Graphing Quadratic Functions Worksheets - Free Printable

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 graph.
3. Sketch the parabola, noting whether it opens up or down and its vertex.

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

We'll compute $ f(x) $ for $ x = -2, -1, 0, 1, 2, 3, 4 $:

- $ f(-2) = -(-2)^2 + 4(-2) - 2 = -4 -8 -2 = -14 $
- $ f(-1) = -(-1)^2 + 4(-1) - 2 = -1 -4 -2 = -7 $
- $ f(0) = -(0)^2 + 4(0) - 2 = -2 $
- $ f(1) = -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1 $
- $ f(2) = -(2)^2 + 4(2) - 2 = -4 + 8 - 2 = 2 $
- $ f(3) = -(3)^2 + 4(3) - 2 = -9 + 12 - 2 = 1 $
- $ f(4) = -(4)^2 + 4(4) - 2 = -16 + 16 - 2 = -2 $

| $ x $ | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|--------|----|----|---|---|---|---|---|
| $ f(x) $ | -14 | -7 | -2 | 1 | 2 | 1 | -2 |

✔ This is a downward-opening parabola (coefficient of $ x^2 $ is negative).
Vertex: Use $ x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2 $.
$ f(2) = 2 $ → Vertex at $ (2, 2) $

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

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

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

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

✔ Opens upward (positive $ x^2 $ coefficient).
Vertex at $ x = -\frac{8}{2(1)} = -4 $, $ f(-4) = -3 $ → vertex $ (-4, -3) $

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

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

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

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

✔ Opens upward.
Vertex at $ x = \frac{2}{2} = 1 $, $ f(1) = -3 $ → vertex $ (1, -3) $

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

First expand: $ f(x) = -x^2 - 2x $

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

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

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

✔ Opens downward.
Vertex at $ x = -\frac{-2}{2(-1)} = -1 $, $ f(-1) = 1 $ → vertex $ (-1, 1) $

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

This is vertex form: vertex at $ (4, 5) $, opens downward.

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

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

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

✔ Downward opening, symmetric about $ x=4 $

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

Simplest quadratic.

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

- $ f(-2) = 4 $
- $ f(-1) = 1 $
- $ f(0) = 0 $
- $ f(1) = 1 $
- $ f(2) = 4 $

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

✔ Opens upward, vertex at $ (0, 0) $

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✔ Summary of Graphing Steps:

For each function:
- Plot the points from the completed table.
- Draw a smooth curve through them.
- Identify vertex and direction (up/down).
- Note symmetry around the axis of symmetry ($ x = -\frac{b}{2a} $).

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Final Answer (Completed Tables):

#### 1) $ f(x) = -x^2 + 4x - 2 $
| $ x $ | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|--------|----|----|---|---|---|---|---|
| $ f(x) $ | -14 | -7 | -2 | 1 | 2 | 1 | -2 |

#### 2) $ f(x) = x^2 + 8x + 13 $
| $ x $ | -6 | -5 | -4 | -3 | -2 |
|--------|----|----|----|----|----|
| $ f(x) $ | 1 | -2 | -3 | -2 | 1 |

#### 3) $ f(x) = x^2 - 2x - 2 $
| $ x $ | -1 | 0 | 1 | 2 | 3 |
|--------|----|---|---|---|---|
| $ f(x) $ | 1 | -2 | -3 | -2 | 1 |

#### 4) $ f(x) = -x(x+2) $
| $ x $ | -3 | -2 | -1 | 0 | 1 |
|--------|----|----|----|---|---|
| $ f(x) $ | -3 | 0 | 1 | 0 | -3 |

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

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

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Now you can plot these points on the provided grids and sketch each parabola accordingly. Let me know if you'd like visual sketches or further explanation!