Find Negative Area: Solving y+1=(x+3)² Parabola Problem

To find the negative area of the given parabola, we need to determine where the function $y = (x + 3)^2 - 1$ is below the x-axis. This corresponds to finding when the parabola is negative.

First, let's set the equation $y = (x + 3)^2 - 1$ equal to zero and solve for $x$ to find the roots:

• Set $(x + 3)^2 - 1 = 0$.

• This simplifies to $(x + 3)^2 = 1$.

• Taking square roots gives $x + 3 = \pm 1$.

• Thus, $x + 3 = 1$ gives $x = -2$, and $x + 3 = -1$ gives $x = -4$.

The roots are $x = -4$ and $x = -2$. The parabola opens upwards since the coefficient of $x^2$ is positive. Therefore, it is negative (below the x-axis) between these roots.

To verify, choose a test point between the roots, say $x = -3$:

• Plug into the equation: $y = ((-3) + 3)^2 - 1 = 0 - 1 = -1$, which is negative.

Therefore, the function is negative on the interval -4 < x < -2 .

The correct answer is -4 < x < -2 .