1. **Problem Statement:** We want to understand why for the function $$y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}$$ with real $x$, the discriminant condition $D \geq 0$ is used and why $y \neq 1$ is considered. 2. **Rewrite the equation:** Multiply both sides by the denominator: $$y(x^2 + 3x + 4) = x^2 - 3x + 4$$ which expands to $$(y-1)x^2 + 3x(y+1) + 4(y-1) = 0$$ 3. **Interpretation as a quadratic in $x$:** For fixed $y$, this is a quadratic equation in $x$: $$a = y-1, \quad b = 3(y+1), \quad c = 4(y-1)$$ 4. **Condition for real $x$:** Since $x$ is real, the quadratic must have real solutions. The discriminant $D$ must satisfy: $$D = b^2 - 4ac \geq 0$$ Substitute $a,b,c$: $$D = [3(y+1)]^2 - 4(y-1) \cdot 4(y-1) = 9(y+1)^2 - 16(y-1)^2 \geq 0$$ 5. **Why $y \neq 1$?** If $y=1$, then $a = y-1 = 0$, and the equation reduces to: $$3x(y+1) + 4(y-1) = 3x(2) + 0 = 6x = 0$$ which implies $x=0$. So $y=1$ corresponds to a single $x$ value, not a quadratic equation. 6. **Summary:** The discriminant condition $D \geq 0$ ensures that for a given $y$, there exists at least one real $x$ satisfying the equation. The restriction $y \neq 1$ is because at $y=1$, the quadratic term vanishes and the equation is linear, so the discriminant condition does not apply. 7. **Final note:** This approach helps find the range of $y$ values for which the original rational function can take real $x$ inputs. **Answer:** The condition $D \geq 0$ ensures real $x$ solutions for fixed $y$, and $y \neq 1$ is required because at $y=1$ the quadratic equation degenerates to linear, so the discriminant test is not valid.