Rational function, vertical asymptotes

Question

Where are the vertical asymptotes of the rational function

g(x) = (x²-9)/(x²+4x-7)

Answer 1

set the denominator = 0 and solve for x

x^2 +4x -7 = 0

x = -4/2 +/-.5sqr(16+28)

x = -2 +/-.5isqr44

x =-2+/-isqr11 are the 2 vertical asymptotes

oddly if you graph the original fraction with Desmos, it gives asymptotes about x= 1 and -5

as if it trreated x as = -2+/-sqr11 = about 1.1 and -5.1

Of course, it's standard to graph imaginary numbers on the vertical axis and reals on the horizontal,

Answer 2

We know that we possibly have vertical asymptotes when the denominator = 0.

Example) (x-a)(x-b) / [(x-a)(x-c)] => x=c is VA but x=a is hole

However, we do not need to worry about this g(x) = (x²-9)/(x²+4x-7) problem.

1. We factor each numerator and denominator.

g(x) = (x-3)(x+3) / [ x^2+4x-7] => denominator is not factorable.

=> we use quadratic formula: a=1, b=4, c= - 7

=> x= (-2 +- sqrt(4+7))/1 = -2+-sqrt(11)

Denominator: (x-(-2+sqrt(11))) * (x-(-2-sqrt(11)))

1. Check holes (anything can we cancel out?) - No
2. VA: x= -2+sqrt(11) and x= -2-sqrt(11)

Answer 3

The problem is to find the vertical asymptotes of g(x) = (x²-9)/(x²+4x-7)

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

Fact 2: A vertical asymptote occurs everywhere the denominator of 'g' 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 often 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.