Solve y = 1/4x² - 9: Finding Values Where Function is Negative

Find the positive and negative domains of the function below:

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

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

$f(x) < 0$

To find where $f(x) = \frac{1}{4}x^2 - 9$ is less than 0, we need to solve the inequality $\frac{1}{4}x^2 - 9 < 0$.

Step-by-step solution:

• Step 1: Rearrange the inequality: $\frac{1}{4}x^2 - 9 < 0$ Add 9 to both sides: $\frac{1}{4}x^2 < 9$ Multiply both sides by 4: $x^2 < 36$
• Step 2: Solve for $x$. Take the square root of both sides: $-6 < x < 6$ This gives us the open interval between -6 and 6.
• Step 3: Confirm by reasoning: The quadratic $\frac{1}{4}x^2$ arises from a parabola opening upwards; thus, it's less than 9 within the interval $-6 < x < 6$.

Therefore, the values for $x$ where $f(x) < 0$ are $-6 < x < 6$.