Analyzing y = (1/3)x² - 3: Domain and Positive Values

Q: Why do I need to find the zeros first?

The zeros (where f(x) = 0) are the boundary points where the function changes from positive to negative or vice versa. They divide the number line into intervals where the sign stays constant.

Q: How do I know which intervals to test?

Once you find the zeros, they create three intervals: before the first zero, between the zeros, and after the second zero. Test one point from each interval.

Q: What if I get confused about which direction the parabola opens?

Look at the coefficient of $ x^2 $! Since $ \frac{1}{3} > 0 $, the parabola opens upward, so it's positive outside the roots and negative between them.

Q: Can I solve this by factoring instead?

Yes! You can write $ \frac{1}{3}x^2 - 3 = \frac{1}{3}(x^2 - 9) = \frac{1}{3}(x-3)(x+3) $. Since $ \frac{1}{3} > 0 $, you need (x-3)(x+3) > 0.

Q: Why is the answer 'x > 3 or x < -3' instead of 'and'?

The function is positive in two separate regions that don't connect. You can't have a single x-value that's both greater than 3 and less than -3 at the same time!

Quadratic Inequalities with Sign Analysis

Find the positive and negative domains of the function below:

$y=\frac{1}{3}x^2-3$

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

$f(x) > 0$

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

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:

$y=\frac{1}{3}x^2-3$

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

$f(x) > 0$

Step-by-step solution

We will solve the quadratic inequality $\frac{1}{3}x^2 - 3 > 0$ .

Step 1: Find roots of the equation $\frac{1}{3}x^2 - 3 = 0$ .

Rewriting, we have:

$\frac{1}{3}x^2 = 3$

Multiply both sides by 3 to clear the fraction:

$x^2 = 9$

Then take the square root of both sides:

$x = \pm 3$

Step 2: Determine the sign of $f(x) = \frac{1}{3}x^2 - 3$ on the intervals determined by the roots:

For $x < -3$ , let $x = -4$ : $f(-4) = \frac{1}{3}(-4)^2 - 3 = \frac{16}{3} - 3 = \frac{7}{3} > 0$ .
For $-3 < x < 3$ , let $x = 0$ : $f(0) = \frac{1}{3}(0)^2 - 3 = -3 < 0$ .
For $x > 3$ , let $x = 4$ : $f(4) = \frac{1}{3}(4)^2 - 3 = \frac{16}{3} - 3 = \frac{7}{3} > 0$ .

Step 3: From the interval test, we find that $f(x) > 0$ when $x < -3$ or $x > 3$ .

Therefore, the solution to the problem is $x > 3$ or $x < -3$ .

Final Answer

$x > 3$ or $x < -3$

Key Points to Remember

Essential concepts to master this topic

Rule: Find zeros first, then test intervals between zeros
Technique: Test x = 0 gives f(0) = -3, so middle interval is negative
Check: Verify x = 4 gives $f(4) = \frac{16}{3} - 3 = \frac{7}{3} > 0$ ✓

Common Mistakes

Avoid these frequent errors

Assuming quadratic is always positive
Don't think $\frac{1}{3}x^2 - 3 > 0$ for all x because it has x² = always positive! The -3 shifts the parabola down, making it negative between the roots. Always find the zeros and test intervals to determine signs.

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 do I need to find the zeros first?

The zeros (where f(x) = 0) are the boundary points where the function changes from positive to negative or vice versa. They divide the number line into intervals where the sign stays constant.

How do I know which intervals to test?

Once you find the zeros, they create three intervals: before the first zero, between the zeros, and after the second zero. Test one point from each interval.

What if I get confused about which direction the parabola opens?

Look at the coefficient of $x^2$ ! Since $\frac{1}{3} > 0$ , the parabola opens upward, so it's positive outside the roots and negative between them.

Can I solve this by factoring instead?

Yes! You can write $\frac{1}{3}x^2 - 3 = \frac{1}{3}(x^2 - 9) = \frac{1}{3}(x-3)(x+3)$ . Since $\frac{1}{3} > 0$ , you need (x-3)(x+3) > 0.

Why is the answer 'x > 3 or x < -3' instead of 'and'?

The function is positive in two separate regions that don't connect. You can't have a single x-value that's both greater than 3 and less than -3 at the same time!

🌟 Unlock Your Math Potential

Get unlimited access to all 18 The Quadratic Function questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime