1. The problem is to find the x-intercepts of the quadratic function $y = x^2 + 7x + 12$. 2. X-intercepts occur where $y=0$, so we solve the equation $x^2 + 7x + 12 = 0$. 3. Factor the quadratic: $x^2 + 7x + 12 = (x + 3)(x + 4)$. 4. Set each factor equal to zero: $x + 3 = 0$ or $x + 4 = 0$. 5. Solve for $x$: $x = -3$ or $x = -4$. 6. Therefore, the x-intercepts are at points $(-3, 0)$ and $(-4, 0)$. Note: The expressions $y = (x+7)(x+12)$ and others given are incorrect factorizations of the original quadratic.