Find the vertical asymptotes. For each vertical asymptote x=a, analyze lim x→a^− f(x) and lim x→a^+f(x).


f(x)=√x^2+2x+6−3 / x−1

Verified step by step guidance

1. Step 1: Identify the points where the function is undefined. The function \( f(x) = \frac{\sqrt{x^2 + 2x + 6} - 3}{x - 1} \) is undefined where the denominator is zero. Set \( x - 1 = 0 \) to find the point where the function is undefined, which gives \( x = 1 \).
2. Step 2: Check if \( x = 1 \) is a vertical asymptote. A vertical asymptote occurs if the function approaches infinity as \( x \) approaches 1 from either side. We need to analyze the limits \( \lim_{x \to 1^-} f(x) \) and \( \lim_{x \to 1^+} f(x) \).
3. Step 3: Simplify the expression \( \sqrt{x^2 + 2x + 6} - 3 \) to check the behavior near \( x = 1 \). Consider rationalizing the numerator by multiplying the numerator and the denominator by the conjugate \( \sqrt{x^2 + 2x + 6} + 3 \).
4. Step 4: After rationalizing, simplify the expression to see if the limit exists or if it approaches infinity. This will help determine the behavior of the function as \( x \to 1^- \) and \( x \to 1^+ \).
5. Step 5: Evaluate the limits \( \lim_{x \to 1^-} f(x) \) and \( \lim_{x \to 1^+} f(x) \) using the simplified expression to confirm the presence of a vertical asymptote at \( x = 1 \) and determine the direction of the asymptote.