Since the $\displaystyle x$-intercept is the point at which the graph of $\displaystyle f$ intersects the $\displaystyle x$-axis, the $\displaystyle y$-coordinate is 0, and the $\displaystyle x$-coordinate can be found by setting $\displaystyle f(x)$ equal to 0 and solving for $\displaystyle y$. Therefore, we need to find $\displaystyle x$ such that 

$\displaystyle 3 ^{x-3}- 9 = 0$. 

$\displaystyle 3 ^{x-3}= 9$

$\displaystyle 3 ^{x-3}= 3 ^{2}$

$\displaystyle x-3 = 2$

$\displaystyle x = 5$

The $\displaystyle x$-intercept is therefore $\displaystyle (5,0)$.

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, $\displaystyle 2 ^{x-1} > 0$ and $\displaystyle y= 2 ^{x-1} + 7 > 7$ for all real values of $\displaystyle x$. The graph will never crosst the line of the equatin $\displaystyle y = 7$, so this is the horizontal asymptote.

First, we rewrite both functions with a common base:

$\displaystyle f(x) = 2^{x +2}$ is left as it is.

$\displaystyle g(x) = 4^{x-1}$ can be rewritten as 

$\displaystyle g(x) = \left (2 ^{2} \right )^{x-1}$

$\displaystyle g(x) = 2 ^{2 (x-1)}$

$\displaystyle g(x) = 2 ^{2x-2}$

To find the point of intersection of the graphs of the functions, set 

$\displaystyle f(x) = g(x)$

$\displaystyle 2^{x +2}= 2 ^{2x-2}$

The powers are equal and the bases are equal, so we can set the exponents equal to each other and solve:

$\displaystyle x+ 2= 2x-2$

$\displaystyle x+ 2 -x +2= 2x-2 -x +2$

$\displaystyle x = 4$

To find the $\displaystyle y\,$-coordinate, substitute 4 for $\displaystyle x$ in either definition:

$\displaystyle f(x) = 2^{x +2}$

$\displaystyle f(4) = 2^{4+2}= 2^{6} = 64$, the correct response.

The $\displaystyle x$-coordinate ofthe $\displaystyle y$-intercept of the graph of $\displaystyle f$ is 0, and its $\displaystyle y$-coordinate is $\displaystyle f(0)$:

$\displaystyle f(x) = 4 ^{x+2} - 3$

$\displaystyle f(0) = 4 ^{0+2} - 3 = 4 ^{2} - 3 = 16 - 3 = 13$

The $\displaystyle y$-intercept is the point $\displaystyle (0,13)$.

First, we rewrite both functions with a common base:

$\displaystyle f(x) = 3^{x +2}$ is left as it is.

$\displaystyle g(x) =\left ( \frac{1}{9} \right )^{x-1}$ can be rewritten as 

$\displaystyle g(x) =\left ( 3^{-2} \right )^{x-1}$

$\displaystyle g(x) = 3^{-2 (x-1)}$

$\displaystyle g(x) = 3^{-2 x+2}$

To find the point of intersection of the graphs of the functions, set 

$\displaystyle f(x) = g(x)$

$\displaystyle 3^{x +2} = 3^{-2 x+2}$

Since the powers of the same base are equal, we can set the exponents equal:

$\displaystyle x+2 = -2x + 2$

$\displaystyle x+2 + 2x -2 = -2x + 2 + 2x -2$

$\displaystyle 3x=0$

$\displaystyle x= 0$

Now substitute in either function:

$\displaystyle f(x) = 3^{x +2}$

$\displaystyle f(0) = 3^{0 +2} = 3^{ 2} = 9$, the correct answer.

Since the $\displaystyle y$-intercept is the point at which the graph of $\displaystyle f$ intersects the $\displaystyle y$-axis, the $\displaystyle x$-coordinate is 0, and the $\displaystyle y$-coordinate is $\displaystyle f(0)$:

$\displaystyle f(x) =5^{x-3}$

$\displaystyle f(0) =5^{0-3}= 5^{-3} = \frac{1}{5^{3}} = \frac{1}{125}$,

The  $\displaystyle y$-intercept is the point $\displaystyle \left ( 0, \frac{1}{125} \right )$.

Rewrite the system as 

$\displaystyle 2 \cdot 2 ^{x} + 3^{y} = 59$

$\displaystyle 2 ^{x } - 3^{y} = -11$

and substitute $\displaystyle u$ and $\displaystyle v$ for $\displaystyle 2 ^{x }$ and $\displaystyle 3^{y}$, respectively, to form the system

$\displaystyle 2u + v= 59$

$\displaystyle u- v= -11$

Add both sides:

$\displaystyle 2u + v= 5$

$\displaystyle \underline{u- v= 43}$

$\displaystyle 3u$        $\displaystyle =48$

$\displaystyle u = 16$.

Now backsolve:

$\displaystyle 16- v= -11$

$\displaystyle - v= - 27$

$\displaystyle v= 27$

Now substitute back:

$\displaystyle u = 16$

$\displaystyle 2 ^{x } = 16$

$\displaystyle x = 4$

and

$\displaystyle v= 27$

$\displaystyle 3^{y}= 27$

$\displaystyle y=3$

$\displaystyle x+y =4+3 = 7$

The $\displaystyle x$-intercept is $\displaystyle (a,0)$, where $\displaystyle f(a)= 0$:

$\displaystyle f(x) = 5 \cdot 4^{x- 3}- 3$

$\displaystyle 5 \cdot 4^{a- 3}- 3 = 0$

$\displaystyle 5 \cdot 4^{a- 3}- 3 + 3= 0 + 3$

$\displaystyle 5 \cdot 4^{a- 3}= 3$

$\displaystyle 5 \cdot 4^{a- 3} \div 5= 3 \div 5$

$\displaystyle 4^{a- 3} = 0.6$

$\displaystyle \ln \left (4^{a- 3} \right )=\ln 0.6$

$\displaystyle (a- 3)\ln 4 =\ln 0.6$

$\displaystyle a- 3=\frac{\ln 0.6}{\ln 4 } \approx \frac{-0.5108}{1.3863}\approx -0.3685$

$\displaystyle a- 3+3 \approx -0.3685 + 3$

$\displaystyle a \approx 2.63$

The $\displaystyle x$-intercept is $\displaystyle (2.63,0)$.

The $\displaystyle y$-intercept is $\displaystyle (0, f(0))$:

$\displaystyle f(x) = 5 \cdot 4^{x- 3}- 3$

$\displaystyle f(0) = 5 \cdot 4^{0- 3}- 3$

$\displaystyle f(0) = 5 \cdot 4^{ - 3}- 3$

$\displaystyle f(0) = 5 \cdot \frac{1}{64}- 3$

$\displaystyle f(0) = \frac{5}{64}- 3$

$\displaystyle f(0) \approx 0.08 - 3 \approx -2.92$

$\displaystyle (0, -2.92)$ is the $\displaystyle y$-intercept.

Since 4 can be raised to the power of any real number, the domain of $\displaystyle f$ is the set of all real numbers. Therefore, there is no vertical asymptote of the graph of $\displaystyle f$.