Calculate Negative Area of y=(x+3)²-36: Below X-Axis Integration

Find the negative area of the function

$y=(x+3)^2-36$

To solve this problem, follow these steps:

• Step 1: Express the function in a more workable form.
• Step 2: Solve for the points where the function equals zero.
• Step 3: Determine where the function is negative within the identified intervals.

Let's begin:

Step 1: The given function is $y = (x+3)^2 - 36$. This can be rewritten as a difference of squares:

$y = \left((x+3) - 6\right)\left((x+3) + 6\right) = (x-3)(x+9)$.

Step 2: Set the function equal to zero to find the roots:

$(x-3)(x+9) = 0$.

The roots are $x = 3$ and $x = -9$.

Step 3: Test the sign of the quadratic in each interval determined by the roots:

- For $x < -9$, choose $x = -10$: $(x-3)(x+9)$ becomes negative, because $(-)(-)$ = positive.

- For $-9 < x < 3$, choose $x = 0$: $(x-3)(x+9)$ becomes negative because $(-)(+)$ = negative.

- For $x > 3$, choose $x = 10$: $(x-3)(x+9)$ becomes positive because $(+)(+)$ = positive.

Therefore, the function is negative within the interval $-9 < x < 3$.

Therefore, the correct answer is $-9 < x < 3$.