Solve y = (1/2)x² - 2: Finding Negative Function Values

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

The zeros are where the function changes sign! For $ y = \frac{1}{2}x^2 - 2 $, the function goes from negative to positive (or vice versa) only at $ x = -2 $ and $ x = 2 $.

Q: What if I get confused about the inequality direction?

Remember: you want \( f(x) , which means negative values. After testing intervals, choose the one where your test point gives a negative result.

Q: Why is the answer an open interval (-2 < x < 2)?

Because we want $ f(x) (strictly less than). At \( x = -2 $ and $ x = 2 $, the function equals zero, not less than zero, so we use open intervals.

Q: Can I solve this by factoring instead?

Yes! $ \frac{1}{2}x^2 - 2 becomes \( \frac{1}{2}(x-2)(x+2) . Since \( \frac{1}{2} > 0 $, you need \( (x-2)(x+2) , which happens when the factors have opposite signs.

Q: What does the graph look like?

It's a parabola opening upward (since the coefficient of $ x^2 $ is positive) with vertex at $ (0, -2) $. The function is negative (below the x-axis) between the roots.

Quadratic Inequalities with Interval Testing

Find the positive and negative domains of the function below:

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

Then 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:

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

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

$f(x) < 0$

Step-by-step solution

The function $y = \frac{1}{2}x^2 - 2$ is quadratic. To find where $f(x) < 0$ , identify where $f(x) = 0$ by setting the equation equal to zero and solving for $x$ .

The equation is $\frac{1}{2}x^2 - 2 = 0$ .

Multiply through by 2 to eliminate the fraction:
$x^2 - 4 = 0$ .

Set the equation as $(x - 2)(x + 2) = 0$ to find the roots.

Solving gives roots $x = 2$ and $x = -2$ .

The function $f(x)$ will change signs at these roots $x = -2$ and $x = 2$ .

Check intervals determined by the roots to find where $f(x) < 0$ :

For $-2 < x < 2$ : Evaluate one test point like $x = 0$ in $f(x)$ :
$f(0) = \frac{1}{2}(0)^2 - 2 = -2$ , indicating $f(x) < 0$ .
For intervals $x < -2$ or $x > 2$ : Select $x = -3$ and $x = 3$ respectively, both yield positive results, therefore $f(x) > 0$ .

Ultimately, the function $f(x)$ is negative only on the interval $-2 < x < 2$ .

Therefore, the values of $x$ for which $f(x) < 0$ are such that $-2 < x < 2$ .

Final Answer

$-2 < x < 2$

Key Points to Remember

Essential concepts to master this topic

Rule: Find zeros first by setting quadratic equal to zero
Technique: Test intervals: $f(0) = \frac{1}{2}(0)^2 - 2 = -2 < 0$
Check: Verify boundary values: $f(-2) = f(2) = 0$ ✓

Common Mistakes

Avoid these frequent errors

Solving inequality without finding zeros first
Don't try to solve $\frac{1}{2}x^2 - 2 < 0$ directly = wrong intervals! This skips the critical step of finding where the function changes sign. Always find zeros first by setting the function equal to zero, then test intervals between 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 function changes sign! For $y = \frac{1}{2}x^2 - 2$ , the function goes from negative to positive (or vice versa) only at $x = -2$ and $x = 2$ .

How do I know which intervals to test?

The zeros divide the number line into sections. For zeros at $x = -2$ and $x = 2$ , test one point in each interval: $x < -2$ , $-2 < x < 2$ , and $x > 2$ .

What if I get confused about the inequality direction?

Remember: you want $f(x) < 0$ , which means negative values. After testing intervals, choose the one where your test point gives a negative result.

Why is the answer an open interval (-2 < x < 2)?

Because we want $f(x) < 0$ (strictly less than). At $x = -2$ and $x = 2$ , the function equals zero, not less than zero, so we use open intervals.

Can I solve this by factoring instead?

Yes! $\frac{1}{2}x^2 - 2 < 0$ becomes $\frac{1}{2}(x-2)(x+2) < 0$ . Since $\frac{1}{2} > 0$ , you need $(x-2)(x+2) < 0$ , which happens when the factors have opposite signs.

What does the graph look like?

It's a parabola opening upward (since the coefficient of $x^2$ is positive) with vertex at $(0, -2)$ . The function is negative (below the x-axis) between the roots.

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