1. **State the problem:** Solve the inequality $$-9x^2 + 12x \leq 4$$ using sign analysis. 2. **Rewrite the inequality:** Move all terms to one side to set the inequality to zero: $$-9x^2 + 12x - 4 \leq 0$$ 3. **Identify the quadratic function:** Let $$f(x) = -9x^2 + 12x - 4$$. 4. **Find the roots of the quadratic equation $$f(x) = 0$$:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = -9$$, $$b = 12$$, and $$c = -4$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 12^2 - 4(-9)(-4) = 144 - 144 = 0$$ Since $$\Delta = 0$$, there is one repeated root: $$x = \frac{-12}{2 \times -9} = \frac{-12}{-18} = \frac{2}{3}$$ 5. **Sign analysis:** Since the quadratic has a repeated root at $$x = \frac{2}{3}$$ and the leading coefficient $$a = -9 < 0$$, the parabola opens downward. - For $$x < \frac{2}{3}$$, $$f(x) < 0$$. - At $$x = \frac{2}{3}$$, $$f(x) = 0$$. - For $$x > \frac{2}{3}$$, $$f(x) < 0$$. 6. **Interpret the inequality $$f(x) \leq 0$$:** Since $$f(x)$$ is less than or equal to zero everywhere except possibly at the vertex, the solution is all real numbers: $$\boxed{(-\infty, \infty)}$$ But let's verify the value at the vertex: $$f\left(\frac{2}{3}\right) = -9\left(\frac{2}{3}\right)^2 + 12\left(\frac{2}{3}\right) - 4 = -9 \times \frac{4}{9} + 8 - 4 = -4 + 8 - 4 = 0$$ Since the parabola opens downward and touches zero at $$x=\frac{2}{3}$$, the function is always less than or equal to zero. **Final answer:** $$x \in (-\infty, \infty)$$