To find the interval where the function $f(x) = 10 - x^2$ is decreasing, we can follow a similar procedure:

1. Find the derivative $f'(x)$: The first derivative of the function will indicate the rate of change of $f(x)$. The function is decreasing where the derivative is negative.

   $f'(x) = \frac{d}{dx} \left( 10 - x^2 \right)$

   $f'(x) = -2x$

2. Determine where $f'(x) < 0$: The function is decreasing where the derivative is less than zero.

   $-2x < 0$

   $x > 0$

Thus, the function $f(x) = 10 - x^2$ is decreasing on the interval $(0, \infty)$.

Would you like any further details or have any questions?

Here are some related questions you might consider:

1. What is the interval where the function is increasing?
2. What are the critical points of the function $f(x) = 10 - x^2$?
3. How do you determine the maximum value of this function?
4. What is the vertex of the parabola described by $f(x) = 10 - x^2$?
5. How can you determine the axis of symmetry of the function $f(x) = 10 - x^2$?

Tip: When analyzing quadratic functions like $f(x) = 10 - x^2$, remember that they represent parabolas. The sign of the $x^2$ term tells you if the parabola opens upwards (minimum point) or downwards (maximum point).