1. **State the problem:** We are given the function $$y = 1 + 3\sqrt{x + 3}$$ and asked to graph it along with its inverse. 2. **Recall the formula and rules:** The function involves a square root, so the domain is restricted to values where the radicand is non-negative: $$x + 3 \geq 0 \Rightarrow x \geq -3$$. 3. **Find the inverse function:** \begin{align*} y &= 1 + 3\sqrt{x + 3} \\ y - 1 &= 3\sqrt{x + 3} \\\frac{y - 1}{3} &= \sqrt{x + 3} \\\\left(\frac{y - 1}{3}\right)^2 &= x + 3 \\ x &= \left(\frac{y - 1}{3}\right)^2 - 3 \end{align*} 4. **Rewrite inverse function:** $$x = \frac{(y - 1)^2}{9} - 3$$ 5. **Domain and range:** - Original function domain: $$x \geq -3$$ - Original function range: $$y \geq 1$$ (since square root is non-negative) - Inverse function domain: $$y \geq 1$$ - Inverse function range: $$x \geq -3$$ 6. **Graphing notes:** - The original function is a square root curve shifted left by 3 and up by 1, scaled vertically by 3. - The inverse is a parabola opening right, shifted left by 3 and up by 1. 7. **Final answer:** - Function: $$y = 1 + 3\sqrt{x + 3}$$ - Inverse: $$x = \frac{(y - 1)^2}{9} - 3$$