By factoring the numerator, we see that this equation is equivalent to

.

That means that we can simplify this equation to .

That means that isn't the slant asymptote, but the equation itself. 

is definitely an asymptote, but a vertical asymptote, not a slant asymptote. 

Be careful not to confuse this equation with the linear slope-intercept form. The y-intercept of an equation is the y-value when the x-value is zero.

Substitute the value of  into the equation.

Simplify the equation.

The y-intercept is:  

To find the horizontal asymptote, take the leading term of the numerator and the denominator and divide. In this case:

The function

is already in simplified form.

To find the vertical asymptotes, we set the denominator equal to  and solve for .

yields the vertical asymptotes

To find the horizontal asymptote, we examine the largest degree of  between the numerator and denominator

Note that

Because the largest degree of  in the numerator is less than the largest degree of  in the denominator, or

we find the horizontal symptote to be 

Factorize both the numerator and denominator.

Notice that one of the binomials will cancel.

The domain of this equation cannot include .

The simplified equation is:

Since the  term canceled, the  term will have a hole instead of an asymptote.   

Set the denominator equal to zero.

Subtract one from both sides.

There will be an asymptote at only:  

The answer is:  

Factor the numerator and denominator.

Rewrite the equation.

Notice that the  will cancel.  This means that the root of  will be a hole instead of an asymptote.

Set the denominator equal to zero and solve for x.

An asymptote is located at:  

The answer is:  

To find the potential rational zeros by using the Rational Zero Theorem, first list the factors of the leading coefficient and the constant term:

Constant 24: 1, 2, 3, 4, 6, 8, 12, 24

Leading coefficient 2: 1, 2

Now we have to divide every factor from the first list by every factor of the second:

Removing duplicates [for example, and are both equivalent to 1] gives us the following list:

The only choice not on this list is .

The potential zeros must have a factor of -15 as their numerator and a factor of 6 as their denominator. This eliminates  as a possibility since 6 is not a factor of -15.

Now we need to test which of these values actually give zero when plugged into the polynomial.

First, :

Now :

Finally :

Since this one doesn't give us zero, it is not a solution of the polynomial.

To use Rational Zeros Theorem, take all factors of the constant term and all factors of the leading coefficient.  That gives you:

Constant Term = 12, so Factors: 1, 2, 3, 4, 6, 12

Leading Coefficient = 3, so Factors: 1, 3

Then divide every number from the first list by every number from the second -- keeping in mind that both positive and negative values are possible. Of course, since this is a multiple-choice question you can also use process-of-elimination to get rid of any values that wouldn't be possible from such division. Note that the only prime factors you have are 2 and 3, meaning that there is no way to produce a 5 or 7. That eliminates (or at least makes very suspect) several choices, leaving the choice:

If you perform division from the two lists you should see how you can arrive at each value from this choice: dividing 12 by 3 would give you 4 (or negative 4, again as both + and - values are possible); dividing 2 by 3 would give you , and dividing 2 by 1 or 6 by 3 would give you 2. Therefore this choice only includes possible zeros.

To use Rational Zeros Theorem, express a polynomial in descending order of its exponents (starting with the biggest exponent and working to the smallest), and then take the constant term (here that's 6) and the coefficient of the leading exponent (here that's 4) and express their factors:

Constant: 6 has as factors 1, 2, 3, and 6

Coefficient: 4 has as factors 1, 2, and 4

Then all possible rational zeros must be formed by dividing a factor from the Constant list by a factor from the Coefficient list (and note that the results should be considered as both positive and negative values).

Here if you then look at the answer choices, note that since the divisor - which must be a factor from the Coefficient list of 1, 2, and 4 - cannot then be a multiple of 3, you know that  is not a possible zero of this polynomial