1. **State the problem:** We are given the rational function $$f(x) = \frac{x+3}{x^2 + 2x - 3}$$ and asked to analyze and graph it, including finding intercepts, domain, vertical asymptotes (VA), removable points of discontinuity (RPOD), and horizontal asymptotes (HA). 2. **Factor the denominator:** $$x^2 + 2x - 3 = (x+3)(x-1)$$ 3. **Simplify the function:** $$f(x) = \frac{x+3}{(x+3)(x-1)}$$ Cancel the common factor $x+3$ (noting $x \neq -3$ to avoid division by zero): $$f(x) = \frac{\cancel{x+3}}{\cancel{x+3}(x-1)} = \frac{1}{x-1} \quad \text{for } x \neq -3$$ 4. **Domain:** The original denominator is zero at $x = -3$ and $x = 1$, so domain is all real numbers except these points: $$\text{Domain} = \mathbb{R} \setminus \{-3, 1\}$$ 5. **Vertical asymptotes (VA):** At $x=1$, the denominator is zero and the factor does not cancel, so there is a vertical asymptote: $$x = 1$$ 6. **Removable point of discontinuity (RPOD):** At $x = -3$, the factor cancels, so there is a hole (removable discontinuity) at $x = -3$. 7. **Horizontal asymptote (HA):** Since the degree of numerator (1) is less than degree of denominator (2), $$\lim_{x \to \pm \infty} f(x) = 0$$ So, $$y = 0$$ 8. **Find x-intercepts:** Set numerator equal to zero: $$x + 3 = 0 \implies x = -3$$ But $x = -3$ is excluded from domain (hole), so no x-intercept. 9. **Find y-intercept:** Evaluate at $x=0$: $$f(0) = \frac{0+3}{0^2 + 2(0) -3} = \frac{3}{-3} = -1$$ 10. **Summary:** - Domain: $x \neq -3, 1$ - VA: $x=1$ - RPOD (hole): $x=-3$ - HA: $y=0$ - x-intercept: none (hole at $x=-3$) - y-intercept: $(0, -1)$ \textbf{Final answers:} - x-int(s): none - y-int: $-1$