Since any number, positive or negative, can appear as an exponent, the domain of the function is the set of all real numbers; in other words, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, and for all real values of . The graph will never crosst the line of the equatin , so this is the horizontal asymptote.

The domain of the function specifies the values that can take. Here, is defined for every value of , so the domain is all real numbers.

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

27 is the only number excluded from the domain.

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

Add to both sides:

Multiply both sides by 2:

,

the correct choice.

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

However, does not have a real value. Therefore, the graph of has no -intercept.

The radius of a circle with area can be found as follows:

The circle, the central angle, and the chord are shown below:

By way of the Isosceles Triangle Theorem, can be proved equilateral, so , the correct response.

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

First, factor the denominator. It is a quadratic trinomial with lead term , so look to "reverse-FOIL" it as

by finding two integers with sum 6 and product 5. By trial and error, these integers can be found to be 1 and 5, so

Therefore, can be rewritten as

Set the denominator equal to 0 and solve for :

By the Zero Factor Principle,

or

Therefore, the binomial factor can be cancelled, and the function can be rewritten as

If , then , so the denominator has only this one zero, and the only vertical asymptote is the line of the equation .

The vertical asymptote(s) of the graph of a rational function such as 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 , so look to factor

by using the grouping technique. We try finding two integers whose sum is and whose product is ; with some trial and error we find that these are and , so:

Break the linear term:

Regroup:

Factor the GCF twice:

Therefore, can be rewritten as

Cancel the common factor from both halves; the function can be rewritten as

Set the denominator equal to 0 and solve for :

The graph of therefore has one vertical asymptotes, the line of the equations . The line of the equation is not a vertical asymptote.

The function is defined for those values of for which the radicand is nonnegative - that is, for which

Subtract 25 from both sides:

Since the square root of a real number is always nonnegative,

for all real numbers . Since the radicand is always positive, this makes the domain of the set of all real numbers.