rational function, finding vertical asymptotes

Question

Find the vertical asymptotes of f(x) = (x²-9)/(x²+4x-21)

Answer 1

x^2+4x-21= (x+7)(x-3) = 0

x=3, -7 are the seemingly 2 vertical asymptotes

but when x=3 the numerator also =0 and you get 0/0 which is undefined

to see what's going on at x=3, graph the fraction with a graphing calculator. It shows 2 solutions -3 and +3

but no -7, but these "solutions" are x intercepts not vertical asymptotes

the x-3 factor in both numerator and denominator can cancel but that conceals a possible point of discontinuity and vertical asymptote. Cancel and that leaves (x+3)/(x-7) with seemingly x=7 as a vertical asymptote and no0/0 problem, and still with y=1 as a vertical asymptote

horizontal axis is y= 1/1=1 the coefficient ratio of highest powered terms in numerator and denominator.

as x approaches infinity and negative infinity

while y also approaches 0 as x approaches -3 and +3 that doesn't mean any vertical asymptotes

final solution, from the graph: No vertical asymptotes

graphing and geometry beat algebra hands down, for asymptotic behavior

You could also use calculus, derivatives, find where f'(x) is undefined for what x values f'(x) involves division by the prohibited zero as in f'(x)= 1/0 or any c/0 where c= any real number

Answer 2

The problem is to find the vertical asymptotes of f(x) = (x²-9)(x²+4x-21)

As your tutor, I'll teach you both some facts that you need to have memorized, and some guiding rules of thumb that you can use to understand this problem and to solve other problems, even entirely new ones.

Over the weeks we work together, we will drill the facts so you have them well memorized. Here are two facts we need:

Fact 1: f is a rational function, meaning it's one polynomial divided by another.

Fact 2: A vertical asymptote occurs everywhere the denominator of f is zero and the numerator is not zero.

Now we need a guiding intuition about manipulating the form of the polynomials. If we are asked to find where a polynomial is zero, it helps to have it factored. If we work together, I will show you some examples of that so you intuitively understand.

So let's factor the numerator. Note that it's a difference of squares. (If you forget factoring patterns like

difference of squares, we will review and drill them.)

x²-9 = (x+3)(x-3)

The zeros are at 3 and -3.

Now we factor the denominator. This is a trinomial with a leading coefficient of 1. Here's the factored form:

x²+4x-21 = (x+7)(x-3)

The zeros of the denominator are at -7 and 3.

But the numerator is also zero at 3, so that just leaves one place where the denominator is zero and the numerator is not zero: at -7.

So this function has one vertical asymptote at -7.