Solve: When is -3x²+12x Less Than Zero? Finding Negative Regions

The given quadratic function is $y = -3x^2 + 12x$. We are tasked with finding for which values of $x$ this function is negative, i.e., $y < 0$.

First, identify the roots of the quadratic by solving the equation:

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

Factor out common terms:

$x(-3x + 12) = 0$

This gives us two solutions or critical points:

$x = 0 \quad \text{and} \quad -3x + 12 = 0$

Solve for $x$ in the second equation:

$-3x + 12 = 0 \Rightarrow -3x = -12 \Rightarrow x = \frac{-12}{-3} \Rightarrow x = 4$

The roots of the quadratic are $x = 0$ and $x = 4$. These roots divide the real number line into three intervals:

• $x < 0$
• $0 < x < 4$
• $x > 4$

To find where the function is negative, evaluate the sign of $y$ in these intervals:

• For $x < 0$: Choose $x = -1$. Then, $-3(-1)^2 + 12(-1) = -3 - 12 = -15$, which is negative.
• For $0 < x < 4$: Choose $x = 2$. Then, $-3(2)^2 + 12(2) = -12 + 24 = 12$, which is positive.
• For $x > 4$: Choose $x = 5$. Then, $-3(5)^2 + 12(5) = -75 + 60 = -15$, which is negative.

The function $y = -3x^2 + 12x$ is negative for $x < 0$ and $x > 4$.

Therefore, the values of $x$ that satisfy $f(x) < 0$ are:

$x < 0$ and $x > 4$.