Solving y = -x² + 1: Finding Values Where Function is Negative

Q: Why do I need to find the roots first before solving the inequality?

The roots divide the number line into intervals where the function doesn't change sign. Finding where $ y = 0 $ gives you the boundary points, then you test each interval to see if it's positive or negative.

Q: How do I know which test points to choose?

Pick any number inside each interval! For $ x , try \( x = -2 $. For $ -1 , try \( x = 0 $. The exact number doesn't matter - just make sure it's clearly in the interval.

Q: What does the negative coefficient tell me about this parabola?

The negative coefficient of $ x^2 $ means this parabola opens downward. It's positive between the roots and negative outside the roots - the opposite of upward-opening parabolas!

Q: Why is the answer written as 'or' instead of 'and'?

We use 'or' because $ x $ can't be both greater than 1 AND less than -1 at the same time! The function is negative in two separate regions, so we need 'or' to include both.

Q: Do I include the boundary points x = 1 and x = -1?

No! At the roots, the function equals zero, not less than zero. Since we want \( f(x) (strictly less than), we use open intervals that don't include the endpoints.

Quadratic Inequalities with Sign Analysis

Look at the following function:

$y=-x^2+1$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=-x^2+1$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

Step-by-step solution

To solve this problem, we need to determine where the quadratic function $y = -x^2 + 1$ is negative.

First, identify the roots of the equation by setting $y = 0$ :
$-x^2 + 1 = 0$ simplifies to $-x^2 = -1$ or $x^2 = 1$ .
Solving $x^2 = 1$ gives us the roots: $x = 1$ and $x = -1$ .
Now, examine the intervals defined by these roots: $(-\infty, -1)$ , $(-1, 1)$ , and $(1, \infty)$ .
Test each interval:
- For $x < -1$ , choose $x = -2$ :
  $y = -(-2)^2 + 1 = -4 + 1 = -3$ , which is negative.
- For $-1 < x < 1$ , choose $x = 0$ :
  $y = -(0)^2 + 1 = 1$ , which is positive.
- For $x > 1$ , choose $x = 2$ :
  $y = -(2)^2 + 1 = -4 + 1 = -3$ , which is negative.
Thus, the function is negative for $x < -1$ and $x > 1$ .

Therefore, the values of $x$ for which $f(x) < 0$ are $x < -1$ or $x > 1$ .

Final Answer

$x > 1$ or $x < -1$

Key Points to Remember

Essential concepts to master this topic

Roots: Find where function equals zero by setting $-x^2 + 1 = 0$
Technique: Test intervals: at $x = 2$ , get $-4 + 1 = -3 < 0$
Check: Verify boundaries: function equals zero at $x = ±1$ , negative outside ✓

Common Mistakes

Avoid these frequent errors

Testing only one interval or guessing the sign pattern
Don't assume the parabola is negative everywhere or guess based on the coefficient! This leads to wrong intervals like $x < 0$ instead of the correct answer. Always find the roots first, then test a point in each interval to determine where the function is actually negative.

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 roots first before solving the inequality?

The roots divide the number line into intervals where the function doesn't change sign. Finding where $y = 0$ gives you the boundary points, then you test each interval to see if it's positive or negative.

How do I know which test points to choose?

Pick any number inside each interval! For $x < -1$ , try $x = -2$ . For $-1 < x < 1$ , try $x = 0$ . The exact number doesn't matter - just make sure it's clearly in the interval.

What does the negative coefficient tell me about this parabola?

The negative coefficient of $x^2$ means this parabola opens downward. It's positive between the roots and negative outside the roots - the opposite of upward-opening parabolas!

Why is the answer written as 'or' instead of 'and'?

We use 'or' because $x$ can't be both greater than 1 AND less than -1 at the same time! The function is negative in two separate regions, so we need 'or' to include both.

Do I include the boundary points x = 1 and x = -1?

No! At the roots, the function equals zero, not less than zero. Since we want $f(x) < 0$ (strictly less than), we use open intervals that don't include the endpoints.

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