
${text{range = }}left{ {begin{array}{*{20}{c}}
{left{ {y in {mathbf{R}};|;y > c{text{ }}} right}{text{ if }}a > 0} \
{left{ {y in {mathbf{R}};|;y < c{text{ }}} right}{text{ if }}a < 0}
end{array}} right.$

Choose the base function as $f(t)=left(dfrac{1}{2}right)^{t}$

horizontal compression by a factor of $30implies left(dfrac{1}{2}right)^{frac{t}{30}}$

vertical stretching by a factor of $80implies 80left(dfrac{1}{2}right)^{frac{t}{30}}$

vertical translation 20 units upward $implies T(t)=80left(dfrac{1}{2}right)^{frac{t}{30}}+20$

$y$-intercept is the value of $T$ when $t=0$ which corresponds to the initial temperature.

The asymptote $y=20$ is the lower limit of the liquid temperature, in other words, the temperature of the liquid would approach $20^circ$ as time increases.

The $y$-intercept is the liquid temperature when $t=0$ (initial temperature). The asymptote $y=20$ is the lower limit of the liquid temperature.

The number of yeast cell doubles every 3 hours and we are given the number cells $N$ after $t$ hours as

$N(t)=N_02^{frac{t}{3}}$

where $N_0$ is the population at $t=0$

a) If doubling time is changed to $9$ h, the new function would be $F(t)=N_02^{frac{t}{9}}$. Notice that the exponent is smaller, so the graph would rise slower than the original function. This also corresponds to horizontal stretching by a factor of 3.

b) The domain is ${ tinbold{R};|;tgeq0}$. The domain is the period of time at which the yeast cells are allowed to grow which can’t be negative.

The range is ${ Ninbold{R};|;N>N_0}$ because $N(t)$ is an increasing function so the population cannot be lower than the initial population $N_0$.

a) The graph would rise slower

b) domain = ${ tinbold{R};|;tgeq 0}$

range = ${ Ninbold{R};|;N>N_0}$

i.e. remains same

To match the equations to their graphs, identify the function transformations involved.

a) $f(x)=-left(dfrac{1}{4}right)^{-x}+3=-(4)^x+3$

The graph of $y=4^x$ lies above $x$-axis and increases from left to right.

$y=-4^x$ is reflected in the $x$-axis, so the function is now below $x$-axis and decreases from left to right.

$y=-4^x+3$ is a vertical translation 3 units upward so the $y$-intercept must be $-(4^0)+3=-1+3=2$

The graph that satisfies this is option (iii)

b) $y=left(dfrac{1}{4}right)^x+3=4^{-x}+3$

The graph of $y=4^x$ lies above $x$-axis and increases from left to right.

The graph of $y=4^{-x}$ is reflected in the $y$-axis so the graph is still above $x$-axis but is now decreasing from the left to right.

The graph of $y=4^{-x}+3$ is vertically translated 3 units upward so the $y$-intercept must be $4^{-0}+3=4$

The graph that satisfies this is option (ii)

c) $g(x)=-left(dfrac{5}{4}right)^x+3$

The graph of $y=left(frac{5}{4}right)^x$ lies above $x$-axis and increasing from left to right.

The graph of $y=left(frac{5}{4}right)^{-x}$ is reflected in the $y$-axis so it’s still above $x$-axis but decreasing from left to right.

The graph of $y=-left(frac{5}{4}right)^{-x}$ is reflected in the $x$-axis so now it is below $x$-axis and increasing from left to right.

The graph is then translated 3 units upward so the $y$-intercept must be $-left(frac{5}{4}right)^{-0}+3=-1+3=2$

The graph that satisfies this is option (iv)

d) $h(x)=2left(dfrac{5}{4}right)^x+3$ lies above $x$-axis and increases from the left to right with $y$-intercept of $2left(frac{5}{4}right)^0+3=2+3=5$. The graph that satisfies this is option (i)

a) The graph of $f(x)=2^{x}$ is above the $x$-axis and is increasing from left to right. If we reflect it in the $x$-axis, we get $y=-2^x$ which lies below $x$-axis and decreases from left to right (similar behavior of the given graph). Since the given graph has asymptote at $y=6$, the graph must have been translated 6 units upward giving $y=-2^{x}+6$. To check whether there is horizontal translation, observe that the $y$-intercept $y=-2^{0}+6=-1+6=5$ which is the same as in the graph, so there is no horizontal translation.
Now, observe that the graph passes through $(2,-10)$

If we use $x=2$, $y=-2^{x}+6=-2^{2}+6=2neq -10$ which implies that there must be horizontal stretching/compression so it must be of the form $y=-2^{kx}+6$

