1. **State the problem:** Find the values of $a$ and $b$ that make the piecewise function $$ f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x < 3 \\ ax^2 - bx + 3 & \text{if } 3 \leq x < 4 \\ 3x - a + b & \text{if } x \geq 4 \end{cases} $$ continuous everywhere. 2. **Recall the continuity condition:** A function $f$ is continuous at a point $c$ if $$\lim_{x \to c^-} f(x) = f(c) = \lim_{x \to c^+} f(x).$$ We need to ensure continuity at the points where the definition changes: $x=3$ and $x=4$. 3. **Continuity at $x=3$:** - Left-hand limit ($x \to 3^-$): Simplify $\frac{x^2 - 9}{x - 3}$. $$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = \cancel{\frac{(x-3)}{(x-3)}}(x+3) = x+3.$$ - So, $$\lim_{x \to 3^-} f(x) = 3 + 3 = 6.$$ - Right-hand limit and value at $x=3$: $$f(3) = a(3)^2 - b(3) + 3 = 9a - 3b + 3.$$ - Set equal for continuity: $$9a - 3b + 3 = 6.$$ - Simplify: $$9a - 3b = 3.$$ 4. **Continuity at $x=4$:** - Left-hand limit ($x \to 4^-$): $$f(4^-) = a(4)^2 - b(4) + 3 = 16a - 4b + 3.$$ - Right-hand limit and value at $x=4$: $$f(4) = 3(4) - a + b = 12 - a + b.$$ - Set equal for continuity: $$16a - 4b + 3 = 12 - a + b.$$ - Rearrange: $$16a + a - 4b - b = 12 - 3,$$ $$17a - 5b = 9.$$ 5. **Solve the system of equations:** $$\begin{cases} 9a - 3b = 3 \\ 17a - 5b = 9 \end{cases}$$ Multiply the first equation by 5 and the second by 3 to eliminate $b$: $$\begin{cases} 45a - 15b = 15 \\ 51a - 15b = 27 \end{cases}$$ Subtract the first from the second: $$51a - 15b - (45a - 15b) = 27 - 15,$$ $$6a = 12,$$ $$a = 2.$$ Substitute $a=2$ into the first equation: $$9(2) - 3b = 3,$$ $$18 - 3b = 3,$$ $$-3b = 3 - 18 = -15,$$ $$b = 5.$$ **Final answer:** $$a = 2, \quad b = 5.$$