Find Domains of y=-(x-1⁴/₅)² + 1: Quadratic Function Analysis

To solve this problem, let's consider the function $y = -\left(x - \frac{9}{5}\right)^2 + 1$ expressed in vertex form as $(x - h)^2 + k$, where $h = \frac{9}{5}$ and $k = 1$. The vertex is at $\left(\frac{9}{5}, 1\right)$.

Since the coefficient of the squared term is negative ($a = -1$), the parabola opens downwards. This means the maximum value of the function is at the vertex $k = 1$ and decreases on either side.

Now, solve for when the function is positive ($y > 0$):

• The parabola is positive when its value is greater than the x-axis ($y = 0$). Set the inequality:

$-\left(x - \frac{9}{5}\right)^2 + 1 > 0$

Simplifying, we get:

$1 > \left(x - \frac{9}{5}\right)^2$

This suggests:

$-1 < \left(x - \frac{9}{5}\right) < 1$

Solving these inequalities:

• For $-1 < \left(x - \frac{9}{5}\right)$:
• $x - \frac{9}{5} > -1$ implies $x > \frac{4}{5}$.
• For $\left(x - \frac{9}{5}\right) < 1$:
• $x - \frac{9}{5} < 1$ implies $x < \frac{14}{5}$.

Combining these results, the function is positive between:

$\frac{4}{5} < x < \frac{14}{5}$

Next, find where $y < 0$:

The parabola is negative outside the interval where it hits the x-axis (the interval where function is 0 or below).

The intervals for which the function is negative are:

$x < \frac{4}{5}$ and $x > \frac{14}{5}$.

Thus, the solution is:

$x > \frac{14}{5}$ or $x < 0 : x < \frac{4}{5}$

$x > 0 : \frac{4}{5} < x < \frac{14}{5}$

Therefore, the correct answer is Choice 2.