Solve y = (1/2)x² - 12½: Finding Values Where Function is Positive

Q: Why do I need to find where the function equals zero first?

The zeros are where the parabola crosses the x-axis! These points divide the number line into regions where the function is either positive or negative. Without finding them, you can't determine the intervals.

Q: How do I know which intervals to test?

The zeros $ x = -5 $ and $ x = 5 $ create three regions: left of -5, between -5 and 5, and right of 5. Pick any point in each region to test!

Q: What if I pick the wrong test point?

Any point within an interval will give the same sign! If $ x = -6 $ gives a positive result, then $ x = -10 $ or $ x = -7 $ will too.

Q: Why is the answer 'x > 5 or x < -5' and not 'and'?

The function is positive in two separate regions that don't connect! Use 'or' when the solution includes multiple disconnected intervals, and 'and' only when x must satisfy both conditions simultaneously.

Q: Do I include the boundary points x = -5 and x = 5?

No! At $ x = ±5 $, the function equals zero, not greater than zero. Use strict inequality symbols () to exclude these boundary points from your solution.

Q: How can I visualize this problem?

Picture a U-shaped parabola opening upward! It touches the x-axis at $ x = -5 $ and $ x = 5 $, and is above the x-axis (positive) outside these points.

Quadratic Inequalities with Interval Testing

Given the function:

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

Determine for which values of X the following holds:

$f(x) > 0$

❤️ 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

Given the function:

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

Determine for which values of X the following holds:

$f(x) > 0$

Step-by-step solution

To solve the problem, let's determine the intervals where the function $y = \frac{1}{2}x^2 - 12.5$ is greater than zero.

Step 1: Solve $\frac{1}{2}x^2 - 12.5 = 0$ .

The equation can be rewritten as:

\frac{1}{2}x^2 = 12.5

Multiply both sides by 2 to eliminate the fraction:

x^2 = 25

Take the square root of both sides to solve for $x$ :

x = \pm 5

Step 2: Analyze the intervals created by the roots $x = -5$ and $x = 5$ .

The roots divide the number line into three intervals: $x < -5$ , $-5 < x < 5$ , and $x > 5$ .

Step 3: Test points within these intervals to check where $\frac{1}{2}x^2 - 12.5 > 0$ .

- For $x < -5$ , pick $x = -6$ :

\frac{1}{2}(-6)^2 - 12.5 = 18 - 12.5 = 5.5 \, (>0)

- For $-5 < x < 5$ , pick $x = 0$ :

\frac{1}{2}(0)^2 - 12.5 = -12.5 \, (<0)

- For $x > 5$ , pick $x = 6$ :

\frac{1}{2}(6)^2 - 12.5 = 18 - 12.5 = 5.5 \, (>0)

This analysis shows the function is positive when:

$x < -5$
$x > 5$

Thus, the solution is:

$x > 5$ or $x < -5$

Final Answer

$x > 5$ or $x < -5$

Key Points to Remember

Essential concepts to master this topic

Rule: Find zeros first by setting quadratic equal to zero
Technique: Test points in intervals: x = -6 gives 18 - 12.5 = 5.5 > 0
Check: Verify boundary values at x = ±5 make function equal zero ✓

Common Mistakes

Avoid these frequent errors

Solving the inequality without finding zeros first
Don't try to solve $\frac{1}{2}x^2 - 12.5 > 0$ directly = confusing mess! This skips the crucial step that reveals where the function changes sign. Always find zeros by setting the function equal to zero, then test intervals between these zeros.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the $$ x $$ -axis.

The parabola's vertex is marked A.

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 where the function equals zero first?

The zeros are where the parabola crosses the x-axis! These points divide the number line into regions where the function is either positive or negative. Without finding them, you can't determine the intervals.

How do I know which intervals to test?

The zeros $x = -5$ and $x = 5$ create three regions: left of -5, between -5 and 5, and right of 5. Pick any point in each region to test!

What if I pick the wrong test point?

Any point within an interval will give the same sign! If $x = -6$ gives a positive result, then $x = -10$ or $x = -7$ will too.

Why is the answer 'x > 5 or x < -5' and not 'and'?

The function is positive in two separate regions that don't connect! Use 'or' when the solution includes multiple disconnected intervals, and 'and' only when x must satisfy both conditions simultaneously.

Do I include the boundary points x = -5 and x = 5?

No! At $x = ±5$ , the function equals zero, not greater than zero. Use strict inequality symbols (<, >) to exclude these boundary points from your solution.

How can I visualize this problem?

Picture a U-shaped parabola opening upward! It touches the x-axis at $x = -5$ and $x = 5$ , and is above the x-axis (positive) outside these points.

🌟 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