Find the axir of symatry Find the vertexy $$ y=3 x^{2}-2 x $$

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It looks like there were a couple of typos in your request. I'll assume you're asking for the **axis of symmetry** and the **vertex** of the quadratic function:

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

### 1. Axis of Symmetry

For any quadratic function of the form $$ y = ax^{2} + bx + c $$, the **axis of symmetry** is a vertical line that passes through the vertex of the parabola. It can be found using the formula:

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

**Given:**
- $$ a = 3 $$
- $$ b = -2 $$

**Calculation:**

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

**So, the axis of symmetry is:**

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

### 2. Vertex

The **vertex** of the parabola is the point where it reaches its maximum or minimum value. Since the coefficient $$ a = 3 $$ is positive, the parabola opens upwards, and the vertex represents the minimum point.

The vertex $$(h, k)$$ can be found using:

$$
h = -\frac{b}{2a}
$$
$$
k = y(h) = 3h^{2} - 2h
$$

**We've already found $$ h = \frac{1}{3} $$. Now, calculate $$ k $$:**

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

**So, the vertex is:**

$$
\left( \frac{1}{3}, -\frac{1}{3} \right)
$$

### Summary

- **Axis of Symmetry:** $$ x = \frac{1}{3} $$
- **Vertex:** $$ \left( \frac{1}{3}, -\frac{1}{3} \right) $$

---

**Graphical Representation:**

![Parabola](https://i.imgur.com/yourImageLink.png) *(Please plot the graph using graphing software or graph paper to visualize the parabola with the axis of symmetry and vertex.)*

If you have any further questions or need additional assistance, feel free to ask!
The axis of symmetry is $$ x = \frac{1}{3} $$ and the vertex is $$ \left( \frac{1}{3}, -\frac{1}{3} \right) $$.

Find the axir of symatry Find the vertexy \( y=3 x^{2}-2 x \)

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