1. **State the problem:** We are given the function $y = (x+3)(x-1)(x-4)$ and asked to analyze it. 2. **Formula and rules:** This is a cubic polynomial expressed in factored form. To understand its behavior, we can expand it and find its roots, intercepts, and critical points. 3. **Expand the expression:** $$y = (x+3)(x-1)(x-4)$$ First, multiply the first two factors: $$ (x+3)(x-1) = x^2 - x + 3x - 3 = x^2 + 2x - 3 $$ Now multiply this result by $(x-4)$: $$ y = (x^2 + 2x - 3)(x - 4) $$ $$ y = x^3 - 4x^2 + 2x^2 - 8x - 3x + 12 $$ $$ y = x^3 - 2x^2 - 11x + 12 $$ 4. **Find the roots:** The roots are the values of $x$ that make $y=0$. From the factored form, the roots are: $$ x = -3, 1, 4 $$ 5. **Find the critical points:** These occur where the derivative $y'$ is zero. $$ y' = \frac{d}{dx}(x^3 - 2x^2 - 11x + 12) = 3x^2 - 4x - 11 $$ Set derivative equal to zero: $$ 3x^2 - 4x - 11 = 0 $$ Use quadratic formula: $$ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-11)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 132}}{6} = \frac{4 \pm \sqrt{148}}{6} $$ Simplify: $$ \sqrt{148} = \sqrt{4 \cdot 37} = 2\sqrt{37} $$ So, $$ x = \frac{4 \pm 2\sqrt{37}}{6} = \frac{2 \pm \sqrt{37}}{3} $$ 6. **Summary:** - Roots at $x = -3, 1, 4$ - Critical points at $x = \frac{2 + \sqrt{37}}{3}$ and $x = \frac{2 - \sqrt{37}}{3}$ This analysis helps understand the shape and key features of the cubic function.