Cubic Coefficients 138C1B

1. **State the problem:** We need to find the coefficients $a$, $b$, $c$, and $d$ of the cubic polynomial $$y = ax^3 + bx^2 + cx + d$$ given that the curve crosses the x-axis at $x = -4$, $x = 2$, and $x = 6$, and crosses the y-axis at $y = -72$. 2. **Use the roots to write the polynomial:** Since the curve crosses the x-axis at these points, these are roots of the polynomial. Therefore, the polynomial can be expressed as: $$y = a(x + 4)(x - 2)(x - 6)$$ where $a$ is a constant to be determined. 3. **Expand the factors:** First, multiply two factors: $$(x + 4)(x - 2) = x^2 - 2x + 4x - 8 = x^2 + 2x - 8$$ Now multiply this result by $(x - 6)$: $$ (x^2 + 2x - 8)(x - 6) = x^3 - 6x^2 + 2x^2 - 12x - 8x + 48 = x^3 - 4x^2 - 20x + 48$$ 4. **Write the polynomial with $a$:** $$y = a(x^3 - 4x^2 - 20x + 48) = a x^3 - 4a x^2 - 20a x + 48a$$ 5. **Use the y-intercept to find $a$:** The curve crosses the y-axis at $y = -72$, which means when $x=0$, $y = -72$. Substitute $x=0$: $$y = 48a = -72$$ Solve for $a$: $$a = \frac{-72}{48} = -\frac{3}{2}$$ 6. **Find $b$, $c$, and $d$ using $a$:** From the expanded form: $$b = -4a = -4 \times -\frac{3}{2} = 6$$ $$c = -20a = -20 \times -\frac{3}{2} = 30$$ $$d = 48a = 48 \times -\frac{3}{2} = -72$$ **Final answer:** $$a = -\frac{3}{2}, \quad b = 6, \quad c = 30, \quad d = -72$$