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.

If we graphed the given complex number on a set of real-imaginary axes, we would plot the real value of the complex number as the x coordinate, and the imaginary value of the complex number as the y coordinate. Because the given complex number is as follows:

We are essentially doing the same as plotting the point on a set of Cartesian axes. We move units right in the x direction, and units down in the y direction, which puts us in the fourth quadrant, or in terms of Roman numerals:

I) The orientation of the parabola is determined solely by the sign of . Since , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is ; since you are not given , you cannot find this. Also, since the line of symmetry has equation , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.

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 -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.

I) The orientation of the parabola is determined solely by the sign of . Since , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is ; since you are not given , you cannot find this. Also, since the line of symmetry has equation , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

IV) The -intercept(s), if any, are the point(s) at which . This is solvable using the quadratic formula

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.

The square root of a real number is defined only for nonnegative radicands; therefore, the domain of is exactly those values for which the radicand is nonnegative. Solve the inequality:

The domain of is .