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function $f(x)=(x^2+x-12)/(x^2-9).$

So for this, I got the deno to $(x-3)(x+3)$ through difference of squares. Doesn't that mean the vert asymptotes are $ x=3 $and $x= -3?$ But when I go to graph it on desmos.com, it shows that $-3 $ is the only vertical asymptote

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1 Answer

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You are right that the values that make the factors in the denominator equal $0$ are the vertical asymptotes.

BUT you forgot that the first step to finding vertical asymptotes is to simplify the problem. You must always simplify the problem. So basically you need to factor the numerator and denominator, and cancel out any common factors.

If you factor the numerator, you should get $(x^{2} + x - 12) = (x - 3)(x + 4)$. So, our problem becomes:

$f(x) = \dfrac{x^{2} + x - 12}{x^{2} - 9} = \dfrac{(x - 3)(x + 4)}{(x - 3)(x + 3)}$

We have a common factor of $(x - 3)$ in the numerator and denominator, so we must cross them out. So our function becomes $f(x) = \dfrac{x + 4}{x + 3}$.

Now you can use the simplified version to find the vertical asymptotes. As you can see, there is only one value of $x$ that makes the denominator equal to $0$, and that is $x = -3$.

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