Worksheet for graphing quadratic functions, including tables and coordinate grids for plotting.

Let's solve each of the six quadratic function problems step by step. For each function:

1. Complete the table by plugging in the given $ x $-values into the function to find $ f(x) $.
2. Plot the points on the coordinate plane.
3. Sketch the graph of the quadratic (which is a parabola).

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1) $ f(x) = -x^2 + 4x - 2 $

We are given $ x = 0, 1, 2, 3, 4 $

| $ x $ | $ f(x) = -x^2 + 4x - 2 $ |
|--------|-----------------------------|
| 0 | $ -(0)^2 + 4(0) - 2 = -2 $ |
| 1 | $ -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1 $ |
| 2 | $ -(2)^2 + 4(2) - 2 = -4 + 8 - 2 = 2 $ |
| 3 | $ -(3)^2 + 4(3) - 2 = -9 + 12 - 2 = 1 $ |
| 4 | $ -(4)^2 + 4(4) - 2 = -16 + 16 - 2 = -2 $ |

Table:
| $ x $ | 0 | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|---|
| $ f(x) $ | -2 | 1 | 2 | 1 | -2 |

Graph: This is a downward-opening parabola (since coefficient of $ x^2 $ is negative). The vertex is at $ x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2 $. So vertex is at $ (2, 2) $.

Plot the points and draw a smooth curve through them.

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2) $ f(x) = x^2 + 8x + 13 $

Given $ x = -6, -5, -4, -3, -2 $

| $ x $ | $ f(x) = x^2 + 8x + 13 $ |
|--------|----------------------------|
| -6 | $ (-6)^2 + 8(-6) + 13 = 36 - 48 + 13 = 1 $ |
| -5 | $ 25 - 40 + 13 = -2 $ |
| -4 | $ 16 - 32 + 13 = -3 $ |
| -3 | $ 9 - 24 + 13 = -2 $ |
| -2 | $ 4 - 16 + 13 = 1 $ |

Table:
| $ x $ | -6 | -5 | -4 | -3 | -2 |
|--------|----|----|----|----|----|
| $ f(x) $ | 1 | -2 | -3 | -2 | 1 |

Graph: Opens upward (positive $ x^2 $). Vertex at $ x = -\frac{8}{2(1)} = -4 $. So vertex is $ (-4, -3) $. Plot the points and sketch the U-shaped parabola.

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3) $ f(x) = x^2 - 2x - 2 $

Given $ x = -1, 0, 1, 2, 3 $

| $ x $ | $ f(x) = x^2 - 2x - 2 $ |
|--------|---------------------------|
| -1 | $ 1 + 2 - 2 = 1 $ |
| 0 | $ 0 - 0 - 2 = -2 $ |
| 1 | $ 1 - 2 - 2 = -3 $ |
| 2 | $ 4 - 4 - 2 = -2 $ |
| 3 | $ 9 - 6 - 2 = 1 $ |

Table:
| $ x $ | -1 | 0 | 1 | 2 | 3 |
|--------|----|---|---|---|---|
| $ f(x) $ | 1 | -2 | -3 | -2 | 1 |

Graph: Opens upward. Vertex at $ x = \frac{2}{2} = 1 $, so vertex is $ (1, -3) $. Symmetric about $ x=1 $. Sketch accordingly.

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4) $ f(x) = -x(x + 2) $

First simplify:
$ f(x) = -x(x + 2) = -x^2 - 2x $

Given $ x = -3, -2, -1, 0, 1 $

| $ x $ | $ f(x) = -x^2 - 2x $ |
|--------|-----------------------|
| -3 | $ -9 + 6 = -3 $ |
| -2 | $ -4 + 4 = 0 $ |
| -1 | $ -1 + 2 = 1 $ |
| 0 | $ 0 + 0 = 0 $ |
| 1 | $ -1 - 2 = -3 $ |

Table:
| $ x $ | -3 | -2 | -1 | 0 | 1 |
|--------|----|----|----|---|---|
| $ f(x) $ | -3 | 0 | 1 | 0 | -3 |

Graph: Opens downward (negative $ x^2 $). Vertex at $ x = -\frac{-2}{2(-1)} = -1 $. So vertex is $ (-1, 1) $. Plot points and sketch.

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5) $ f(x) = -(x - 4)^2 + 5 $

This is in vertex form: $ f(x) = -(x - h)^2 + k $, so vertex is $ (4, 5) $

Given $ x = 2, 3, 4, 5, 6 $

| $ x $ | $ f(x) = -(x - 4)^2 + 5 $ |
|--------|-----------------------------|
| 2 | $ -( -2)^2 + 5 = -4 + 5 = 1 $ |
| 3 | $ -( -1)^2 + 5 = -1 + 5 = 4 $ |
| 4 | $ -(0)^2 + 5 = 5 $ |
| 5 | $ -(1)^2 + 5 = -1 + 5 = 4 $ |
| 6 | $ -(2)^2 + 5 = -4 + 5 = 1 $ |

Table:
| $ x $ | 2 | 3 | 4 | 5 | 6 |
|--------|---|---|---|---|---|
| $ f(x) $ | 1 | 4 | 5 | 4 | 1 |

Graph: Opens downward (negative sign), vertex at $ (4, 5) $. Symmetric around $ x = 4 $. Points form a symmetric "hill".

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6) $ f(x) = x^2 $

Simplest quadratic. Given $ x = -2, -1, 0, 1, 2 $

| $ x $ | $ f(x) = x^2 $ |
|--------|------------------|
| -2 | 4 |
| -1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |

Table:
| $ x $ | -2 | -1 | 0 | 1 | 2 |
|--------|----|----|---|---|---|
| $ f(x) $ | 4 | 1 | 0 | 1 | 4 |

Graph: Standard upward-opening parabola with vertex at $ (0, 0) $. Symmetric about y-axis.

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✔ Summary of Key Features

| Problem | Function | Opens | Vertex | Axis of Symmetry |
|--------|---------|-------|--------|------------------|
| 1 | $ -x^2 + 4x - 2 $ | Down | (2, 2) | $ x = 2 $ |
| 2 | $ x^2 + 8x + 13 $ | Up | (-4, -3) | $ x = -4 $ |
| 3 | $ x^2 - 2x - 2 $ | Up | (1, -3) | $ x = 1 $ |
| 4 | $ -x^2 - 2x $ | Down | (-1, 1) | $ x = -1 $ |
| 5 | $ -(x - 4)^2 + 5 $ | Down | (4, 5) | $ x = 4 $ |
| 6 | $ x^2 $ | Up | (0, 0) | $ x = 0 $ |

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📌 Instructions for Graphing:

For each problem:
1. Fill in the table using the function.
2. Plot each point on the grid.
3. Connect the points smoothly with a parabola.
4. Ensure it opens correctly (up or down).
5. Mark the vertex and axis of symmetry if possible.

Let me know if you'd like me to generate the actual graphs or provide a printable version!