h\left(2 + \frac{6}{x}\right) = 4 + \frac{36}{x^2}, find  h(x)
 

To find the function \( h(x) \) given that:

\[
h\left(2 + \frac{6}{x}\right) = 4 + \frac{36}{x^2}
\]

we need to express \( h(x) \) as a function of \( x \) directly. Here's how to approach this:

1. **Identify the expression for \( u = 2 + \frac{6}{x} \)****:  
   Let \( u = 2 + \frac{6}{x} \), so we have:
   \[
   h(u) = 4 + \frac{36}{x^2}
   \]

2. **Express \( x \) in terms of \( u \)**:  
   From \( u = 2 + \frac{6}{x} \), solve for \( x \) in terms of \( u \):
   \[
   u - 2 = \frac{6}{x}
   \]
   \[
   x = \frac{6}{u - 2}
   \]

3. **Substitute \( x \) into the expression for \( h(u) \)**:  
   Now substitute \( x = \frac{6}{u - 2} \) into the equation \( h(u) = 4 + \frac{36}{x^2} \):
   \[
   h(u) = 4 + \frac{36}{\left(\frac{6}{u - 2}\right)^2}
   \]

4. **Simplify the expression**:  
   Simplify the equation:
   \[
   h(u) = 4 + \frac{36}{\frac{36}{(u - 2)^2}} = 4 + (u - 2)^2
   \]

So, the function \( h(x) \) is:
\[
h(x) = 4 + (x - 2)^2
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you find the inverse of the function \( h(x) = 4 + (x - 2)^2 \)?
2. What is the vertex of the quadratic function \( h(x) \)?
3. How does the transformation of the basic quadratic function \( x^2 \) result in \( h(x) = 4 + (x - 2)^2 \)?
4. What is the range of the function \( h(x) \)?
5. How does the function \( h(x) \) behave as \( x \) approaches infinity or negative infinity?

**Tip:** Recognize patterns in functions, such as \( h(x) \), to quickly identify transformations like shifts and stretches.

Discover how to find the function h(x) from the equation h(2 + 6/x) = 4 + 36/x^2 with detailed steps. This problem involves functions, quadratic equations, and inverse functions, suitable for Grades 10-12.

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