The function you provided is:

$f(t) = t^2 + 9$

This represents a quadratic function, where the variable $t$ is squared and then added to 9. Here are some useful insights:

1. Domain: The domain of the function is all real numbers, since there are no restrictions on $t$.

   $\text{Domain}: (-\infty, \infty)$

2. Range: Since $t^2$ is always non-negative (i.e., $t^2 \geq 0$), the minimum value of the function is when $t = 0$. At that point, $f(t) = 9$, and as $t^2$ increases, so does $f(t)$.

   $\text{Range}: [9, \infty)$

3. Vertex: The vertex of this parabola is at $t = 0$, since the function takes the form $f(t) = t^2 + 9$, which is already in vertex form $f(t) = (t - 0)^2 + 9$.

   $\text{Vertex}: (0, 9)$

4. Shape and Direction: The parabola opens upwards since the coefficient of $t^2$ is positive.

5. Y-Intercept: The y-intercept occurs when $t = 0$, so:

   $f(0) = 0^2 + 9 = 9$

Would you like to go into more details or have any questions?

Here are 5 relative questions:

1. What happens to $f(t)$ as $t$ becomes very large or very negative?
2. Can you find the x-intercepts of the function?
3. How would you differentiate $f(t) = t^2 + 9$?
4. What transformations could change the graph of $f(t)$?
5. What would happen if we added a linear term to the function, say $f(t) = t^2 + 2t + 9$?

Tip: The vertex form of a quadratic equation helps easily identify the minimum or maximum point of a parabola.