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How do you find the vertex of a parabola #f(x) = -3x^2 + 6x -2#?

Answer 1

Vertex (1, 1)

#f(x) = - 3x^2 + 6x - 2# x-coordinate of vertex: #x = -b/(2a) = -6/-6 = 1# y-coordinate of vertex: #f(1) = - 3 + 6 - 2 = 1# Vertex (1, 1) graph{-3x^2 + 6x - 2 [-5, 5, -2.5, 2.5]}

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Answer 2

Jace Anderson

To find the vertex of a parabola in the form (f(x) = ax^2 + bx + c), you can use the formula for the x-coordinate of the vertex: (x = -\frac{b}{2a}). Then, plug this value of (x) into the equation to find the corresponding (y)-coordinate. So, for the given equation (f(x) = -3x^2 + 6x - 2), the x-coordinate of the vertex is (x = -\frac{6}{2(-3)} = 1). Plugging (x = 1) into the equation gives (f(1) = -3(1)^2 + 6(1) - 2 = 1). Therefore, the vertex is (1, 1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.