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.

Understand the quadratic function f(t) = t^2 + 9, covering concepts like domain, range, vertex, and the shape of the parabola. Learn how to analyze this function's key properties, suitable for students in Grades 8-10.

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