1. **State the problem:** Expand the function $y=2(x-3)^2+6$ to general form and find its range. 2. **Recall the formula:** The general form of a quadratic function is $y=ax^2+bx+c$. 3. **Expand the squared term:** $$ (x-3)^2 = x^2 - 2\cdot3\cdot x + 3^2 = x^2 - 6x + 9 $$ 4. **Substitute back into the function:** $$ y = 2(x^2 - 6x + 9) + 6 $$ 5. **Distribute the 2:** $$ y = 2x^2 - 12x + 18 + 6 $$ 6. **Combine like terms:** $$ y = 2x^2 - 12x + 24 $$ 7. **General form:** $$ y = 2x^2 - 12x + 24 $$ 8. **Find the range:** Since $a=2 > 0$, the parabola opens upwards, so the vertex is the minimum point. 9. **Find vertex $x$-coordinate:** $$ x = -\frac{b}{2a} = -\frac{-12}{2\cdot2} = \frac{12}{4} = 3 $$ 10. **Find vertex $y$-coordinate:** Substitute $x=3$ into the original function: $$ y = 2(3-3)^2 + 6 = 2(0)^2 + 6 = 6 $$ 11. **Range in interval notation:** Since the minimum value is 6 and the parabola opens upwards, the range is: $$ [6, \infty) $$ **Final answers:** - Expanded form: $y=2x^2 - 12x + 24$ - Range: $[6, \infty)$