what did i do wrong?? [algebra 2..graphing functions]?

Question

the equation:

f(x)= (x^2 + x - 2) / (x^2 - x - 6)
to get the VA i set the denominator equal to zero...

x^2 - x - 6 = 0
(x -3) ( x +2) = 0
3, -2.
so there is a VA at 3 and a VA at -2. but when i graphed it on my calculator...the graph went through the VA. is this becasue there is a hole at -2?

Answer 1

Both the numerator and denominator factorize, and as shown below, they have a common factor (x + 2). For all practical purposes, that means that it can be divided out.

Thus (x^2 + x - 2) = (x - 1)(x + 2), while

(x^2 - x - 6) = (x - 3)(x + 2).

So f(x)= (x^2 + x - 2) / (x^2 - x - 6) = [(x - 1)(x + 2)] / [(x - 3)(x + 2)]

= (x - 1) / (x - 3).

So the function simply has just ONE vertical asymptote at x = 3. [On either side of x = 2 it's perfectly well behaved and has the same limit. With the possible exception of just that ONE point (definitely a set of measure zero), the graph is that of the original expression with (x + 2) divided out of both numerator and denominator.

Answer 2

There is no vertical assymptote at x = -2. Notice that the function is (x - 1)(x + 2)/(x - 3)(x + 2). Clearly the function is not defined at x = -2 because the denominator vanishes there. But notice when x is not -2, we can cancel the factor x + 2, and f behaves exactly like F(x) = (x - 1)/(x - 3). As x -> -2 this function has the limit 3/5. Thus, the "hole" you see results from x = -2 not being in the domain of f, but as we approach
-2 either from the left or the right, the graph gets close to 3/5.

Cute question.

Answer 3

Hi,
The numerator factors into (x+2)(x-1) therefore

f(x)= (x+2)(x-1)/(x+2)(x-3) = F(x) = (x-1)/(x-3) for all x not equal to -2. So x= -2 isn't a v.a.
The important thing is to realize that dividing out the x+2 is perfectly okay as long as x is not -2 where the function f (and F) doesn't exist anyway.