Quadratic Function Fe8F3F

1. **State the problem:** We need to find the quadratic function with x-intercepts at $-2$ and $3$, and y-intercept at $-12$. 2. **Recall the form of a quadratic function given roots:** If the roots (x-intercepts) are $r_1$ and $r_2$, the quadratic can be written as $$y = a(x - r_1)(x - r_2)$$ where $a$ is a constant. 3. **Substitute the roots:** Here, $r_1 = -2$ and $r_2 = 3$, so $$y = a(x + 2)(x - 3)$$ 4. **Use the y-intercept to find $a$:** The y-intercept occurs at $x=0$, and $y = -12$. Substitute $x=0$: $$-12 = a(0 + 2)(0 - 3) = a(2)(-3) = -6a$$ 5. **Solve for $a$:** $$-12 = -6a$$ $$\cancel{-12} = \cancel{-6}a$$ $$a = \frac{-12}{-6} = 2$$ 6. **Write the full equation:** $$y = 2(x + 2)(x - 3)$$ 7. **Expand the equation:** $$y = 2(x^2 - 3x + 2x - 6) = 2(x^2 - x - 6) = 2x^2 - 2x - 12$$ **Final answer:** The quadratic function is $$y = 2x^2 - 2x - 12$$. This function has x-intercepts at $-2$ and $3$, and y-intercept at $-12$ as required.