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

To solve this problem, we will apply the following steps:

• Step 1: Find the roots of the given quadratic function.
• Step 2: Determine the sign of the quadratic between and beyond the roots.
• Step 3: Conclude which values of $x$ make the quadratic positive.

Let's start with Step 1: Find the roots of the function $y = x^2 - 4x - 4$.
We use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 1, b = -4, c = -4$, we get:
$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-4) = 16 + 16 = 32$

Thus, the roots are:
$x = \frac{-(-4) \pm \sqrt{32}}{2(1)} = \frac{4 \pm \sqrt{32}}{2}$

Simplifying further gives us $\sqrt{32} = 4\sqrt{2}$, so:
$x = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2}$

The roots are $x = 2 + 2\sqrt{2}$ and $x = 2 - 2\sqrt{2}$.

Step 2: Determine the sign of the quadratic.
Since the parabola opens upward (coefficient of $x^2$ is positive), it is below the x-axis between the roots and above the x-axis outside the roots.

Step 3: Conclude values for which $f(x) > 0$.
$f(x) > 0$ for $x < 2 - 2\sqrt{2}$ or $x > 2 + 2\sqrt{2}$.

Finally, the solution to the problem is: $x > 2 + 2\sqrt{2}$ or $x < 2 - 2\sqrt{2}$.