Solving y = (1/2)x² - 8: Finding Where Function is Positive

Find the positive and negative domains of the function below:

$y=\frac{1}{2}x^2-8$

Determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we need to find when the quadratic function $y = \frac{1}{2}x^2 - 8$ is positive.

We begin by solving the equation $\frac{1}{2}x^2 - 8 = 0$ to find the critical points where the function changes sign.

• Step 1: Solve $\frac{1}{2}x^2 - 8 = 0$.
  Multiply through by 2 to clear the fraction: $x^2 - 16 = 0$.
  Factor or solve: $x^2 = 16$.
  Taking square roots gives $x = 4$ and $x = -4$.
• Step 2: Determine the sign of the quadratic in the intervals: $(-\infty, -4)$, $(-4, 4)$, and $(4, \infty)$.
• Step 3: Check each interval by selecting test points:
  • Interval $(-\infty, -4)$: Choose a test point, say $x = -5$:
    $y = \frac{1}{2}(-5)^2 - 8 = \frac{1}{2}(25) - 8 = 12.5 - 8 = 4.5$, which is positive.
  • Interval $(-4, 4)$: Choose a test point, say $x = 0$:
    $y = \frac{1}{2}(0)^2 - 8 = 0 - 8 = -8$, which is negative.
  • Interval $(4, \infty)$: Choose a test point, say $x = 5$:
    $y = \frac{1}{2}(5)^2 - 8 = \frac{1}{2}(25) - 8 = 12.5 - 8 = 4.5$, which is positive.
• Conclusion: $y > 0$ for $x \in (-\infty, -4) \cup (4, \infty)$.
  Thus, the solution is: $x < -4$ or $x > 4$.

Therefore, the values of $x$ for which $f(x) > 0$ are $x > 4$ or $x < -4$.