Answer:
A parabola has an axis of symmetry at x = 3 if the x-value of the vertex is 3.
f(x) = x^2 + 3x + 1 = x^2 + 3x + 9/4 - 5/4 = (x + 3/2)^2 - 5/4 => vertex = (-3/2, -5/4)
f(x) = x^2 - 3x - 3 = x^2 - 3x + 9/4 - 21/4 = (x - 3/2)^2 - 21/4 => vertex = (3/2, -21/4)
f(x) = x^2 + 6x + 3 = x^2 + 6x + 9 - 6 = (x + 3)^2 - 6 => vertex = (-3, -6)
f(x) = x^2 - 6x - 1 = x^2 - 6x + 9 - 10 = (x - 3)^2 - 10 => vertex = (3, -10)
Therefore, f(x) = x^2 - 6x - 1 has an axis of symmetry at x = 3.
Step-by-step explanation:
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