1. **Problem:** Determine if $g(x) = \frac{x^2 + 6x + 8}{x + 4}$ is continuous at $x = -4$. 2. **Formula and rules:** A function is continuous at $x = a$ if: 1. $g(a)$ is defined. 2. $\lim_{x \to a} g(x)$ exists. 3. $\lim_{x \to a} g(x) = g(a)$. 3. **Evaluate $g(-4)$:** Substitute $x = -4$ into $g(x)$: $$g(-4) = \frac{(-4)^2 + 6(-4) + 8}{-4 + 4} = \frac{16 - 24 + 8}{0} = \frac{0}{0} \text{ (undefined)}$$ So, $g(-4)$ is not defined. 4. **Simplify $g(x)$:** Factor numerator: $$x^2 + 6x + 8 = (x + 4)(x + 2)$$ So, $$g(x) = \frac{(x + 4)(x + 2)}{x + 4}$$ For $x \neq -4$, this simplifies to: $$g(x) = x + 2$$ 5. **Find $\lim_{x \to -4} g(x)$:** $$\lim_{x \to -4} g(x) = \lim_{x \to -4} (x + 2) = -4 + 2 = -2$$ 6. **Conclusion:** Since $g(-4)$ is undefined but the limit exists and equals $-2$, $g(x)$ is **not continuous** at $x = -4$. **Final answer:** $g(x)$ is not continuous at $x = -4$ because $g(-4)$ is undefined but the limit exists.