Find the Domain of y=-3x²+4 11/16: Quadratic Function Analysis

To find the positive and negative domains of the function $y = -3x^2 + 4\frac{11}{16}$, we need to first find the roots of the equation.

Step 1: Set the function equal to zero to find the roots:

$-3x^2 + 4\frac{11}{16} = 0$

Simplify the equation:

First, express $4\frac{11}{16}$ as a fraction:

$4\frac{11}{16} = \frac{75}{16}$

Substitute into the equation:

$-3x^2 + \frac{75}{16} = 0$

Multiply through by 16 to clear the fraction:

$-48x^2 + 75 = 0$ $48x^2 = 75$

Divide both sides by 48:

$x^2 = \frac{75}{48}$

Simplify the fraction:

$x^2 = \frac{25}{16}$

Take the square root of both sides:

$x = \pm \frac{5}{4}$

We have two roots: $x = \frac{5}{4}$ and $x = -\frac{5}{4}$.

Step 2: Identify the intervals to test for positivity and negativity.

The critical points create intervals: $(-\infty, -\frac{5}{4})$, $(-\frac{5}{4}, \frac{5}{4})$, and $(\frac{5}{4}, \infty)$.

Step 3: Determine the sign of $y$ in each interval by testing sample points:

• For $x \in (-\infty, -\frac{5}{4})$ (e.g., $x = -2$): $y = -3(-2)^2 + \frac{75}{16}$ yields a negative value.
• For $x \in (-\frac{5}{4}, \frac{5}{4})$ (e.g., $x = 0$): $y = -3(0)^2 + \frac{75}{16} = \frac{75}{16}$, which is positive.
• For $x \in (\frac{5}{4}, \infty)$ (e.g., $x = 2$): $y = -3(2)^2 + \frac{75}{16}$ yields a negative value.

Thus, the function is positive for $x \in \left(-\frac{5}{4}, \frac{5}{4}\right)$ and negative for $x < -\frac{5}{4}$ and $x > \frac{5}{4}$.

Therefore, the solution to the problem is:

x > 1\frac{1}{4} or x < 0 : x < -1\frac{1}{4}

x > 0 : -1\frac{1}{4} < x < 1\frac{1}{4}

This matches choice 4.