Solve y = (1/2)x² - 12½: Finding Values Where Function is Positive

To solve the problem, let's determine the intervals where the function $y = \frac{1}{2}x^2 - 12.5$ is greater than zero.

• Step 1: Solve $\frac{1}{2}x^2 - 12.5 = 0$.

The equation can be rewritten as:

$\frac{1}{2}x^2 = 12.5$

Multiply both sides by 2 to eliminate the fraction:

$x^2 = 25$

Take the square root of both sides to solve for $x$:

$x = \pm 5$

• Step 2: Analyze the intervals created by the roots $x = -5$ and $x = 5$.

The roots divide the number line into three intervals: $x < -5$, $-5 < x < 5$, and $x > 5$.

• Step 3: Test points within these intervals to check where $\frac{1}{2}x^2 - 12.5 > 0$.

- For $x < -5$, pick $x = -6$:

$\frac{1}{2}(-6)^2 - 12.5 = 18 - 12.5 = 5.5 \, (>0)$

- For $-5 < x < 5$, pick $x = 0$:

$\frac{1}{2}(0)^2 - 12.5 = -12.5 \, (<0)$

- For $x > 5$, pick $x = 6$:

$\frac{1}{2}(6)^2 - 12.5 = 18 - 12.5 = 5.5 \, (>0)$

This analysis shows the function is positive when:

• $x < -5$
• $x > 5$

Thus, the solution is:

$x > 5$ or $x < -5$