evaluate u squared minus two for the function g(t)= square root of t square plus 4t  pus 4.

To evaluate \( u^2 - 2 \) for the function \( g(t) = \sqrt{t^2 + 4t + 4} \), we first need to find the expression for \( u \) when \( u = g(t) \).

1. **Simplify the expression for \( g(t) \)**:
   \[
   g(t) = \sqrt{t^2 + 4t + 4}
   \]
   Notice that \( t^2 + 4t + 4 \) is a perfect square trinomial. It can be factored as:
   \[
   t^2 + 4t + 4 = (t + 2)^2
   \]
   Therefore, 
   \[
   g(t) = \sqrt{(t + 2)^2}
   \]

2. **Simplify \( g(t) \)**:
   \[
   g(t) = |t + 2|
   \]
   Here, \( g(t) \) equals the absolute value of \( t + 2 \) because the square root of a square returns the absolute value of the expression.

3. **Substitute \( u = g(t) = |t + 2| \) into \( u^2 - 2 \)**:
   \[
   u^2 - 2 = (|t + 2|)^2 - 2
   \]

4. **Simplify further**:
   Since \( (|t + 2|)^2 = (t + 2)^2 \), the expression becomes:
   \[
   u^2 - 2 = (t + 2)^2 - 2
   \]
   Expand the square:
   \[
   (t + 2)^2 = t^2 + 4t + 4
   \]
   So,
   \[
   u^2 - 2 = t^2 + 4t + 4 - 2 = t^2 + 4t + 2
   \]

Thus, the expression \( u^2 - 2 \) for the given function \( g(t) \) simplifies to \( t^2 + 4t + 2 \).

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Would you like further details or have any questions? Here are some related problems you might consider:

1. What is the derivative of \( g(t) = \sqrt{t^2 + 4t + 4} \)?
2. Evaluate \( u^2 - 2 \) for a specific value of \( t \) in the function \( g(t) \).
3. How does the expression change if \( g(t) \) is \( \sqrt{t^2 - 4t + 4} \) instead?
4. What are the roots of the expression \( t^2 + 4t + 2 \)?
5. How does \( u^2 - 2 \) behave as \( t \) approaches infinity?

**Tip:** Whenever dealing with square roots of squared terms, remember to consider the absolute value function, as it represents the distance from zero.

Learn how to evaluate u^2 - 2 for the function g(t) = square root of t square plus 4t plus 4 with a detailed step-by-step solution. This problem covers key algebraic concepts, including square roots, absolute values, and quadratic equations. Suitable for high school students and those studying advanced algebra.

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