Find the Domain of Function f(x)=(x+3)²-5: Positive and Negative Regions

To find the positive and negative domains of the function $y = (x+3)^2 - 5$, we first identify the roots by setting $y = 0$ and solving for $x$.

Let's solve $(x+3)^2 - 5 = 0$:

1. Start with the equation: $(x+3)^2 - 5 = 0$
2. Add 5 to both sides: $(x+3)^2 = 5$
3. Take the square root of both sides: $x+3 = \pm \sqrt{5}$
4. Solve for $x$:
• $x = -3 + \sqrt{5}$
• $x = -3 - \sqrt{5}$

Thus, the roots of the function are $x = -3 + \sqrt{5}$ and $x = -3 - \sqrt{5}$.

Since the parabola opens upwards (the coefficient of $(x+3)^2$ is positive), the function $y$ is:

• Negative between the roots: $-3 - \sqrt{5} < x < -3 + \sqrt{5}$
• Positive outside these roots: $x < -3 - \sqrt{5}$ and $x > -3 + \sqrt{5}$

Therefore, the positive and negative domains are:

• Negative domain: $-3 - \sqrt{5} < x < -3 + \sqrt{5}$
• Positive domain: $x < -3 - \sqrt{5}$ or $x > -3 + \sqrt{5}$

Upon reviewing the multiple choice options, the correct answer that corresponds to this solution is:

x < 0 : -3-\sqrt{5} < x < -3+\sqrt{5}

x>-3+\sqrt{5} or x > 0 : x < -3-\sqrt{5}