1. **State the problem:** We want to find values of $a$ and $b$ such that the piecewise function $$f(x) = \begin{cases} x^2 + a, & x \geq 3 \\ bx + a, & -3 < x < 3 \\ \sqrt{-b - x}, & x \leq -3 \end{cases}$$ is continuous everywhere. 2. **Continuity conditions:** The function must be continuous at the boundaries $x = -3$ and $x = 3$ where the definition changes. 3. **Continuity at $x=3$:** Left limit (linear) must equal right limit (quadratic) and the function value at $x=3$. \- From linear side: $\lim_{x\to 3^-} f(x) = 3b + a$ \- From quadratic side: $f(3) = 3^2 + a = 9 + a$ Setting these equal: $$3b + a = 9 + a \implies 3b = 9 \implies b = 3$$ 4. **Continuity at $x = -3$:** Left limit and function value from the square root side equals right limit from the linear side. \- From square root side (at $x = -3$): $$f(-3) = \sqrt{-b - (-3)} = \sqrt{-b + 3}$$ \- From linear side (approaching $-3$ from right): $$\lim_{x \to -3^+} f(x) = b(-3) + a = -3b + a$$ Equate these for continuity: $$\sqrt{-b + 3} = -3b + a$$ 5. **Substitute $b=3$ found from step 3:** $$\sqrt{-3 + 3} = -3(3) + a \implies \sqrt{0} = -9 + a \implies 0 = -9 + a \implies a = 9$$ 6. **Summary:** $$b=3, \quad a=9$$ These values make the function continuous at $x = -3$ and $x = 3$. **Final answer:** $$\boxed{a = 9, \quad b = 3}$$