Quadratic Function Analysis: Finding Domains and Negative Regions of (1/2)x² - 12½

Q: Why does the parabola open upward if the coefficient of x² is positive?

Since the coefficient of $ x^2 $ is $ \frac{1}{2} > 0 $, this parabola opens upward. This means it's negative between the roots and positive outside them.

Q: How do I convert the mixed number 12½ to work with?

Convert $ 12\frac{1}{2} $ to $ \frac{25}{2} $ by doing 12 × 2 + 1 = 25, then put over 2. This makes calculations much easier!

Q: What if I get confused about which intervals to test?

The roots divide the number line into sections. With roots at -5 and 5, you get three intervals: $ (-\infty, -5) $, $ (-5, 5) $, and $ (5, \infty) $. Pick any number from each section to test.

Q: Why do I need to test points instead of just looking at the graph?

Testing points gives you exact mathematical proof! While sketching helps visualize, substituting actual values like x = 0 shows definitively that \( y = -12\frac{1}{2} in the middle interval.

Q: Can I solve this by factoring first?

Absolutely! Setting $ \frac{1}{2}x^2 - \frac{25}{2} = 0 $ leads to $ x^2 - 25 = 0 $, which factors as $ (x-5)(x+5) = 0 $. This is often the fastest approach.

Q: What does f(x) < 0 mean in practical terms?

It means you're looking for x-values where the parabola is below the x-axis. Since this parabola opens upward, it dips below zero between its two roots at x = -5 and x = 5.

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:

$y=\frac{1}{2}x^2-12\frac{1}{2}$

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

$f(x) < 0$

2

Step-by-step solution

To find the values of $x$ for which the function $y = \frac{1}{2}x^2 - 12\frac{1}{2}$ is negative, follow these steps:

Step 1: Solve the equation
$y = \frac{1}{2}x^2 - \frac{25}{2} = 0$ . Multiply through by 2 for simplicity:
$x^2 - 25 = 0$

Step 2: Factor the quadratic expression:
$(x - 5)(x + 5) = 0$

Step 3: Find the roots:
$x = 5$ and $x = -5$

Step 4: Test intervals around these roots to find where $y < 0$ .
The intervals to test are $(-\infty, -5)$ , $(-5, 5)$ , and $(5, \infty)$ .
Step 5: Evaluate a test point within each interval:

For $x = 0$ in interval $(-5, 5)$ :
$y = \frac{1}{2}(0)^2 - \frac{25}{2} = -\frac{25}{2}$ , which is less than 0.

For $x = 6$ in interval $(5, \infty)$ :
$y = \frac{1}{2}(6)^2 - \frac{25}{2} = \frac{11}{2}$ , which is greater than 0.

For $x = -6$ in interval $(-\infty, -5)$ :
$y = \frac{1}{2}(-6)^2 - \frac{25}{2} = \frac{11}{2}$ , which is greater than 0.

Thus, the function is negative in the interval $-5 < x < 5$ .

Therefore, the solution is $-5 < x < 5$ .

3

Final Answer

$-5 < x < 5$

Key Points to Remember

Essential concepts to master this topic

Zeros First: Find x-intercepts by setting quadratic equal to zero
Test Method: Evaluate y = ½(0)² - 12½ = -12½ in middle interval
Verify Signs: Check intervals (-∞, -5), (-5, 5), (5, ∞) for negative values ✓

Common Mistakes

Avoid these frequent errors

Forgetting to test intervals between roots
Don't just find the roots x = ±5 and assume the function is always negative between them = wrong conclusion! Parabolas change signs at their roots, but you must test actual values to determine which intervals are positive or negative. Always substitute test points from each interval into the original function.

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 does the parabola open upward if the coefficient of x² is positive?

+

Since the coefficient of $x^2$ is $\frac{1}{2} > 0$ , this parabola opens upward. This means it's negative between the roots and positive outside them.

How do I convert the mixed number 12½ to work with?

+

Convert $12\frac{1}{2}$ to $\frac{25}{2}$ by doing 12 × 2 + 1 = 25, then put over 2. This makes calculations much easier!

What if I get confused about which intervals to test?

+

The roots divide the number line into sections. With roots at -5 and 5, you get three intervals: $(-\infty, -5)$ , $(-5, 5)$ , and $(5, \infty)$ . Pick any number from each section to test.

Why do I need to test points instead of just looking at the graph?

+

Testing points gives you exact mathematical proof! While sketching helps visualize, substituting actual values like x = 0 shows definitively that $y = -12\frac{1}{2} < 0$ in the middle interval.

Can I solve this by factoring first?

+

Absolutely! Setting $\frac{1}{2}x^2 - \frac{25}{2} = 0$ leads to $x^2 - 25 = 0$ , which factors as $(x-5)(x+5) = 0$ . This is often the fastest approach.

What does f(x) < 0 mean in practical terms?

+

It means you're looking for x-values where the parabola is below the x-axis. Since this parabola opens upward, it dips below zero between its two roots at x = -5 and x = 5.

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Quadratic Function Analysis: Finding Domains and Negative Regions of (1/2)x² - 12½

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