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

To find the vertex of the quadratic function \( y = 3x^2 - 2x \), we can use the vertex formula \( x = -\frac{b}{2a} \). Here, \( a = 3 \) and \( b = -2 \). Plugging these values into the formula gives us: \[ x = -\frac{-2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3} \] Now, substituting \( x = \frac{1}{3} \) back into the original equation to find \( y \): \[ y = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) = 3\left(\frac{1}{9}\right) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3} \] Thus, the vertex is at the point \( \left( \frac{1}{3}, -\frac{1}{3} \right) \). To transform this quadratic into vertex form \( y = a(x-h)^2 + k \), you can complete the square. First, factor out the 3 from the first two terms: \[ y = 3\left(x^2 - \frac{2}{3}x\right) \] Next, take half of \(-\frac{2}{3}\), square it to get \(\frac{1}{9}\), and add and subtract this term inside the parentheses: \[ y = 3\left(x^2 - \frac{2}{3}x + \frac{1}{9} - \frac{1}{9}\right) \] This gives us: \[ y = 3\left(\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right) \] Distributing the 3: \[ y = 3\left(x - \frac{1}{3}\right)^2 - \frac{1}{3} \] Now, you can see it clearly: the vertex form is \( y = 3\left(x - \frac{1}{3}\right)^2 - \frac{1}{3} \) and the vertex is indeed \( \left( \frac{1}{3}, -\frac{1}{3} \right) \)!