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

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 determine for which values of $x$ the function $y = -\left(4x + 32\right)^2$ is positive, we must analyze the behavior of this function.

The function $y = -\left(4x + 32\right)^2$ is a quadratic function with a negative leading coefficient (the negative sign outside the squared term). This indicates that the parabola opens downwards. Let's break down the expression:

• The expression inside the parentheses is $4x + 32$.
• Squaring any real number $\left( 4x + 32 \right)^2$ always results in a non-negative value (it is zero or positive).
• Multiplying this non-negative result by $-1$ makes $y$ non-positive (either zero or negative).

Since the smallest value $\left( 4x + 32 \right)^2$ can be is zero (when $x = -8$), and all other values will just make it negative when multiplied by $-1$, $y$ can never be greater than zero for any real number $x$.

Thus, there are no values of $x$ for which $f(x) > 0$. Consequently, the answer is:

True for no values of $x$.