Solve y = -0.9x²: Finding Values Where Function is Positive

Given the function:

$y=-0.9x^2$

Determine for which values of x $f(x) > 0$ holds

To solve this problem, let's apply the analysis and reasoning as follows:

Step 1: Analyze the function $y = -0.9x^2$. This is a quadratic function of the form $ax^2$ where $a = -0.9$. Since $a < 0$, the parabola opens downwards.

Step 2: Consider the values of $y$. For a parabola opening downwards, the peak (vertex) is at its maximum, and from this point, the parabola decreases, stretching indefinitely in the negative direction of $y$.

Step 3: Determine the maximum value. In the quadratic function $y = -0.9x^2$, the vertex at $x = 0$ gives the maximum value of $y$, which is $y = 0$ since $y = -0.9 \times 0^2 = 0$.

Step 4: Examine the entire function's range. Since beyond the vertex $y = 0$, the values of $y$ are strictly negative, there are no values of $x$ for which $f(x) > 0$.

Conclusion: Because the function $y = -0.9x^2$ has its only non-negative point at $x = 0$ (where it equals zero) and decreases for all other values of $x$, there are no $x$-values that make the function positive (i.e., $f(x) > 0$ is never true). Therefore, no $x$ satisfies the condition $f(x) > 0$.

The correct choice is 3: No x.