Substitute the point $(2,-10)$ to find $k$

$-10=-2^{k(2)}+6$

$2^{2k}=16$

$2^{2k}=2^{4}$

$2k=4$

$k=2$

Therefore, the equation is

$$
y=-2^{2x}+6
$$

b) The graph is decreasing from left to right, so $f(x)=2^{x}$ must have been reflected in the $y$-axis $(y=2^{-x})$ and since the asymptote is at $y=-2$, it must have been translated $2$ units downward $(y=2^{-x}-2)$ which would result in a $y$-intercept of $2^0-2=1-2=-1$. However, the $y$-value of $-1$ occurs at $x=-3$ implying that the graph is shifted 3 units to the left. Therefore, the equation must be

$y=2^{-(x+3)}-2$

a) $y=-2^{2x}+6$

b) $y=2^{-(x+3)}-2$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Here are the necessary transformation to convert

$f(x)=2^{x+1}+5$ $implies$ $g(x)=dfrac{1}{4}(2^x)$

(1) Vertical translation 5 units downward

$f(x)-5implies y=2^{x+1}$

(2) Horizontal translation 1 unit to the right

$f(x-1)-5 implies y=2^{x}$

(3) Vertical compression by a factor of $dfrac{1}{4}$

$$
dfrac{1}{4}[f(x-1)-5] implies g(x)=dfrac{1}{4}(2^x)
$$

(1) vertical translation 5 units downward

(2) horizontal translation 1 unit to the right

(3) vertical compression by a factor of $frac{1}{4}$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

1) Use the base function $y=2^x$ then reflect in the $y$-axis
to obtain $y=2^{-x}=dfrac{1}{2^x}$

2) Reflect $y=dfrac{1}{2^x}$ in the $x$-axis to obtain $y=dfrac{-1}{2^x}$

With the base function $y=2^x$, perform the following transformation.

1) Reflect in the $y$-axis to obtain $y=2^{-x}=dfrac{1}{2^x}$

2) Reflect in the $x$-axis to obtain $y=dfrac{-1}{2^x}$

3) Vertically stretch by a factor of 3 to obtain $y=dfrac{-3}{2^x}$

4) Translate 2 units to the left and 1 unit downwards $y=dfrac{-3}{2^{x+2}}-1$

Remember that $dfrac{1}{a}=a^{-1}$ so we can rewrite $g(x)$ as

$g(x)=4-2left( dfrac{1}{3}right)^{-0.5x+1}=4-2(3^{-1})^{-0.5x+1}=4-2(3^{0.5x-1})$

$$
g(x)=4-2left(3^{0.5(x-2)}right)
$$

(1) Start from the function $y=3^{x}$ and reflect in the $x$-axis to obtain $y=-3^{x}$

(2) Horizontally compress by a factor of $dfrac{1}{0.5}=2$ to obtain $y=-3^{0.5x}$

(3) Vertically stretch by a factor of 2 to obtain $y=-2left(3^{0.5x}right)$

(4) Translate 2 units to the right and 4 units upward to obtain
$y=-2(3)^{0.5(x-2)}+4$

*Note that if we used $y=left(dfrac{1}{3}right)^x$ instead of $y=3^x$ as the base function, you need to reflect in the $y$-axis, that is, change $f(x)$ to $f(-x)$.

(1) Start from the function $y=3^{x}$ and reflect in the $x$-axis to obtain $y=-3^{x}$

(2) Horizontally compress by a factor of $dfrac{1}{0.5}=2$ to obtain $y=-3^{0.5x}$

(3) Vertically stretch by a factor of 2 to obtain $y=-2left(3^{0.5x}right)$

(4) Translate 2 units to the right and 4 units upward to obtain
$y=-2(3)^{0.5(x-2)}+4$ which is just the same as $y=-2left(dfrac{1}{3}right)^{-0.5(x-2)}+4$

We need to transform

$m(x)=-left(dfrac{3}{2}right)^{2x-2}$ to $n(x)=-left(dfrac{9}{4}right)^{-x+1}+2$

Note that they have different base function so we need to rewrite $m(x)$ such that it has a base of $dfrac{9}{4}$

$m(x)=-left[left(dfrac{3}{2}right)^2right]^{2x-2}$

$m(x)=-left(dfrac{9}{4}right)^{frac{2x-2}{2}}$

$m(x)=-left(dfrac{9}{4}right)^{x-1}$

Thus, this transformation simplifies as

$$
m(x)=-left(dfrac{9}{4}right)^{x-1} text{ to } n(x)=-left(dfrac{9}{4}right)^{-(x-1)}+2
$$

To transform $m(x)$ to $n(x)$,

(1) Reflect in the $y$-axis to obtain $y=-left(dfrac{9}{4}right)^{-(x-1)}=-left(dfrac{9}{4}right)^{-x+1}$

(2) Vertically translate 2 units upward to obtain $n(x)=-left(dfrac{9}{4}right)^{-x+1}+2$

Note that no vertical or horizontal scaling are needed.

(1) reflect in the $y$-axis

(2) vertically translate 2 units upward