Solve: When is -(1/3x + 15)² Greater Than Zero?

Look at the function below:

$y=-\left(\frac{1}{3}x+15\right)^2$

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

$f(x) > 0$

To solve the problem, we must analyze the given function $y = -\left(\frac{1}{3}x + 15\right)^2$.

Firstly, as a general rule of algebra, the square of any real number is non-negative. Therefore, $\left(\frac{1}{3}x + 15\right)^2 \geq 0$ for all real values of $x$.

Secondly, the function is $y = -\left(\frac{1}{3}x + 15\right)^2$. The negative sign in front affects the entire expression, making the range of $y$ non-positive ($y \leq 0$) since the expression within the square is always non-negative. This implies every $y$ is either zero or negative.

Thus, the function $y$ will never be greater than zero because multiplying any non-negative number by $(-1)$ results in a non-positive number.

Conclusion: The function $f(x) > 0$ is true for no values of $x$.

Therefore, the correct answer choice is: True for no values of $x$.