Solve y=-(x-16)²: Finding Values Where Function is Positive

Look at the function below:

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

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

$f(x) > 0$

To solve this problem, let's analyze the function $y = -\left(x-16\right)^2$.

• The function is in vertex form $y = - (x - 16)^2$, which suggests it is a quadratic function opening downwards because the coefficient of the squared term is negative.
• The vertex of the function is at the point $(16, 0)$, meaning the maximum point of the parabola is at $y = 0$.
• For a parabola that opens downward, the value of $y$ is always less than or equal to the value at the vertex. Here, since the vertex itself is at 0, there are no values of $x$ for which $y = - (x - 16)^2$ is greater than zero.

Therefore, there are no values of $x$ for which $f(x) > 0$.

In conclusion, the solution to the problem is: True for no values of $x$.