1. **State the problem:** Find the domain and range of the function $$f(x) = x^2 + 6x + 5$$. 2. **Domain:** The domain of a polynomial function like this is all real numbers because you can plug any real number into $$x$$ and get a valid output. So, $$\text{Domain} = (-\infty, \infty)$$. 3. **Range:** To find the range, we need to find the minimum or maximum value of the quadratic function. Since the coefficient of $$x^2$$ is positive ($$1 > 0$$), the parabola opens upwards, so it has a minimum value. 4. **Find the vertex:** The vertex form of a quadratic is $$f(x) = a(x-h)^2 + k$$ where $$(h,k)$$ is the vertex. The $$x$$-coordinate of the vertex is given by $$h = -\frac{b}{2a}$$. Here, $$a=1$$ and $$b=6$$, so $$h = -\frac{6}{2 \times 1} = -3$$. 5. **Find the $$y$$-coordinate of the vertex:** $$k = f(-3) = (-3)^2 + 6(-3) + 5 = 9 - 18 + 5 = -4$$. 6. **Interpretation:** The vertex is at $$(-3, -4)$$, which is the minimum point. Therefore, the range is all $$y$$ values greater than or equal to $$-4$$. So, $$\text{Range} = [-4, \infty)$$. **Final answer:** - Domain: $$(-\infty, \infty)$$ - Range: $$[-4, \infty)$$