Calculate Area Under y=-(x+3)²+25: Quadratic Function Analysis

To solve this problem, follow these steps:

• Step 1: Find the x-intercepts by solving the equation $-(x+3)^2 + 25 = 0$.
• Step 2: Determine when the expression $-(x+3)^2 + 25$ is positive.
• Step 3: Verify the correct answer choice based on the solution.

Now let's work through each step:

Step 1: Solve $-(x+3)^2 + 25 = 0$.

Rewriting, we have:

$-(x+3)^2 = -25$

$(x+3)^2 = 25$

This gives $x+3 = \pm 5$.

Solving these, we find $x = 2$ and $x = -8$.

Step 2: Determine when $-(x+3)^2 + 25 > 0$.

We know that the parabola is downward opening, so it will be positive between the roots found, $-8$ and $2$.

Thus, the positive interval is $-8 < x < 2$.

Step 3: Verify the correct answer choice.

The correct answer, matching the solution above, is choice 1: $-8 < x < 2$.

Therefore, the positive area of the function occurs in the interval $-8 < x < 2$.