QUESTION IMAGE
Question
find the values of x where the given function has vertical asymptotes. list them in increasing order.
f(x) = \frac{1}{x^{2}+2x - 3}
x = ?,
Step1: Factor the denominator
We factor $x^{2}+2x - 3=(x + 3)(x - 1)$.
Step2: Set the factored - denominator equal to zero
Set $(x + 3)(x - 1)=0$.
Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x+3 = 0$ gives $x=-3$ and $x - 1=0$ gives $x = 1$.
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$x=-3,1$