Quadratic Function Analysis: Finding Domains and Positive Values of (1/4)x² - 9

Find the positive and negative domains of the function below:

$y=\frac{1}{4}x^2-9$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we'll follow these steps:

• Step 1: Find where the function $y = \frac{1}{4}x^2 - 9$ is greater than zero.
• Step 2: Set the inequality $\frac{1}{4}x^2 - 9 > 0$ and solve.
• Step 3: Identify regions by testing intervals derived from critical points.

Now, let's work through each step:
Step 1: The function is already given, and we need to solve where $\frac{1}{4}x^2 - 9 > 0$.
Step 2: Set the inequality: $\frac{1}{4}x^2 - 9 > 0$.
Start by multiplying through by 4 to clear the fraction: $x^2 - 36 > 0$.
Step 3: Solve the equation $x^2 - 36 = 0$ to find the zeros.
Solve for $x$: $x^2 = 36$, which gives $x = 6$ and $x = -6$.

The critical points divide the number line into intervals: $x < -6$, $-6 < x < 6$, and $x > 6$.

Test an x-value in each region to determine if $f(x) > 0$:
- For $x < -6$, choose $x = -7$: $(-7)^2 - 36 > 0$ gives $49 - 36 = 13$, so positive.
- For $-6 < x < 6$, choose $x = 0$: $0^2 - 36 = -36$, so negative.
- For $x > 6$, choose $x = 7$: $7^2 - 36 > 0$ gives $49 - 36 = 13$, so positive.

Therefore, the solution for the values where the function is greater than zero is when $x > 6$ or $x < -6$.

Thus, the positive domain is when $x > 6$ or $x < -6$.

The correct choice is : $x > 6$ or $x < -6$