For parabola y = 3x^2 + 6x + 7, what is the x-coordinate of the vertex?

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Multiple Choice

For parabola y = 3x^2 + 6x + 7, what is the x-coordinate of the vertex?

Explanation:
For a parabola in standard form y = ax^2 + bx + c, the x-coordinate of the vertex occurs where the derivative is zero, which leads to the formula x = -b/(2a). This is the point where the parabola changes direction, its axis of symmetry. Here, a = 3 and b = 6. Plugging into the formula gives x = -6 / (2·3) = -6 / 6 = -1. So the x-coordinate of the vertex is -1. If you want the full vertex, you can substitute x = -1 back into the equation: y = 3(-1)^2 + 6(-1) + 7 = 3 - 6 + 7 = 4, so the vertex is (-1, 4).

For a parabola in standard form y = ax^2 + bx + c, the x-coordinate of the vertex occurs where the derivative is zero, which leads to the formula x = -b/(2a). This is the point where the parabola changes direction, its axis of symmetry.

Here, a = 3 and b = 6. Plugging into the formula gives x = -6 / (2·3) = -6 / 6 = -1. So the x-coordinate of the vertex is -1. If you want the full vertex, you can substitute x = -1 back into the equation: y = 3(-1)^2 + 6(-1) + 7 = 3 - 6 + 7 = 4, so the vertex is (-1, 4).

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