Solve y=-7x² > 0: Finding Values Where Function is Positive

To solve this problem, let's work through the following steps:

Step 1: Analyze the inequality $-7x^2 > 0$.
Since $x^2 \geq 0$ for all real $x$ and $x^2 = 0$ when $x = 0$, we know $x^2 > 0$ when $x \neq 0$. However, the expression $-7x^2$ is always less than or equal to zero because multiplying a non-negative $x^2$ by $-7$ gives a non-positive result.
Therefore, $-7x^2 > 0$ cannot be true for any real $x$.

Step 2: Conclude based on this analysis.
The only scenario where $-7x^2$ could have been greater than zero is if we had a positive term offsetting it, which is not the case.
Hence, there are no values of $x$ for which $f(x) > 0$.

The correct answer based on the provided choices is No x.