1Cademy - Finding the Axis of Symmetry and Vertex of $$y = 3x^2

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Parabola

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Finding the Axis of Symmetry and Vertex of $y = 3x^2 - 6x + 2$

To find the axis of symmetry and the vertex for the parabola $y = 3x^2 - 6x + 2$ :

Step 1: Find the axis of symmetry. Identify the coefficients $a = 3$ and $b = -6$ . Substitute them into the formula $x = -\frac{b}{2a}$ :

$x = -\frac{-6}{2(3)}$ $x = \frac{6}{6}$ $x = 1$

The axis of symmetry is the line $x = 1$ .

Step 2: Find the vertex. The vertex lies on the axis of symmetry, so its x-coordinate is $1$ . To find the y-coordinate, substitute $x = 1$ into the original equation:

$y = 3(1)^2 - 6(1) + 2$ $y = 3(1) - 6 + 2$ $y = 3 - 6 + 2$ $y = -1$

The y-coordinate is $-1$ . Therefore, the vertex is the point $(1, -1)$ .

Updated 2026-04-21

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OpenStax Elementary Algebra

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