Question 1199700
**a. Find the expression for h(x)**

1. **Isolate y^2:**
   - (x^2)/9 - (y^2)/16 = 1
   - (y^2)/16 = (x^2)/9 - 1
   - y^2 = 16 * [(x^2)/9 - 1]
   - y^2 = (16x^2)/9 - 16

2. **Solve for y:**
   - y = ±√[(16x^2)/9 - 16] 

Since we're considering only the portion of the hyperbola in the first quadrant (where both x and y are positive), we take the positive square root:

   - h(x) = √[(16x^2)/9 - 16] 

**b. Justify why h(x) lies below the line y = 4x/3**

1. **Consider the equation of the asymptote:**
   - The asymptote in the first quadrant is y = 4x/3.

2. **Analyze the behavior of h(x):**
   - h(x) represents the y-coordinate of the hyperbola for a given x-value.
   - As x increases, the value of (16x^2)/9 increases significantly.
   - However, the square root function grows relatively slowly compared to a linear function like y = 4x/3.

3. **Conclusion:**
   - The square root in h(x) will cause the y-values of the hyperbola to increase at a slower rate than the linear increase of the asymptote y = 4x/3. 
   - Therefore, the graph of h(x) will always remain below the line y = 4x/3 in the first quadrant.

**In essence:**

* The hyperbola approaches the asymptote y = 4x/3 as x increases, but it never touches or crosses it.
* The square root in the expression for h(x) limits the growth of the y-values compared to the linear growth of the asymptote.

This analysis demonstrates why the portion of the hyperbola in the first quadrant (h(x)) lies below the asymptote y = 4x/3.