Check Bijective Condition

🎥 Video Explanation

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📝 Question

Given \( f(x) = -x^2 + 6x – 8 \), find the correct domain \(A\) and codomain \(B\) such that \(f\) is bijective.

• A. \(A = (-\infty, 5]\), \(B = (-\infty, 1]\)
• B. \(A = [-3, \infty)\), \(B = (-\infty, 1]\)
• C. \(A = (-\infty, 3]\), \(B = [1, \infty)\)
• D. \(A = [3, \infty)\), \(B = [1, \infty)\)

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✅ Solution

🔹 Step 1: Convert to Vertex Form

\[ f(x) = -x^2 + 6x – 8 \]

\[ = -(x^2 – 6x + 9) + 1 \]

\[ = -(x-3)^2 + 1 \]

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🔹 Step 2: Nature of Graph

This is a downward opening parabola.

Vertex at: \[ (3, 1) \]

Maximum value = 1

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🔹 Step 3: Condition for Bijective

To make function one-one, restrict domain to one side of vertex:

• \((-\infty, 3]\) OR \([3, \infty)\)

Range will be: \[ (-\infty, 1] \]

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🔹 Step 4: Match Options

Only option with correct domain and range:

• Domain: \((-\infty, 3]\)
• Range: \((-\infty, 1]\)

✔️ Correct option: A

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🔹 Final Answer

\[ \boxed{\text{Option A}} \]