Digital ACT Math Practice Question #2111
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Hole vs Asymptote rule: Factors that cancel completely from the denominator leave holes, NOT vertical asymptotes. Only $(x+3)$ remains $\implies x = -3$.
Factor the denominator: $f(x) = \frac{x - 3}{(x - 3)(x + 3)}$. For $x \ne 3$, $f(x) = \frac{1}{x + 3}$. The factor $(x - 3)$ creates a removable discontinuity (hole) at $x = 3$, while $(x + 3)$ produces a true vertical asymptote at $x = -3$...
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