The purpose of this question is to understand when x values will yield y values. The term inside of a square root can be positive or equal to zero in order to yield a value. This means that x can be equal to all real numbers that are -9 or higher, which shows that the domain of the function is all real numbers that are at least -9.

For the square root of a function to be defined on the real numbers, the radicand must be nonnegative. Therefore, 

, or .

Any nonnegative number can be the radicand, so  has no lower bound. This makes the natural domain 

The key here is to factor the denominator, bearing in mind that once we do, we can find the values for which the denominator will be and therefore the values for which the function will not be valid.

We can thus deduce from those three factors that the function will not be valid when .

The domain of a function is all values that you can put in for x without breaking any rules. When first approaching this problem, you must realize that when dealing with a fraction, the denominator can never be 0. Thus, any x value that makes the denominator 0 must be removed from our domain set. Thus,

Since x=3 will make our denominator 0, it must be removed. All other values are permitted, so our answer is

To find the domain, you must find all the values you can put in for x. Thus, you must figure out what values would "break" your function and give you something unable to be computed. 

First we know that the number inside a square root must be positive. Thus, we can set the inside greater than or equal to 0 and solve.

We also know that the denominator of a fraction must never be 0. Thus, if we find out when it is 0, we can exclude that x value from our domain.

If we combine both of these for x, we can create an interval for our domain.

Consider the function . Then 

.

 is an absolute value of a linear function. Since the value of this function cannot be less than 0, its graph changes direction at the value of  at which . This value can be found by setting 

and solving for :

. Also, , so . Since the absolute value of an expression cannot be negative, 0 is the minimum end of the range of .

Also, if 

, 

then, using the properties of inequality,

Therefore, if , then . A function of the form  is a linear function and is either constantly increasing or constantly decreasing; since  in this case,  increases, and as  goes to infinity, so does . Therefore, the range of  has no upper bound, and the correct choice is .

Since the piecewise-defined function  is defined two different ways, one for nonpositive numbers and one for positive numbers, examine both definitions and determine each partial range separately;  the union of the partial ranges will be the overall range.

If , then 

Since 

,

applying the properties of inequality,

Therefore, on the portion of the domain comprising nonpositive numbers, the partial range of  is the set .

If , then 

Since 

,

applying the properties of inequality,

Therefore, on the portion of the domain comprising positive numbers, the partial range of  is the set .

The overall range is the union of these partial ranges, which is .

A quadratic function has a parabola as its graph; this graph changes direction (downward to upward, or vice versa) at a given point called the vertex.

 exists on a given domain interval if and only if there does not exist  and  on this domain such that , but . This will happen if the graph changes direction on the domain interval. The key is therefore to identify the interval that contains the vertex.

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

Of the five intervals in the choices, 

,

so  cannot exist if the domain of  is restricted to this interval. This is the correct choice.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place. 

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

, so the vertex is on the domain.  The maximum and the minimum of  must occur at the vertex and one endpoint, so evaluate , , and .

The minimum and maximum values of  are  and 12, respectively, so the correct range is .

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place. 

The -coordinate of the vertex of the parabola of the function

is .

The -coordinate of the vertex of the parabola of  can be found by setting :

.

, so the vertex is not within the interval to which the domain is restricted. Therefore,  increases or decreases constantly on , and its maximum and minimum on this interval will be found on the endpoints. These values are  and , which can be evaluated using substitution:

The range is .