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 \(\displaystyle f\).

We can rewrite this as follows:

\(\displaystyle g(x) = 9 \cdot 3^{x}- 3\)

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

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

This is a translation of the graph of \(\displaystyle f(x) = 3^{x}\), which has \(\displaystyle y = 0\) as its horizontal asymptote, to the right two units and down three units. Because of the latter translation, the horizontal asymptote is \(\displaystyle y = -3\).

We can rewrite the statements using \(\displaystyle y\) for both \(\displaystyle f(x)\) and \(\displaystyle g(x)\) as follows:

\(\displaystyle y = e^{x}+ 9\)

\(\displaystyle y = 4e^{x}- 7\)

To solve this, we can multiply the first equation by \(\displaystyle -4\), then add:

\(\displaystyle -4y = -4e^{x}-36\)

      \(\displaystyle \underline{y = 4e^{x} \; \; - 7}\)

\(\displaystyle -3y =\)            \(\displaystyle -43\)

\(\displaystyle -3y \div (-3) = -43 \div (-3 )\)

\(\displaystyle y \approx 14.3\)

We can rewrite the statements using \(\displaystyle y\) for both \(\displaystyle f(x)\) and \(\displaystyle g(x)\) as follows:

\(\displaystyle y = e^{x}+ 9\)

\(\displaystyle y = 4e^{x}- 7\)

To solve this, we can set the expressions equal, as follows:

\(\displaystyle 4e^{x}- 7 = e^{x}+ 9\)

\(\displaystyle 4e^{x}- 7 - e^{x} + 7 = e^{x}+ 9 - e^{x} + 7\)

\(\displaystyle 3e^{x} = 16\)

\(\displaystyle 3e^{x}\div 3 = 16 \div 3\)

\(\displaystyle e^{x} = \approc 5.3333\)

\(\displaystyle x \approx \ln 5.333 \approx 1.7\)