Calculate Area Under y=-(x-4)²+1: Quadratic Function Problem

To solve this problem, follow these steps:

• Step 1: Identify the vertex form of the parabola, which is given as $y = -(x-4)^2 + 1$.
• Step 2: Convert the equation to find the x-intercepts (roots), set $y = 0$:

We set the equation equal to zero:

$-(x-4)^2 + 1 = 0$

$-(x-4)^2 = -1$

$(x-4)^2 = 1$

• Step 3: Solve for $x$ by taking the square root of both sides:

$x-4 = \pm 1$

This gives two solutions for $x$:

$x = 4 + 1 = 5$

$x = 4 - 1 = 3$

• Step 4: Determine the interval where $y$ is positive by analyzing intervals around the roots 3 and 5.
• Check intervals: $x < 3$, $3 < x < 5$, and $x > 5$.
• Since the vertex (highest point) occurs at $x = 4$, we know the parabola is positive between its roots from 3 to 5.

Thus, the segment of the domain where the parabola lies above the x-axis is:

$3 < x < 5$

The correct choice is: $3 < x < 5$.