Solution: Since the difference $ f(x+1) - f(x) $ is linear, $ f(x) $ must be a quadratic polynomial. Let $ f(x) = ax^2 + bx + c $. Then:
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- Now compute $ \frac{a^2 + 4b^2}{a^2 - 4b^2} $. Since $ b = 0 $, this becomes:
- \frac{a^2}{a^2} = \boxed{1}
- Question: A polynomial $ f(x) $ satisfies $ f(x + 1) - f(x) = 6x + 4 $. Find the leading coefficient of $ f(x) $.
- f(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 + 2x + 1) + b(x + 1) + c = ax^2 + 2ax + a + bx + b + c
- f(x+1) - f(x) = (ax^2 + 2ax + a + bx + b + c) - (ax^2 + bx + c) = 2ax + a + b
- f(x+1) - f(x) = 6x + 4
