Solve y=-(4x+32)²: Finding Values Where Function is Negative

Look at the function below:

$y=-\left(4x+32\right)^2$

Then determine for which values of $x$ the following is true:

$f(x) < 0$

To solve this problem, we'll follow these steps:

• Step 1: Analyze the structure of the function.
• Step 2: Determine when the squared term is zero.
• Step 3: Apply the negative sign.
• Step 4: Conclude when the function is negative.

Now, let's work through each step:

Step 1: Analyze the function $y = -\left(4x + 32\right)^2$. Note that the expression $(4x + 32)$ is squared, and any square of a real number is non-negative.

Step 2: For $(4x + 32)^2$ to equal zero, solve $4x + 32 = 0$.

Solve $4x + 32 = 0$:

$4x = -32 \quad \Rightarrow \quad x = -8$

Step 3: Apply the negative sign: Since $(4x + 32)^2$ can only be zero or positive, applying the negative sign results in:

$-\left(4x + 32\right)^2 \leq 0$

Step 4: Conclude that $f(x) < 0$ for all values except when $(4x + 32)^2$ is zero. Thus, the function is negative for all $x$ except $x = -8$.

Therefore, the solution is $x \ne -8$.