ℎ(𝑡) = √𝑥2 + 2𝑥 + 4 ; 𝑓𝑖𝑛𝑑 ℎ(2)

To find \( h(2) \) for the given function \( h(t) = \sqrt{x^2 + 2x + 4} \), we need to substitute \( t = 2 \) into the function. However, there seems to be a small mistake in the function notation. The variable inside the square root is \( x \), but the function is written in terms of \( t \). Let's assume the correct function is:

\[
h(x) = \sqrt{x^2 + 2x + 4}
\]

Now, we substitute \( x = 2 \):

\[
h(2) = \sqrt{2^2 + 2(2) + 4}
\]

Calculate the expression inside the square root:

\[
h(2) = \sqrt{4 + 4 + 4} = \sqrt{12}
\]

Now simplify \( \sqrt{12} \):

\[
h(2) = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}
\]

So, \( h(2) = 2\sqrt{3} \).

Would you like further details on this calculation or any other assistance?

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Here are 8 related questions you might find useful:
1. How do you simplify a square root?
2. What is the domain of the function \( h(x) = \sqrt{x^2 + 2x + 4} \)?
3. How do you find the vertex of the quadratic expression inside the square root?
4. What is the derivative of \( h(x) = \sqrt{x^2 + 2x + 4} \)?
5. How would you evaluate \( h(-2) \) for this function?
6. Can \( h(x) = \sqrt{x^2 + 2x + 4} \) ever equal zero?
7. What is the general approach to evaluating a square root function at a given value?
8. How does the quadratic formula relate to the function inside the square root?

**Tip:** When simplifying square roots, look for perfect squares as factors to simplify your expression more easily.

Learn how to evaluate h(2) for the square root function h(x) = sqrt(x^2 + 2x + 4) with a step-by-step solution. This problem involves substituting x = 2 into the function and simplifying the expression under the square root. Suitable for students in Grades 10-12.

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