Examine the quadratic -x² + 9: Find x-values where f(x) < 0

Q: Why can't I just solve -x² + 9 < 0 like a regular equation?

Inequalities with quadratic expressions need special treatment! The parabola changes from positive to negative (or vice versa) at its zeros. You must find these boundary points first, then test which side of each boundary satisfies your inequality.

Q: How do I know which intervals to test?

The zeros divide the number line into separate regions. For $ x = ±3 $, you get three intervals: x , -3 , and x > 3. Pick any test point in each interval.

Q: What if I get the wrong sign when testing?

Double-check your arithmetic! For $ f(x) = -x^2 + 9 $, remember that negative times negative equals positive. At x = -4: $ f(-4) = -(-4)^2 + 9 = -16 + 9 = -7 $.

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

Think about it logically! A number can't be both greater than 3 and less than -3 at the same time. We use or because x can be in either region where f(x)

Q: Do I include the boundary points x = 3 and x = -3?

No! At the boundary points, $ f(3) = f(-3) = 0 $. Since we want f(x) (strictly less than), we don't include points where f(x) equals zero.

Quadratic Inequalities with Sign Analysis

Look at the following function:

$y=-x^2+9$

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:

$y=-x^2+9$

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

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

Step-by-step solution

To determine where the function $f(x) = -x^2 + 9$ is less than 0, we first need to find the points where it equals 0.

We solve the equation:

$-x^2 + 9 = 0$

This can be rearranged to:

$x^2 = 9$

Taking the square root of both sides, we get:

$x = 3$ or $x = -3$

The roots of the equation are $x = 3$ and $x = -3$ . These points divide the x-axis into three intervals: $x < -3$ , $-3 < x < 3$ , and $x > 3$ .

Next, we determine the sign of $f(x)$ in each interval:

For $x < -3$ , pick a test point like $x = -4$ :
$f(-4) = -(-4)^2 + 9 = -16 + 9 = -7$ . Here, $f(x) < 0$ .
For $-3 < x < 3$ , pick a test point like $x = 0$ :
$f(0) = -(0)^2 + 9 = 9$ . Here, $f(x) > 0$ .
For $x > 3$ , pick a test point like $x = 4$ :
$f(4) = -(4)^2 + 9 = -16 + 9 = -7$ . Here, $f(x) < 0$ .

Therefore, the function $f(x) = -x^2 + 9$ is negative for $x < -3$ and $x > 3$ .

The correct answer 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 signs in each interval
Technique: Set f(x) = 0: -x² + 9 = 0 gives x = ±3
Check: Test x = -4 and x = 4: f(-4) = -7 < 0 ✓

Common Mistakes

Avoid these frequent errors

Solving -x² + 9 < 0 algebraically without finding zeros first
Don't try to solve -x² + 9 < 0 directly by moving terms around = confusing sign errors! This skips the critical step of finding where the parabola crosses the x-axis. Always find the zeros first, then use sign analysis to determine which intervals make f(x) < 0.

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 can't I just solve -x² + 9 < 0 like a regular equation?

Inequalities with quadratic expressions need special treatment! The parabola changes from positive to negative (or vice versa) at its zeros. You must find these boundary points first, then test which side of each boundary satisfies your inequality.

How do I know which intervals to test?

The zeros divide the number line into separate regions. For $x = ±3$ , you get three intervals: x < -3, -3 < x < 3, and x > 3. Pick any test point in each interval.

What if I get the wrong sign when testing?

Double-check your arithmetic! For $f(x) = -x^2 + 9$ , remember that negative times negative equals positive. At x = -4: $f(-4) = -(-4)^2 + 9 = -16 + 9 = -7$ .

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

Think about it logically! A number can't be both greater than 3 and less than -3 at the same time. We use or because x can be in either region where f(x) < 0.

Do I include the boundary points x = 3 and x = -3?

No! At the boundary points, $f(3) = f(-3) = 0$ . Since we want f(x) < 0 (strictly less than), we don't include points where f(x) equals zero.

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