Find the range of the quadratic function. Give your answer using interval notation. \[ y=3 x^{2}+18 x+9 \]

The given function is \[ y = 3x^{2} + 18x + 9. \] **Step 1: Write the quadratic in vertex form** Factor out the coefficient of \(x^2\) from the first two terms: \[ y = 3\left(x^2 + 6x\right) + 9. \] Next, complete the square inside the parentheses. Take half of the coefficient of \(x\), which is \(\frac{6}{2} = 3\), and square it to get \(3^2 = 9\). Add and subtract this value inside the parentheses: \[ y = 3\left(x^2 + 6x + 9 - 9\right) + 9. \] Group the perfect square trinomial: \[ y = 3\left((x + 3)^2 - 9\right) + 9. \] Distribute the 3: \[ y = 3(x + 3)^2 - 27 + 9. \] Combine the constant terms: \[ y = 3(x + 3)^2 - 18. \] **Step 2: Determine the range** Since the coefficient of the squared term is positive (\(3 > 0\)), the parabola opens upward, meaning the vertex represents the minimum value of the function. The vertex is at \((-3, -18)\), so the minimum value of \(y\) is \(-18\). Thus, the range of \(y\) is all values greater than or equal to \(-18\): \[ [-18, \infty). \]