1. **Stating the problem:** We have a piecewise function defined as: $$f(x) = \begin{cases} \sqrt[3]{x^2(x-1)} & x \geq 0 \\ \frac{x+3}{x^2 + 2x - 3} + 2 & x < 0 \\ a^2 x^2 - 10x - 4 & -1 < x \leq 3 \end{cases}$$ We want to understand the behavior and graph of this function. 2. **Important notes:** - The function has three branches with different domains. - The rational expression denominator can be factored to find domain restrictions. - The quadratic branch depends on parameter $a$. 3. **Analyze each branch:** **Branch 1: $x \geq 0$** $$f(x) = \sqrt[3]{x^2(x-1)} = \sqrt[3]{x^3 - x^2}$$ This is defined for all $x \geq 0$ since cube roots are defined for all real numbers. **Branch 2: $x < 0$** $$f(x) = \frac{x+3}{x^2 + 2x - 3} + 2$$ Factor denominator: $$x^2 + 2x - 3 = (x+3)(x-1)$$ So, $$f(x) = \frac{x+3}{(x+3)(x-1)} + 2 = \frac{\cancel{x+3}}{\cancel{x+3}(x-1)} + 2 = \frac{1}{x-1} + 2$$ Note: $x \neq -3$ to avoid division by zero, but since $x < 0$, $x = -3$ is in domain and must be excluded. **Branch 3: $-1 < x \leq 3$** $$f(x) = a^2 x^2 - 10x - 4$$ This is a quadratic polynomial defined on $(-1,3]$. 4. **Summary of domain:** - Branch 1: $[0, \infty)$ - Branch 2: $(-\infty, 0)$ excluding $x = -3$ - Branch 3: $(-1, 3]$ 5. **Graphing notes:** - Branch 2 has a vertical asymptote at $x=1$ (outside domain $x<0$ so no issue here). - Branch 2 is undefined at $x=-3$ (hole). - Branch 3 overlaps partially with branch 1 domain; the function is piecewise defined so values depend on the branch. 6. **Final function description:** $$f(x) = \begin{cases} \sqrt[3]{x^2(x-1)} & x \geq 0 \\ \frac{1}{x-1} + 2 & x < 0, x \neq -3 \\ a^2 x^2 - 10x - 4 & -1 < x \leq 3 \end{cases}$$ This piecewise function combines a cube root, a rational function shifted by 2, and a quadratic polynomial on specified intervals.