Question 747339
A. It has two vertical asymptotes, x = 4 and x = -3.


According to the rules of graphing rational functions, vertical asymptotes are defined by the values of x that make the function undefined; that is, these values will cause the fraction to have a denominator value of 0. Let us first set the parameters of the denominator by figuring out an expression that makes the denominator undefined when x = 4 and x = -3. We can use the Zero-Product Property to find this expression:


x = 4 --> x - 4 = 0,
x = -3 --> x + 3 = 0.

 
The denominator must have the product of (x-4) and (x+3) in it. Multiplying these two expressions, we receive the polynomial (x^2 - x - 12) which, when x = 4 and x = -3, has a value of 0. (The roots of this polynomial are -3 and 4.) Since this value of 0 would make the function undefined, this polynomial is what we need in the denominator.


To check, we can plug -3 and 4 into our polynomial:


f(-3) = (-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0,
f (4) = (4)^2 - (4) - 12 = 16 - 4 - 12  = 0.


B. It has one horizontal asymptote y = 2.


To get a horizontal asymptote of y = 2, we need the numerator and denominator to have the same power, and we need the coefficient of the highest power of x in the  numerator to be such that when it is divided by the coefficient of the highest power of x in the denominator, the remainder is 2. The highest power in the numerator must be the same power in the denominator. 


Since the highest power of x in the fraction is 2 (x^2 in the denominator), we need a matching power in the numerator. (If there is a higher power in the denominator, then the horizontal asymptote is y = 0, which is not what we need. If the higher power of x is in the numerator, then there is no horizontal asymptote.) So we know that there is at least a (x^2) in the numerator. However, we need the coefficient of this to be such that when it is divided by the coefficient of the (x^2) in the denominator, the result is 2. Since the coefficient of (x^2) in the denominator is 1 (1x^2 = x^2), then we know that we have (2x^2) in the numerator.


2x^2 matches the highest power of x in the denominator, and 2/1 equals 2. We now know that the function looks somewhat like (2x^2)/(x^2-x-12). However, we will see that simply doing this is not enough.


The way to check whether or not we have the horizontal asymptote y = 2 is by trying to make y = 2 by solving algebraically for x. (The graph should not cross at y = 2.) So, 


2 = (2x^2)/(x^2 - x - 12)
2x^2 - 2x - 24 = 2x^2 
-2x - 24 = 0
-2x = 24
x = -11.


This is not what we want. The numerator can't be 2x^2 alone! If it were, the function would be defined at y = 2 and the graph would cross this horizontal asymptote. We need the case where the graph does not cross y = 2. We can do this by simply making the numerator so that all powers of x cancel out between the numerator and the denominator. Let's add (-2x) to our numerator and try to solve for x again:


2 = (2x^2 - 2x)/(x^2 - x - 12)
2x^2 - 2x - 24 = 2x^2 - 2x
-2x - 24 = -2x
-24 = 0.


Now we see that the function doesn't make sense when y = 2. Our new numerator is (2x^2 - 2x). We now can fit the final piece into our numerator by solving step c.


C. It has one x-intercept x = -1.


This means the graph crosses point (-1, 0) on the x-axis. So, we need to make the function such that when x = -1, y = 0. (We need to make the fraction equal 0 when x = -1.) This shouldn't be hard because in order to make a fraction equal 0, the 0 must be in the numerator. If the zero was in the denominator, then the fraction would be undefined. So, with our numerator (2x^2-2x), we need to add another term so that this equals 0 when x = -1. Right now, we can see that if we plug -1 into the expression, we get 4. We just need to add (-4) into our numerator to cancel out the positive 4 and make the numerator 0.


In short: Our rational graph has the function f(x) = (2x^2-2x-4)/(x^2-x-12).

1. Does this graph have vertical asymptotes x = -3 and x = 4?

Yes, when x = -3 and x = 4, f(x) is undefined because the function would equal 23/0 and 20/0 respectively. Thus, the graph cannot cross the lines x = -3 and x = 4.


2. Does this graph have horizontal asymptote y = 2?


Yes, when y = 2, we arrive at a contradiction 20 = 0 or -20 = 0. This does not make sense, so the graph must be broken at the line y = 2. 


3. Does this graph cross point (-1, 0) so that it has an x-intercept at this point?


Yes, when x = -1, y = 0 and we have an x-intercept.


Hope this helps!