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The image you uploaded contains a problem involving a quadratic function and a table of values. Let's break it down and provide assistance step-by-step:

### Problem Breakdown:
1. **Table of values**: 
   - A table is given with \( x \) and \( y \) values for the function \( y = 4 + 2x - x^2 \).
   - \( x \) ranges from -2 to 4, and some \( y \)-values are missing.
   
2. **Tasks**:
   a. **Find the value of \( y \) when \( x = 2 \)**.
   
   b. **Using the table, write the interval where \( 1 < y < 4 \)**.
   
   c. **Determine where the function is decreasing in the interval**.
   
   d. **Express the function as \( y = b - (a - x)^2 \)** and solve the quadratic equation to find a value for \( \sqrt{5} \).

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### Let's start solving the sub-parts:

#### (a) Find the value of \( y \) when \( x = 2 \):
The function is \( y = 4 + 2x - x^2 \).
Plug in \( x = 2 \):

\[
y = 4 + 2(2) - (2)^2 = 4 + 4 - 4 = 4
\]

Thus, the value of \( y \) when \( x = 2 \) is \( y = 4 \).

#### (b) Write the interval for \( 1 < y < 4 \):
We need to examine the values of \( y \) from the table and find where \( y \) is between 1 and 4. Let me write down the \( y \)-values first based on the function:
For each \( x \):
- \( x = -2 \), \( y = 4 + 2(-2) - (-2)^2 = -4 \)
- \( x = -1 \), \( y = 4 + 2(-1) - (-1)^2 = 1 \)
- \( x = 0 \), \( y = 4 + 2(0) - 0^2 = 4 \)
- \( x = 1 \), \( y = 4 + 2(1) - (1)^2 = 5 \)
- \( x = 2 \), \( y = 4 + 2(2) - 2^2 = 4 \)
- \( x = 3 \), \( y = 4 + 2(3) - 3^2 = 1 \)
- \( x = 4 \), \( y = 4 + 2(4) - 4^2 = -4 \)

We observe that \( y \) lies in the interval \( 1 < y < 4 \) when \( x = -1 \) and \( x = 3 \), as \( y = 1 \) at those points. Therefore, there is no exact interval where \( 1 < y < 4 \), but the values hover around this range.

#### (c) Find the interval where the function is decreasing:
The derivative of the function \( y = 4 + 2x - x^2 \) is:
\[
\frac{dy}{dx} = 2 - 2x
\]

The function is decreasing where the derivative is negative, i.e., when:
\[
2 - 2x < 0 \quad \Rightarrow \quad x > 1
\]

Thus, the function is decreasing when \( x > 1 \).

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Would you like further clarifications or need me to continue solving other parts? Here are 5 related questions for more practice:

1. What happens to the function as \( x \to \infty \)?
2. How do you find the vertex of the quadratic function \( y = 4 + 2x - x^2 \)?
3. Can you sketch the graph of \( y = 4 + 2x - x^2 \)?
4. How do you find where the function crosses the x-axis?
5. What are the maximum and minimum values of the function?

### Tip:
When finding intervals where a function is increasing or decreasing, always differentiate the function and look for where the derivative is positive (increasing) or negative (decreasing).

An incomplete table of y-values corresponding to several x-values of the function y = 4 + 2x - x^2 is given below. (a) Find the value of y when x = 2. (b) Using the table, write the interval 1 < y < 4. (c) Using the quadratic function, determine the interval where the function is decreasing. Express the function in the form y = b - (a - x)^2, and solve the quadratic equation to find a value for √5.

This problem covers how to solve the quadratic function y = 4 + 2x - x^2, determine where the function is decreasing, and express it in a different form. It also includes the use of derivatives and identifying specific intervals. Suitable for Grades 10-12.

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