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The problem provided involves filling out tables based on the function \( f(x) = -x^2 + 4x - 2 \).

Let's break it down step by step.

### Part (i): Table 1
You need to compute \( f(x) = -x^2 + 4x - 2 \) for different values of \( x \) and fill in the missing values.

Given table:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
f(x) &  & 1 &  &  &  & -2 &  \\
\hline
\end{array}
\]

Let's compute:

1. For \( x = -1 \):
   \[
   f(-1) = -(-1)^2 + 4(-1) - 2 = -1 - 4 - 2 = -7
   \]

2. For \( x = 1 \):
   \[
   f(1) = -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1
   \]

3. For \( x = 2 \):
   \[
   f(2) = -(2)^2 + 4(2) - 2 = -4 + 8 - 2 = 2
   \]

4. For \( x = 3 \):
   \[
   f(3) = -(3)^2 + 4(3) - 2 = -9 + 12 - 2 = 1
   \]

5. For \( x = 5 \):
   \[
   f(5) = -(5)^2 + 4(5) - 2 = -25 + 20 - 2 = -7
   \]

Thus, the completed table becomes:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
f(x) & -7 & 1 & 1 & 2 & 1 & -2 & -7 \\
\hline
\end{array}
\]

### Part (ii): Table 2
In this table, we need to compute \( f(x) + 2 \) for different values of \( x \).

Given table:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
f(x) + 2 &  &  &  &  & 3 &  &  \\
\hline
\end{array}
\]

Using values from the completed first table:
1. \( f(-1) = -7 \), so \( f(-1) + 2 = -7 + 2 = -5 \).
2. \( f(0) = 1 \), so \( f(0) + 2 = 1 + 2 = 3 \).
3. \( f(1) = 1 \), so \( f(1) + 2 = 1 + 2 = 3 \).
4. \( f(2) = 2 \), so \( f(2) + 2 = 2 + 2 = 4 \).
5. \( f(4) = -2 \), so \( f(4) + 2 = -2 + 2 = 0 \).
6. \( f(5) = -7 \), so \( f(5) + 2 = -7 + 2 = -5 \).

Thus, the completed table becomes:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
f(x) + 2 & -5 & 3 & 3 & 4 & 3 & 0 & -5 \\
\hline
\end{array}
\]

### Part (iii): Table 3
In this table, we need to compute \( f(x) - 4 \) for different values of \( x \).

Given table:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
f(x) - 4 &  &  &  &  &  & -11 &  \\
\hline
\end{array}
\]

Using values from the completed first table:
1. \( f(-1) = -7 \), so \( f(-1) - 4 = -7 - 4 = -11 \).
2. \( f(0) = 1 \), so \( f(0) - 4 = 1 - 4 = -3 \).
3. \( f(1) = 1 \), so \( f(1) - 4 = 1 - 4 = -3 \).
4. \( f(2) = 2 \), so \( f(2) - 4 = 2 - 4 = -2 \).
5. \( f(3) = 1 \), so \( f(3) - 4 = 1 - 4 = -3 \).
6. \( f(5) = -7 \), so \( f(5) - 4 = -7 - 4 = -11 \).

Thus, the completed table becomes:
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
f(x) - 4 & -11 & -3 & -3 & -2 & -3 & -11 & -11 \\
\hline
\end{array}
\]

---

Would you like further details or have any questions?

Here are 5 additional related questions:
1. How do you find the vertex of the quadratic function \( f(x) = -x^2 + 4x - 2 \)?
2. What is the range of the function \( f(x) = -x^2 + 4x - 2 \)?
3. How do you graph a quadratic function?
4. How does the transformation \( f(x) + k \) shift the graph of a quadratic function?
5. Can you explain how completing the square can help in understanding the quadratic function?

**Tip:** For any quadratic function of the form \( ax^2 + bx + c \), the axis of symmetry is given by \( x = -\frac{b}{2a} \), which is helpful in finding the vertex.

In parts (i) and (ii) let f(x) = -x^2 + 4x - 2. Complete the following tables of values.

Learn how to evaluate the quadratic function f(x) = -x^2 + 4x - 2 to complete multiple tables of values. This detailed guide is ideal for students studying algebra and functions at the Grades 9-11 level.

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