Solve y = x²-4x-4: Finding Values Where Function is Negative

We are tasked with finding the values of $x$ for which the function $y = x^2 - 4x - 4$ satisfies $f(x) < 0$.

To solve this, follow these steps:

• Find the roots of the function:
  The roots of a quadratic equation $ax^2 + bx + c = 0$ can be found using the quadratic formula: $\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For our function, $a = 1$, $b = -4$, and $c = -4$. Plug these into the quadratic formula: $\ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-4)}}{2 \times 1}$ $\ x = \frac{4 \pm \sqrt{16 + 16}}{2}$ $\ x = \frac{4 \pm \sqrt{32}}{2}$ $\ x = \frac{4 \pm 4\sqrt{2}}{2}$ $\ x = 2 \pm 2\sqrt{2}$ Thus, the roots are $x = 2 + 2\sqrt{2}$ and $x = 2 - 2\sqrt{2}$.
• Determine intervals:
  The function changes sign at these roots. The intervals to consider are $(-\infty, 2 - 2\sqrt{2})$, $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$, and $(2 + 2\sqrt{2}, \infty)$.
• Test each interval for $f(x) < 0$:
  Since the parabola opens upwards (as $a > 0$), the function will be negative between the roots: - In $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$, the function is less than zero. Thus, $f(x) < 0$ for $2 - 2\sqrt{2} < x < 2 + 2\sqrt{2}$.

Therefore, the solution is $\boxed{2 - 2\sqrt{2} < x < 2 + 2\sqrt{2}}$.