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

Quadratic Inequalities with Interval Testing

Find the positive and negative domains of the function below:

y=14x29 y=\frac{1}{4}x^2-9

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

f(x)>0 f(x) > 0

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the positive and negative domains of the function below:

y=14x29 y=\frac{1}{4}x^2-9

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

f(x)>0 f(x) > 0

2

Step-by-step solution

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

  • Step 1: Find where the function y=14x29 y = \frac{1}{4}x^2 - 9 is greater than zero.
  • Step 2: Set the inequality 14x29>0\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 14x29>0 \frac{1}{4}x^2 - 9 > 0 .
Step 2: Set the inequality: 14x29>0\frac{1}{4}x^2 - 9 > 0.
Start by multiplying through by 4 to clear the fraction: x236>0x^2 - 36 > 0.
Step 3: Solve the equation x236=0x^2 - 36 = 0 to find the zeros.
Solve for xx: x2=36x^2 = 36, which gives x=6x = 6 and x=6x = -6.

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

Test an x-value in each region to determine if f(x)>0f(x) > 0:
- For x<6x < -6, choose x=7x = -7: (7)236>0(-7)^2 - 36 > 0 gives 4936=1349 - 36 = 13, so positive.
- For 6<x<6-6 < x < 6, choose x=0x = 0: 0236=360^2 - 36 = -36, so negative.
- For x>6x > 6, choose x=7x = 7: 7236>07^2 - 36 > 0 gives 4936=1349 - 36 = 13, so positive.

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

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

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

.

3

Final Answer

x>6 x > 6 or x<6 x < -6

Key Points to Remember

Essential concepts to master this topic
  • Rule: Find zeros first, then test intervals between critical points
  • Technique: For 14x29>0 \frac{1}{4}x^2 - 9 > 0 , multiply by 4: x236>0 x^2 - 36 > 0
  • Check: Test x=7 x = 7 : 7236=13>0 7^2 - 36 = 13 > 0

Common Mistakes

Avoid these frequent errors
  • Testing only one interval or ignoring critical points
    Don't just solve x2>36 x^2 > 36 and assume x>6 x > 6 = wrong answer! This misses the negative solution. Always find both zeros (x=±6 x = ±6 ) and test all three intervals: x<6 x < -6 , 6<x<6 -6 < x < 6 , and x>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 \( f\left(x\right) > 0 \).

AAABBBCCCX

FAQ

Everything you need to know about this question

Why do I need to test intervals instead of just solving the inequality?

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Because quadratic functions change from positive to negative (or vice versa) at their zeros! Testing intervals tells you exactly where the function is positive or negative, not just where it equals zero.

How do I know which test values to pick?

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Pick any value inside each interval. For x<6 x < -6 , try x=7 x = -7 . For 6<x<6 -6 < x < 6 , try x=0 x = 0 . The exact number doesn't matter as long as it's in the right interval!

What if I get confused about the inequality direction?

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Always substitute your test values back into the original inequality 14x29>0 \frac{1}{4}x^2 - 9 > 0 . If the result is positive, that interval works. If negative, it doesn't!

Why multiply by 4 to clear the fraction?

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Multiplying by 4 makes the numbers easier to work with! x236=0 x^2 - 36 = 0 is much simpler than 14x29=0 \frac{1}{4}x^2 - 9 = 0 , and since 4 is positive, it doesn't change the inequality direction.

Can the answer include the boundary points x = 6 and x = -6?

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Not for this problem! Since we need f(x)>0 f(x) > 0 (strictly greater), and f(6)=f(6)=0 f(6) = f(-6) = 0 , these boundary points are excluded from our solution.

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