Rewrite the quadratic function \( y=3 x^{2}+9 x-1 \) in vertex form. Select one: \( \begin{array}{l}y=9(x+3)^{2}-1 \\ y=3(x+1.5)^{2}-7.75 \\ y=3(x+9)^{2}-1 \\ y=3(x-1)^{2}+9 \\ \text { Clear my choice }\end{array} \)

To rewrite the quadratic function \( y=3 x^{2}+9 x-1 \) in vertex form, we can use the method of completing the square. First, factor out the coefficient of \( x^2 \) (which is 3) from the first two terms: \[ y = 3(x^2 + 3x) - 1. \] Next, complete the square inside the parentheses. Take half of the coefficient of \( x \) (which is 3), square it (which is \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} \)), and add and subtract this value inside the parentheses: \[ y = 3\left(x^2 + 3x + \frac{9}{4} - \frac{9}{4}\right) - 1. \] Now rewrite it: \[ y = 3\left((x + \frac{3}{2})^2 - \frac{9}{4}\right) - 1. \] Distribute the 3: \[ y = 3(x + \frac{3}{2})^2 - \frac{27}{4} - 1. \] Convert -1 to a fraction: \[ y = 3(x + \frac{3}{2})^2 - \frac{27}{4} - \frac{4}{4}. \] Combine the constants: \[ y = 3(x + \frac{3}{2})^2 - \frac{31}{4}. \] Thus, the vertex form is: \[ y = 3\left(x + \frac{3}{2}\right)^2 - \frac{31}{4}. \] However, this isn't one of the options provided. By checking them: If we rewrite \( y = 3(x + 1.5)^2 - 7.75 \), we see that it's equivalent to the simplified and adjusted version because \( -7.75 \) is \( -\frac{31}{4} \). Thus, the answer closely matching our completion is: **Correct option:** \( y = 3(x+1.5)^{2}-7.75 \)