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:

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

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

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

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

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

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

\(\displaystyle The\ graph\ of\ the\ function\ y=3^{x}-2\ has\ an\ asymptote\ at\ y=-2.\)

\(\displaystyle An\ asymptote\ on\ a\ graph\ is\ a\ value\ that\ a \ function\ approaches\ but\ never\ touches.\)

\(\displaystyle Since\ this\ graph\ approaches\ but\ never\ touches\ -2\ we\ use\ the >\ symbol.\)

\(\displaystyle \\Also\ because\ the\ 3\ is\ positive\ that\ means\ the\ graph\ is\ concave\ up\ so\ >\ is\ chosen\ instead\ of\ < .\)

\(\displaystyle \\The\ concavity\ of\ an\ exponential\ graph\ can\ be\ determined\ by\ its\ a-value.\)

\(\displaystyle This\ function\ is\ in\ the\ form\ y=a(b)^{x}+k.\)

\(\displaystyle \ So\ the\ a\ value\ is\ 2, based\ on\ the\ function\ we\ were\ given\ y=2(3)^{x}-5\)

\(\displaystyle If\ the\ a\ value\ is\ positive,\ the\ graph\ will\ be\ concave\ up\).

\(\displaystyle If\ the\ a\ value\ is\ negative,\ the\ graph\ will\ be\ concave\ down.\)

\(\displaystyle To\ find\ the\ y-intercept,\ plug\ in\ x=0, and\ keep\ in\ mind\ order\ of\ operations,\)

\(\displaystyle \ Exponent first, then\ multiply.\)

\(\displaystyle y=2(3)^{0}-5\)

\(\displaystyle y=2(1)-5=-3\)

\(\displaystyle The\ y-intercept\ is\ (0,-3).\)

Define \(\displaystyle g(x) = e^{x}\). In terms of \(\displaystyle g(x)\), \(\displaystyle f(x)\) can be restated as

\(\displaystyle f(x) = 2g( x-3 )+ 7\)

The graph of \(\displaystyle f(x)\) is a transformation of that of \(\displaystyle g(x)\). As an exponential function, \(\displaystyle g(x)\) has a graph that has no vertical asymptote, as \(\displaystyle g(x)\) is defined for all real values of \(\displaystyle x\); it follows that being a transformation of this function, \(\displaystyle f(x)\) also has a graph with no vertical asymptote as well.

Define \(\displaystyle g(x) = e ^{x}\). In terms of \(\displaystyle g(x)\), \(\displaystyle f(x)\) can be restated as

\(\displaystyle f(x) = 2e^{x-3}- 7\).

The graph of \(\displaystyle g(x)\) has as its horizontal asymptote the line of the equation \(\displaystyle y = 0\). The graph of \(\displaystyle f(x)\) is a transformation of that of \(\displaystyle g(x)\) - a right shift of 3 units ( \(\displaystyle x-3\) ), a vertical stretch ( \(\displaystyle 2g\) ), and a downward shift of 7 units ( \(\displaystyle -7\) ). The right shift and the vertical stretch do not affect the position of the horizontal asymptote, but the downward shift moves the asymptote to the line of the equation \(\displaystyle y = -7\). This is the correct response.

The \(\displaystyle y\)-intercept of the graph of \(\displaystyle f(x)\) is the point at which it intersects the \(\displaystyle y\)-axis. Its \(\displaystyle x\)-coordinate is 0; its \(\displaystyle y\)-coordinate is \(\displaystyle f(0)\), which can be found by substituting 0 for \(\displaystyle x\) in the definition:

\(\displaystyle f(x) = 2 \cdot 10 ^{x} - 8\)

\(\displaystyle f(0) = 2 \cdot 10 ^{0} - 8 = 2 \cdot 1 - 8 = 2 - 8 = -6\)

The \(\displaystyle y\)-intercept of the graph of \(\displaystyle f(x)\) is the point at which it intersects the \(\displaystyle y\)-axis. Its \(\displaystyle x\)-coordinate is 0; its \(\displaystyle y\)-coordinate is \(\displaystyle f(0)\), which can be found by substituting 0 for \(\displaystyle x\) in the definition:

\(\displaystyle f(x) = 2e^{x}+ 7\)

\(\displaystyle f(0) = 2e^{0}+ 7 = 2 \cdot 1 + 7 = 2+7 = 9\),

the correct choice.

The \(\displaystyle x\)-intercept(s) of the graph of \(\displaystyle f(x)\) are the point(s) at which it intersects the \(\displaystyle x\)-axis. The \(\displaystyle y\)-coordinate of each is 0,; their \(\displaystyle x\)-coordinate(s) are those value(s) of \(\displaystyle x\) for which \(\displaystyle f(x) = 0\), so set up, and solve for \(\displaystyle x\), the equation:

\(\displaystyle f(x)= 0\)

\(\displaystyle 2e^{x}+ 7 = 0\)

Subtract 7 from both sides:

\(\displaystyle 2e^{x}+ 7 - 7 = 0 - 7\)

\(\displaystyle 2e^{x}= - 7\)

Divide both sides by 2:

\(\displaystyle \frac{2e^{x}}{2}=\frac{ - 7}{2}\)

\(\displaystyle e^{x} =-\frac{ 7}{2}\)

The next step would normally be to take the natural logarithm of both sides in order to eliminate the exponent. However, the negative number \(\displaystyle -\frac{ 7}{2}\) does not have a natural logarithm. Therefore, this equation has no solution, and the graph of \(\displaystyle f(x)\) has no \(\displaystyle x\)-intercept.