Solve (2x-16)² > 0: Finding Values Where Function is Positive

Look at the function below:

$y=\left(2x-16\right)^2$

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

$f(x) > 0$

To determine when the function $y = (2x - 16)^2$ is greater than zero, we observe the following:

• The function's expression, $(2x - 16)^2$, is a square and thus always non-negative ($\geq 0$).
• The expression will equate to zero when the inside term is zero: $2x - 16 = 0$.
• Solve the equation $2x - 16 = 0$ to find $x = 8$.
• Therefore, $(2x - 16)^2 = 0$ only at $x = 8$.
• For $y > 0$, $(2x - 16)^2$ must be greater than zero, which occurs for all $x$ except $x = 8$.

The solution is $x \ne 8$, which means $y$ is positive for all $x$ except $x = 8$.