The domain is defined as the set of all possible values of the independent variable . First,  must be greater than or equal to zero, because if , thenwill be undefined. In addition, the total expression under the radical, i.e.  must be greater than or equal to zero:

That means that the expression under the radical is always positive and therefore is defined. Hence, the domain of the function is .

A function's symmetry is related to its classification as even, odd, or neither.

Even functions obey the following rule:

Because of this, even functions are symmetric about the y-axis.

Odd functions obey the following rule:

Because of this, odd functions are symmetric about the origin.

If a function does not obey either rule, it is neither odd nor even. (A graph that is symmetric about the x-axis is not a function, because it does not pass the vertical line test.)

To test for symmetry, simply substitute into the original equation.

Thus, this equation is even and therefore symmetric about the y-axis.

To find the horizontal asymptote, compare the degrees of the top and bottom polynomials. In this case, the two degrees are the same (1), which means that the equation of the horizontal asymptote is equal to the ratio of the leading coefficients (top : bottom). Since the numerator's leading coefficient is 1, and the denominator's leading coefficient is 2, the equation of the horizontal asymptote is .

To find the vertical asymptote, set the denominator equal to zero to find when the entire function is undefined:

This question is asking for the equation's slant asymptote. To find the slant asymptote, divide the numerator by the denominator. Long division gives us the following:

However, because we are considering  as it approaches infinity, the effect that the last term has on the overall linear equation quickly becomes negligible (tends to zero). Thus, it can be ignored.

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression fits the description of the zeroes.

Now we need to check the answer.

We are able to get back to the original expression, meaning that the answer is .

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is , which we can expland using the cube of a binomial pattern.

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is . To put this in expanded form:

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of  such that the output will be zero. Thus, we first factor the equation. 

Then, we are looking for the values where each of these factors are equal to zero. 

 implies 

and  implies 

Thus, these are our solutions. 

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if  is a zero:

Since the function equals zero when  is , one of the factors of the polynomial is . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

Now we can factor the function this way:

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

When we factor using synthetic substitution for , we get the following result:

Using our quadratic factoring rules, we can factor completely:

Thus, the zeroes of  are

To simplify the polynomial, begin by combining like terms: