Quadratic X Intercepts Ede3Fa

1. **State the problem:** We need to find which quadratic equation produces a parabola with x-intercepts at $(-1, 0)$ and $(3, 0)$. 2. **Recall the factored form of a quadratic:** A quadratic with roots $r_1$ and $r_2$ can be written as $$y = a(x - r_1)(x - r_2)$$ where $a$ is a nonzero constant. 3. **Apply the roots:** Given roots $-1$ and $3$, the factors are $(x - (-1)) = (x + 1)$ and $(x - 3)$. 4. **Form the general equation:** $$y = a(x + 1)(x - 3)$$ 5. **Check the options:** - $y = -4(x - 1)(x + 3)$ has roots $1$ and $-3$, not correct. - $y = -4(x + 1)^2 - 3$ has a repeated root at $-1$ but shifted down by $3$, so x-intercepts are not at $-1$ and $3$. - $y = 4(x - 1)(x + 3)$ has roots $1$ and $-3$, not correct. - $y = 4(x + 1)(x - 3)$ has roots $-1$ and $3$, which matches the required intercepts. 6. **Conclusion:** The correct equation is $$y = 4(x + 1)(x - 3)$$. **Final answer:** $y = 4(x + 1)(x - 3)$