Find the Domain of y=-(x-4/9)²+1: Quadratic Function Analysis

The function given is $y = -\left(x-\frac{4}{9}\right)^2 + 1$, which is a downward-opening parabola because the coefficient of the squared term ($a = -1$) is negative.

The vertex form tells us the vertex of the parabola is at $\left(\frac{4}{9}, 1\right)$.

The function will be zero where $-\left(x-\frac{4}{9}\right)^2 + 1 = 0$. Solving this equation, we set:

$-\left(x-\frac{4}{9}\right)^2 + 1 = 0$ $\left(x-\frac{4}{9}\right)^2 = 1$

Taking the square root of both sides gives:

$x - \frac{4}{9} = \pm 1$

Thus, $x = \frac{4}{9} + 1 = \frac{13}{9}$ and $x = \frac{4}{9} - 1 = -\frac{5}{9}$.

These are the roots of the quadratic, splitting the domain into three intervals: $(-\infty, -\frac{5}{9})$, $(-\frac{5}{9}, \frac{13}{9})$, and $(\frac{13}{9}, \infty)$.

We need to test the sign of $y$ in each interval:

• For $x < -\frac{5}{9}$, choose a test point like $x = -1$. Substituting into the function yields a negative value, hence the function is negative in this interval.
• For $-\frac{5}{9} < x < \frac{13}{9}$, choose a test point like $x = 0$. Substituting into the function yields a positive value, hence the function is positive in this interval.
• For $x > \frac{13}{9}$, choose a test point like $x = 2$. Substituting into the function yields a negative value, hence the function is negative in this interval.

After analyzing these intervals, the function is positive for $-\frac{5}{9} < x < \frac{13}{9}$ and negative otherwise.

Therefore, the positive and negative domains of the function are as follows:

$x > \frac{13}{9}$ or $x < 0 : x < -\frac{5}{9}$

$x > 0 : -\frac{5}{9} < x < \frac{13}{9}$