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

Q: Why is the answer an interval instead of specific values?

Because we're solving an inequality (less than), not an equation! The parabola $ y = \frac{1}{4}x^2 - 9 $ is negative for all values between -6 and 6, creating a continuous interval.

Q: How do I know if the endpoints are included?

Check what happens at $ x = 6 $ and $ x = -6 $: $ f(6) = \frac{1}{4}(36) - 9 = 0 $. Since we want f(x) (not ≤), the endpoints are not included.

Q: What if I get confused about which way the inequality goes?

Always test a point! Pick $ x = 0 $: \( f(0) = -9 ✓. Since 0 is between -6 and 6, and f(0) -6 is correct.

Q: Why does the parabola go below zero between the roots?

This parabola opens upward (positive coefficient of $ x^2 $) and has roots at x = -6 and x = 6. Between the roots, it dips below the x-axis, making f(x)

Q: Can I solve this by factoring instead?

Yes! Factor as $ \frac{1}{4}(x^2 - 36) = \frac{1}{4}(x-6)(x+6) . Since \( \frac{1}{4} > 0 $, you need \( (x-6)(x+6) , which happens between the roots.

Quadratic Inequalities with Interval Solutions

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:

$f(x) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

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:

$f(x) < 0$

Step-by-step solution

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 $-6 < x < 6$ .

Final Answer

$-6 < x < 6$

Key Points to Remember

Essential concepts to master this topic

Setup: Rearrange inequality to get quadratic expression less than zero
Technique: Solve $x^2 < 36$ gives $-6 < x < 6$
Check: Test x = 0: $\frac{1}{4}(0)^2 - 9 = -9 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting the negative solution when taking square roots
Don't solve $x^2 < 36$ as just $x < 6$ = missing half the answer! This ignores negative values that also satisfy the inequality. Always remember $x^2 < 36$ means $-6 < x < 6$ .

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of $$ x $$ where $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why is the answer an interval instead of specific values?

Because we're solving an inequality (less than), not an equation! The parabola $y = \frac{1}{4}x^2 - 9$ is negative for all values between -6 and 6, creating a continuous interval.

How do I know if the endpoints are included?

Check what happens at $x = 6$ and $x = -6$ : $f(6) = \frac{1}{4}(36) - 9 = 0$ . Since we want f(x) < 0 (not ≤), the endpoints are not included.

What if I get confused about which way the inequality goes?

Always test a point! Pick $x = 0$ : $f(0) = -9 < 0$ ✓. Since 0 is between -6 and 6, and f(0) < 0, the interval -6 < x < 6 is correct.

Why does the parabola go below zero between the roots?

This parabola opens upward (positive coefficient of $x^2$ ) and has roots at x = -6 and x = 6. Between the roots, it dips below the x-axis, making f(x) < 0.

Can I solve this by factoring instead?

Yes! Factor as $\frac{1}{4}(x^2 - 36) = \frac{1}{4}(x-6)(x+6) < 0$ . Since $\frac{1}{4} > 0$ , you need $(x-6)(x+6) < 0$ , which happens between the roots.

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