Find Domains of y=-3x²+13: Analyzing a Quadratic Function

To solve this problem, we first determine where the quadratic function $y = -3x^2 + 13$ is equal to zero. Setting $-3x^2 + 13 = 0$ yields:

• Rearrange the equation: $-3x^2 = -13$, leading to $x^2 = \frac{13}{3}$.

• Solve for $x$: $x = \pm \sqrt{\frac{13}{3}}$.

The roots of the equation are $x = \sqrt{\frac{13}{3}}$ and $x = -\sqrt{\frac{13}{3}}$. These roots divide the number line into three intervals: $x < -\sqrt{\frac{13}{3}}$, $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, and $x > \sqrt{\frac{13}{3}}$.

Determine the sign of the function in each interval:

• For $x < -\sqrt{\frac{13}{3}}$, select a test point (e.g., $x = -3$), the function value is negative because $-3(-3^2) + 13 = -27 + 13 = -14$.

• For $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, select a test point (e.g., $x = 0$), the function value is positive because $-3(0)^2 + 13 = 13$.

• For $x > \sqrt{\frac{13}{3}}$, select a test point (e.g., $x = 3$), the function value is negative because $-3(3^2) + 13 = -27 + 13 = -14$.

Therefore, the positive domain of the function is $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, and the negative domain is $x < -\sqrt{\frac{13}{3}}$ or $x > \sqrt{\frac{13}{3}}$.

The answer matches choice 3:

$x > \sqrt{\frac{13}{3}}$ or $x < 0 : x < -\sqrt{\frac{13}{3}}$

$x > 0 : -\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$