The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle x+ 8 = 0\)

\(\displaystyle x+ 8 - 8= 0 - 8\)

\(\displaystyle x = -8\)

The graph of the line \(\displaystyle x = -8\) is the only vertical asymptote of the graph of \(\displaystyle f(x)\).

The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator.

Set the denominator equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle x - 2 = 0\)

\(\displaystyle x- 2 + 2 = 0 + 2\)

\(\displaystyle x = 2\)

The only vertical asymptote of the graph is the line of the equation \(\displaystyle x = 2\).

The \(\displaystyle y\)-intercept of the graph of a function is the point at which it intersects the \(\displaystyle y\)-axis. The \(\displaystyle x\)-coordinate is 0, so the \(\displaystyle y\)-coordinate can be found by evaluating \(\displaystyle f(0)\). This is done by substitution, as follows:

\(\displaystyle f(x) = \frac{x^{2}- 5x}{x}\)

\(\displaystyle f(0) = \frac{0^{2}- 5 \cdot 0}{ 0}\)

The value of the denominator here is 0, so \(\displaystyle f(0)\) is an undefined quantity. Consequently, the graph of \(\displaystyle f(x)\) has no \(\displaystyle y\)-intercept.

The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. 

First, factor the numerator and denominator.

The numerator is a perfect square trinomial and can be factored as such:

\(\displaystyle x^{2}+12x+36 = x^{2}+2 \cdot x \cdot 6 + 6^{2} = (x+6)^{2} = (x+6)(x+6)\)

The denominator can be factored as the difference of squares:

\(\displaystyle x^{2}-36 = x^{2}-6 ^{2} = (x+6) (x-6)\)

Rewrite

\(\displaystyle f(x) = \frac{x^{2}+12x+36}{x^{2}-36}\)

as

\(\displaystyle f(x) = \frac{ (x+6)(x+6)}{(x+6) (x-6)}\)

The expression can be reduced by cancelling \(\displaystyle x+6\) in both halves:

\(\displaystyle f(x) = \frac{ x+6}{x-6}\)

Set the denominator equal to 0 and solve:

\(\displaystyle x-6 = 0\)

\(\displaystyle x-6+ 6 = 0 + 6\)

\(\displaystyle x = 6\)

The only vertical asymptote is therefore the line of the equation \(\displaystyle x = 6\).

The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. 

First, factor the numerator. It is a quadratic trinomial with lead term \(\displaystyle x^{2}\), so look to "reverse-FOIL" it as

\(\displaystyle x^{2}-7x - 18 = (x- a)(x-b)\)

We seek two integers whose sum is \(\displaystyle -7\) and whose product is \(\displaystyle -18\); through trial and error, we find \(\displaystyle -9\) and 2, so

\(\displaystyle x^{2}-7x - 18 = (x- 9)(x+2)\)

Therefore, \(\displaystyle f(x)\) can be rewritten as 

\(\displaystyle f(x) = \frac{ (x- 9)(x+2)}{x-9}\)

Cancelling \(\displaystyle x -9\), this can be seen to be essentially a polynomial function:

\(\displaystyle f(x) = x+2\)

which does not have a vertical asymptote.

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:

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

\(\displaystyle \frac{x^{2}- 5x}{x} = 0\)

A fraction is equal to 0 if and only if the numerator is equal to 0, so set 

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

Factor out \(\displaystyle x\):

\(\displaystyle x ( x - 5 ) = 0\)

By the Zero Product Property, one of the factors must be equal to 0, so either

\(\displaystyle x = 0\)

or 

\(\displaystyle x-5= 0\)

in which case

\(\displaystyle x= 5\). 

However, setting \(\displaystyle x = 0\) in the definition, we see that

\(\displaystyle f(0) = \frac{0^{2}- 5 \cdot 0}{0}\), an undefined expression due to the zero denominator. 5 cannot be eliminated similarly. Therefore, \(\displaystyle (5,0)\) is the only \(\displaystyle x\)-intercept.

The inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) can be found as follows:

Replace \(\displaystyle f(x)\) with \(\displaystyle y\):

\(\displaystyle f(x) = \sqrt{4x- 9}\)

\(\displaystyle y= \sqrt{4x- 9}\)

Switch the positions of \(\displaystyle y\) and \(\displaystyle x\):

\(\displaystyle x= \sqrt{4y- 9}\)

or

\(\displaystyle \sqrt{4y- 9} = x\)

Solve for \(\displaystyle y\). This can be done as follows:

Square both sides:

\(\displaystyle \left (\sqrt{4y- 9} \right ) ^{2}= x ^{2}\)

\(\displaystyle 4y- 9 = x ^{2}\)

Add 9 to both sides:

\(\displaystyle 4y- 9+9 = x ^{2}+9\)

\(\displaystyle 4y = x ^{2}+9\)

Multiply both sides by \(\displaystyle \frac{1}{4}\), distributing on the right:

\(\displaystyle \frac{1}{4} \cdot 4y = \frac{1}{4} \cdot \left (x ^{2}+9 \right )\)

\(\displaystyle y = \frac{1}{4} \cdot x ^{2}+ \frac{1}{4} \cdot 9\)

\(\displaystyle y = \frac{1}{4} x ^{2}+ \frac{9}{4}\)

Replace \(\displaystyle y\) with \(\displaystyle f^{-1}(x)\):

\(\displaystyle f^{-1}(x)= \frac{1}{4} x ^{2}+ \frac{9}{4}\)

The statement is false. Look for the point on the graph of \(\displaystyle f(x)\) with \(\displaystyle x\)-coordinate \(\displaystyle \frac{1}{2}\) by going right \(\displaystyle \frac{1}{2}\) unit, then moving up and noting the \(\displaystyle y\)-value, as follows:

\(\displaystyle f \left ( \frac{1}{2} \right ) = \frac{3}{2}\), so the statement is false.

The \(\displaystyle y\)-intercept of the graph of a function is the point at which it intersects the \(\displaystyle y\)-axis (the vertical axis). That point is marked on the diagram below:

The point is about one and three-fourths units above the origin, making the coordinates of the \(\displaystyle y\)-intercept \(\displaystyle \left ( 0, 1\frac{3}{4}\right )\).

This is the composition of two functions. By definition, \(\displaystyle (g \circ f )(x) = g \left [ f(x)\right ]\).  To find the range of \(\displaystyle (g \circ f )(x)\), we need to find the values of this function for each value in the domain of \(\displaystyle f(x)\). Since \(\displaystyle (g \circ f )(x) = g \left [ f(x)\right ]\), this is equivalent to evaluating \(\displaystyle g(x)\) for each value in the range of \(\displaystyle f(x)\), as follows:

\(\displaystyle g(x) = 3x- 5\)

Range value: 3 

\(\displaystyle g(3) = 3 (3)- 5 = 9 - 5 = 4\)

Range value: 5

\(\displaystyle g(5) = 3 (5)- 5 = 15 - 5 = 10\)

Range value: 8

\(\displaystyle g(8) = 3 (8)- 5 = 24 - 5 = 19\)

Range value: 13

\(\displaystyle g(13) = 3 (13)- 5 = 39 - 5 = 34\)

Range value: 21

\(\displaystyle g(21) = 3 (21)- 5 = 63 - 5 = 58\)

The range of \(\displaystyle g(x)\) on the set of range values of \(\displaystyle f(x)\) - and consequently, the range of \(\displaystyle \left (g \circ f \right ) (x)\) - is the set \(\displaystyle \left \{4 , 10, 19, 34, 58\right \}\). Of  the five choices, only 45 does not appear in this set; this is the correct choice.