Question: Put the equation $y = x^2 + 12x + 27$ into the form $y = (x - h)^2 + k$
1. **State the problem:** We want to rewrite the quadratic equation $$y = x^2 + 12x + 27$$ in vertex form $$y = (x - h)^2 + k$$ where $(h, k)$ is the vertex of the parabola.
2. **Recall the formula:** The vertex form of a quadratic is $$y = (x - h)^2 + k$$ where $$h = -\frac{b}{2a}$$ for a quadratic $$ax^2 + bx + c$$.
3. **Identify coefficients:** Here, $$a = 1$$, $$b = 12$$, and $$c = 27$$.
4. **Calculate $h$:**
$$h = -\frac{b}{2a} = -\frac{12}{2 \times 1} = -\frac{12}{2} = -6$$
5. **Complete the square:**
Start with $$y = x^2 + 12x + 27$$
Add and subtract $$\left(\frac{b}{2}\right)^2 = \left(\frac{12}{2}\right)^2 = 6^2 = 36$$ inside the equation:
$$y = x^2 + 12x + 36 - 36 + 27$$
6. **Rewrite as a perfect square:**
$$y = (x + 6)^2 - 36 + 27$$
7. **Simplify constants:**
$$y = (x + 6)^2 - 9$$
8. **Final vertex form:**
$$y = (x - (-6))^2 - 9$$
This means the vertex is at $$(-6, -9)$$.
**Answer:** $$y = (x + 6)^2 - 9$$