1. **Problem:** Analyze the function $$f(x) = \frac{x^2 - 16}{x^2 + 10x + 24}$$ to find its x-intercepts, vertical asymptotes (VA), horizontal asymptote (HA), holes, domain, and range. 2. **Formula and rules:** - X-intercepts occur where the numerator equals zero and the denominator is not zero. - Vertical asymptotes occur where the denominator equals zero and the numerator is not zero. - Holes occur where numerator and denominator share a common factor. - Horizontal asymptote depends on degrees of numerator and denominator. - Domain excludes values making denominator zero. 3. **Factor numerator and denominator:** $$x^2 - 16 = (x - 4)(x + 4)$$ $$x^2 + 10x + 24 = (x + 6)(x + 4)$$ 4. **Simplify function and identify holes:** $$f(x) = \frac{(x - 4)\cancel{(x + 4)}}{(x + 6)\cancel{(x + 4)}} = \frac{x - 4}{x + 6}, \quad x \neq -4$$ There is a hole at $$x = -4$$ because the factor $$(x + 4)$$ cancels. 5. **Find x-intercepts:** Set numerator zero: $$x - 4 = 0 \implies x = 4$$ Check denominator at $x=4$: $$4 + 6 = 10 \neq 0$$ So, x-intercept at $$(4, 0)$$. 6. **Find vertical asymptotes:** Set denominator zero: $$x + 6 = 0 \implies x = -6$$ Since numerator is not zero at $x=-6$, vertical asymptote at $$x = -6$$. 7. **Find horizontal asymptote:** Degrees numerator and denominator are both 2. Leading coefficients are both 1. So, $$\text{HA} = \frac{1}{1} = 1$$ 8. **Domain:** Exclude values making denominator zero: $$x \neq -6, -4$$ 9. **Range:** Since the simplified function is $$\frac{x - 4}{x + 6}$$ with a hole at $x=-4$, the range is all real numbers except the value at the hole. Find hole's y-value: $$\lim_{x \to -4} f(x) = \frac{-4 - 4}{-4 + 6} = \frac{-8}{2} = -4$$ So, range excludes $$-4$$. **Final answers:** - x-intercept: $$(4, 0)$$ - Vertical asymptote: $$x = -6$$ - Horizontal asymptote: $$y = 1$$ - Hole: at $$x = -4$$ with $$y = -4$$ - Domain: $$(-\infty, -6) \cup (-6, -4) \cup (-4, \infty)$$ - Range: $$(-\infty, -4) \cup (-4, \infty)$$