Find the axs of symmetry $$ y=3 x^{2}-2 x $$

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To find the **axis of symmetry** for the quadratic function:

$$
y = 3x^{2} - 2x
$$

we can use the formula for the axis of symmetry of a parabola in the form $$ y = ax^{2} + bx + c $$:

$$
x = -\frac{b}{2a}
$$

**Step-by-Step Calculation:**

1. **Identify the coefficients**:
   - $$ a = 3 $$
   - $$ b = -2 $$
   - $$ c = 0 $$ (since there's no constant term)

2. **Apply the formula**:
   $$
   x = -\frac{b}{2a} = -\frac{-2}{2 \times 3} = \frac{2}{6} = \frac{1}{3}
   $$

**Conclusion:**

The **axis of symmetry** for the given quadratic function is the vertical line:

$$
x = \frac{1}{3}
$$

This line divides the parabola into two mirror-image halves.

**Graphical Interpretation:**

If you were to graph the equation $$ y = 3x^{2} - 2x $$, the parabola would open upwards (since $$ a = 3 > 0 $$) and be symmetric about the line $$ x = \frac{1}{3} $$.

---

**Additional Information:**

- **Vertex**: The axis of symmetry also passes through the vertex of the parabola. You can find the y-coordinate of the vertex by substituting $$ x = \frac{1}{3} $$ back into the original equation:

$$
  y = 3\left(\frac{1}{3}\right)^{2} - 2\left(\frac{1}{3}\right) = 3 \times \frac{1}{9} - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}
  $$

So, the vertex is at $$ \left(\frac{1}{3}, -\frac{1}{3}\right) $$.

- **Graph Sketch**:
  
  ```
        |
      y |          *
        |        *   *
        |      *       *
        |    *           *
        |  *               *
        |-------------------------
        | 
        |
  ```
  
  The asterisk (*) represents points on the parabola, symmetric about the vertical line $$ x = \frac{1}{3} $$.
The axis of symmetry for the equation $$ y = 3x^{2} - 2x $$ is $$ x = \frac{1}{3} $$.

Find the axs of symmetry \[ y=3 x^{2}-2 x \]

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