Answer:
The vertex of the graph of the equation $y = 3x^2 - 6x + 7$ is the point $(1, 4)$.
Explanation:
The vertex of the graph of a parabolic function $y = ax^2 + bx + c$ is given by the formula:
$x = \frac{-b}{2a}$
where a, b, and c are constants and x is the x-coordinate of the vertex.
For the given equation $y = 3x^2 - 6x + 7$, we have a = 3, b = -6, and c = 7. So, we can use the formula to find the x-coordinate of the vertex:
$x = \frac{-b}{2a} = \frac{-(-6)}{2(3)} = \frac{6}{6} = 1$
So, the x-coordinate of the vertex is 1. To find the y-coordinate, we substitute the value of x back into the original equation:
$y = 3(1)^2 - 6(1) + 7 = 3 - 6 + 7 = 4$
So, the vertex of the graph of the equation $y = 3x^2 - 6x + 7$ is the point $(1, 4)$.