Solve y = 3/4x²: Finding Positive Value Domains

Even though $\frac{3}{4}$ is positive, we need f(x) > 0, not just ≥ 0. When $x = 0$, we get $\frac{3}{4}(0)^2 = 0$, which is not greater than zero!

$x^2 = 0$ only when x = 0. For any other real number (positive or negative), $x^2 > 0$. For example: $(-3)^2 = 9 > 0$ and $5^2 = 25 > 0$.

> 0 means 'greater than zero' (positive), while ≥ 0 means 'greater than or equal to zero' (non-negative). Since we want f(x) > 0, we exclude x = 0 where the function equals zero.

The coefficient $\frac{3}{4}$ is positive, so it doesn't change the sign of $x^2$. If the coefficient were negative, like $-\frac{3}{4}x^2$, then the function would be negative (except at x = 0).

Absolutely! Try x = 1: $\frac{3}{4}(1)^2 = \frac{3}{4} > 0 ✓$
Try x = -2: $\frac{3}{4}(-2)^2 = \frac{3}{4}(4) = 3 > 0 ✓$
Try x = 0: $\frac{3}{4}(0)^2 = 0$ (not > 0)