Solve -3x² + 12x > 0: Finding Positive Function Values

To determine where the function $f(x) = -3x^2 + 12x$ is greater than zero, we need to find the roots of the equation.

Step 1: Set the function equal to zero to find the zeros or roots:

$-3x^2 + 12x = 0$

Step 2: Factor the equation:

$-3x(x - 4) = 0$

Setting each factor equal to zero gives us the roots:

• $-3x = 0 \rightarrow x = 0$
• $x - 4 = 0 \rightarrow x = 4$

Step 3: Since the parabola opens downwards (as the coefficient of $x^2$ is negative), $f(x) > 0$ for $x$ values between the roots. Thus, the function is positive between $x = 0$ and $x = 4$.

Therefore, the solution is the interval $0 < x < 4$.

In conclusion, the values of $x$ for which the function is greater than zero are $0 < x < 4$.