1. **Problem Statement:** Determine if the function $f(x) = \frac{x+4}{x-2}$ is continuous at $x = -2$ and $x = 0$. 2. **Recall the definition of continuity at a point $c$: ** A function $f$ is continuous at $x = c$ if: $$\lim_{x \to c} f(x) = f(c)$$ This means three things: - $f(c)$ is defined. - The limit $\lim_{x \to c} f(x)$ exists. - The limit equals the function value: $\lim_{x \to c} f(x) = f(c)$. 3. **Check continuity at $x = -2$: ** - Calculate $f(-2)$: $$f(-2) = \frac{-2 + 4}{-2 - 2} = \frac{2}{-4} = -\frac{1}{2}$$ - Calculate the limit $\lim_{x \to -2} f(x)$: Since $f(x)$ is a rational function and the denominator $x-2$ is not zero at $x = -2$, the function is defined and continuous at $x = -2$. So, $$\lim_{x \to -2} f(x) = f(-2) = -\frac{1}{2}$$ 4. **Check continuity at $x = 0$: ** - Calculate $f(0)$: $$f(0) = \frac{0 + 4}{0 - 2} = \frac{4}{-2} = -2$$ - Calculate the limit $\lim_{x \to 0} f(x)$: Again, denominator is not zero at $x=0$, so function is continuous there. $$\lim_{x \to 0} f(x) = f(0) = -2$$ 5. **Conclusion:** The function $f(x) = \frac{x+4}{x-2}$ is continuous at $x = -2$ and $x = 0$ because the function is defined at these points and the limits equal the function values. **Final answer:** - Continuous at $x = -2$. - Continuous at $x = 0$.