Match each quadratic function with its graph and complete the table for vertex, axis of symmetry, maximum/minimum, x-intercept, and range.

Problem Analysis:

The task involves matching quadratic functions with their corresponding graphs and completing a table for one of the functions. Quadratic functions are of the form \( y = ax^2 + bx + c \), and their graphs are parabolas. The key characteristics to consider are:

1. Direction of Opening:
- If \( a > 0 \), the parabola opens upwards.
- If \( a < 0 \), the parabola opens downwards.

2. Vertex:
- The vertex is the highest or lowest point of the parabola, depending on the direction of opening.
- The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).

3. Axis of Symmetry:
- The axis of symmetry is a vertical line passing through the vertex, given by \( x = -\frac{b}{2a} \).

4. y-Intercept:
- The y-intercept is the point where the graph intersects the y-axis, which occurs when \( x = 0 \).

5. Domain and Range:
- The domain of a quadratic function is all real numbers, \( (-\infty, \infty) \).
- The range depends on the direction of opening and the vertex:
- If the parabola opens upwards, the range is \( [y_{\text{vertex}}, \infty) \).
- If the parabola opens downwards, the range is \( (-\infty, y_{\text{vertex}}] \).

Step-by-Step Solution:

#### Part 1: Matching Quadratic Functions with Graphs
We are given five quadratic functions and five graphs. Let's analyze each function and match it with the appropriate graph.

1. Function (i): \( y = -x^2 + 2x \)
- Coefficient of \( x^2 \): \( a = -1 \) (parabola opens downwards).
- Vertex: \( x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1 \). Substitute \( x = 1 \) into the function to find \( y \):
\[
y = -(1)^2 + 2(1) = -1 + 2 = 1.
\]
So, the vertex is \( (1, 1) \).
- y-Intercept: When \( x = 0 \), \( y = 0 \).
- Graph: This matches Graph (c), which is a downward-opening parabola with vertex at \( (1, 1) \) and y-intercept at \( (0, 0) \).

2. Function (ii): \( y = \frac{1}{2}x^2 + 2 \)
- Coefficient of \( x^2 \): \( a = \frac{1}{2} \) (parabola opens upwards).
- Vertex: Since there is no \( x \)-term, the vertex is at \( x = 0 \). Substitute \( x = 0 \) into the function:
\[
y = \frac{1}{2}(0)^2 + 2 = 2.
\]
So, the vertex is \( (0, 2) \).
- y-Intercept: When \( x = 0 \), \( y = 2 \).
- Graph: This matches Graph (e), which is an upward-opening parabola with vertex at \( (0, 2) \).

3. Function (iii): \( y = -x^2 + 2 \)
- Coefficient of \( x^2 \): \( a = -1 \) (parabola opens downwards).
- Vertex: Since there is no \( x \)-term, the vertex is at \( x = 0 \). Substitute \( x = 0 \) into the function:
\[
y = -(0)^2 + 2 = 2.
\]
So, the vertex is \( (0, 2) \).
- y-Intercept: When \( x = 0 \), \( y = 2 \).
- Graph: This matches Graph (d), which is a downward-opening parabola with vertex at \( (0, 2) \).

4. Function (iv): \( y = x^2 - 2x + 2 \)
- Coefficient of \( x^2 \): \( a = 1 \) (parabola opens upwards).
- Vertex: \( x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1 \). Substitute \( x = 1 \) into the function to find \( y \):
\[
y = (1)^2 - 2(1) + 2 = 1 - 2 + 2 = 1.
\]
So, the vertex is \( (1, 1) \).
- y-Intercept: When \( x = 0 \), \( y = 2 \).
- Graph: This matches Graph (a), which is an upward-opening parabola with vertex at \( (1, 1) \) and y-intercept at \( (0, 2) \).

5. Function (v): \( y = x^2 - 2x + 1 \)
- Coefficient of \( x^2 \): \( a = 1 \) (parabola opens upwards).
- Vertex: \( x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1 \). Substitute \( x = 1 \) into the function to find \( y \):
\[
y = (1)^2 - 2(1) + 1 = 1 - 2 + 1 = 0.
\]
So, the vertex is \( (1, 0) \).
- y-Intercept: When \( x = 0 \), \( y = 1 \).
- Graph: This matches Graph (b), which is an upward-opening parabola with vertex at \( (1, 0) \) and y-intercept at \( (0, 1) \).

#### Part 2: Completing the Table for \( y = x^2 + 6x + 8 \)
We need to complete the following table for the function \( y = x^2 + 6x + 8 \):

| Feature | Value |
|------------------------|---------------|
| Direction of Opening | |
| Vertex | |
| Axis of Symmetry | |
| y-Intercept | |
| Domain | |
| Range | |

1. Direction of Opening:
- The coefficient of \( x^2 \) is \( a = 1 \), so the parabola opens upwards.

2. Vertex:
- The x-coordinate of the vertex is \( x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3 \).
- Substitute \( x = -3 \) into the function to find \( y \):
\[
y = (-3)^2 + 6(-3) + 8 = 9 - 18 + 8 = -1.
\]
So, the vertex is \( (-3, -1) \).

3. Axis of Symmetry:
- The axis of symmetry is \( x = -3 \).

4. y-Intercept:
- When \( x = 0 \), \( y = 8 \). So, the y-intercept is \( (0, 8) \).

5. Domain:
- The domain of a quadratic function is all real numbers, \( (-\infty, \infty) \).

6. Range:
- Since the parabola opens upwards and the vertex is at \( (-3, -1) \), the range is \( [-1, \infty) \).

#### Final Answer:
The completed table is:

| Feature | Value |
|------------------------|---------------|
| Direction of Opening | Upwards |
| Vertex | \( (-3, -1) \) |
| Axis of Symmetry | \( x = -3 \) |
| y-Intercept | \( (0, 8) \) |
| Domain | \( (-\infty, \infty) \) |
| Range | \( [-1, \infty) \) |

The matched functions and graphs are:
- (i) \( y = -x^2 + 2x \) → Graph (c)
- (ii) \( y = \frac{1}{2}x^2 + 2 \) → Graph (e)
- (iii) \( y = -x^2 + 2 \) → Graph (d)
- (iv) \( y = x^2 - 2x + 2 \) → Graph (a)
- (v) \( y = x^2 - 2x + 1 \) → Graph (b)

The final answer is:
\[
\boxed{
\begin{array}{|c|c|}
\hline
\text{Feature} & \text{Value} \\
\hline
\text{Direction of Opening} & \text{Upwards} \\
\hline
\text{Vertex} & (-3, -1) \\
\hline
\text{Axis of Symmetry} & x = -3 \\
\hline
\text{y-Intercept} & (0, 8) \\
\hline
\text{Domain} & (-\infty, \infty) \\
\hline
\text{Range} & [-1, \infty) \\
\hline
\end{array}
}
\]