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

Look at the following function:

$y=-x^2+9$

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

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

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 $x < -3$.