1. **State the problem:** We need to find the equation of a rational function with a point of discontinuity (hole) at $(-3,-5)$, a $y$-intercept at $(0,4)$, and an $x$-intercept at $(2,0)$. The graph behavior near the hole is going right and down on the left side and up on the right side, and the curve goes up to the left. 2. **Recall the form of rational functions with holes:** A hole at $x = -3$ means the function has a factor $(x+3)$ in both numerator and denominator that cancels out. So the function can be written as: $$f(x) = \frac{(x+3)g(x)}{(x+3)h(x)} = \frac{g(x)}{h(x)}$$ where $g(x)$ and $h(x)$ are polynomials with no $(x+3)$ factor. 3. **Use the hole coordinates:** The hole at $(-3,-5)$ means that after canceling $(x+3)$, the simplified function $f(x)$ approaches $-5$ as $x \to -3$. So: $$f(-3) = -5$$ 4. **Use the $x$-intercept:** The function is zero at $x=2$, so numerator zero at $x=2$: $$g(2) = 0$$ 5. **Use the $y$-intercept:** At $x=0$, $f(0) = 4$: $$f(0) = \frac{g(0)}{h(0)} = 4$$ 6. **Choose simplest polynomials:** Let numerator after canceling be $g(x) = a(x-2)$ (since zero at 2) and denominator $h(x) = bx + c$ (linear for simplicity). 7. **Apply hole condition:** $$f(-3) = \frac{a(-3-2)}{b(-3) + c} = \frac{a(-5)}{-3b + c} = -5$$ 8. **Apply $y$-intercept condition:** $$f(0) = \frac{a(0-2)}{b(0) + c} = \frac{-2a}{c} = 4$$ 9. **Solve for $a$ and $c$ from $y$-intercept:** $$\frac{-2a}{c} = 4 \implies -2a = 4c \implies a = -2c$$ 10. **Substitute $a$ into hole condition:** $$\frac{-2c \times (-5)}{-3b + c} = -5 \implies \frac{10c}{-3b + c} = -5$$ 11. **Solve for $b$ and $c$:** Multiply both sides: $$10c = -5(-3b + c) = 15b - 5c$$ Bring terms together: $$10c + 5c = 15b \implies 15c = 15b \implies b = c$$ 12. **Choose $c=1$ for simplicity:** Then $b=1$, $a = -2(1) = -2$. 13. **Write the function before canceling:** $$f(x) = \frac{(x+3)(-2)(x-2)}{(x+3)(1 \cdot x + 1)} = \frac{-2(x+3)(x-2)}{(x+3)(x+1)}$$ 14. **Simplify by canceling $(x+3)$:** $$f(x) = \frac{-2(x-2)}{x+1}$$ 15. **Check behavior near hole $x=-3$:** As $x \to -3$, denominator $x+1 = -2$ (negative), numerator $-2(x-2) = -2(-3-2) = -2(-5) = 10$ (positive), so $f(x) \to 10/-2 = -5$ consistent. 16. **Check $y$-intercept:** $$f(0) = \frac{-2(0-2)}{0+1} = \frac{-2(-2)}{1} = 4$$ correct. 17. **Check $x$-intercept:** $$f(2) = \frac{-2(2-2)}{2+1} = 0$$ correct. 18. **Graph behavior:** - For $x < -3$, numerator positive, denominator negative, function negative and goes down to the right. - For $x > -3$, numerator and denominator signs change, function goes up. - As $x \to -\infty$, $f(x) \approx \frac{-2x}{x} = -2$, so curve goes up to the left. **Final answer:** $$\boxed{f(x) = \frac{-2(x-2)}{x+1}}$$