Worksheet for graphing quadratic functions with tables and coordinate grids.

Let’s solve each problem step by step. We’ll plug in the given x-values into each function to find f(x), then you can use those points to sketch the graph.

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Problem 1: f(x) = -x² + 4x - 2

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

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

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

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Problem 2: f(x) = x² + 8x + 13

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

- x = -6 → (-6)² + 8(-6) + 13 = 36 - 48 + 13 = (36+13) - 48 = 49 - 48 = 1
- x = -5 → 25 + 8(-5) + 13 = 25 - 40 + 13 = (25+13) - 40 = 38 - 40 = -2
- x = -4 → 16 + 8(-4) + 13 = 16 - 32 + 13 = (16+13) - 32 = 29 - 32 = -3
- x = -3 → 9 + 8(-3) + 13 = 9 - 24 + 13 = (9+13) - 24 = 22 - 24 = -2
- x = -2 → 4 + 8(-2) + 13 = 4 - 16 + 13 = (4+13) - 16 = 17 - 16 = 1

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

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Problem 3: f(x) = x² - 2x - 2

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

- x = -1 → (-1)² - 2(-1) - 2 = 1 + 2 - 2 = 1
- x = 0 → 0 - 0 - 2 = -2
- x = 1 → 1 - 2 - 2 = -3
- x = 2 → 4 - 4 - 2 = -2
- x = 3 → 9 - 6 - 2 = 1

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

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Problem 4: f(x) = -x(x + 2)

First, expand if needed: f(x) = -x² - 2x
But we can also plug in directly.

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

- x = -3 → -(-3)(-3 + 2) = -(-3)(-1) = -(3) = -3
Wait — let’s do it carefully:
f(-3) = -(-3) * (-3 + 2) = -(-3)*(-1) = -[ (-3)*(-1) ] = -[3] = -3

Actually, better to compute as:
f(x) = -x(x+2)

So:

- x = -3 → -(-3)(-3+2) = -(-3)(-1) = -[3] = -3
- x = -2 → -(-2)(-2+2) = -(-2)(0) = -[0] = 0
- x = -1 → -(-1)(-1+2) = -(-1)(1) = -[-1] = 1
- x = 0 → -(0)(0+2) = 0 → 0
- x = 1 → -(1)(1+2) = -(1)(3) = -3

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

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Problem 5: f(x) = -(x - 4)² + 5

This is vertex form. Plug in x = 2, 3, 4, 5, 6

- x = 2 → -(2-4)² + 5 = -(-2)² + 5 = -4 + 5 = 1
- x = 3 → -(3-4)² + 5 = -(-1)² + 5 = -1 + 5 = 4
- x = 4 → -(0)² + 5 = 0 + 5 = 5
- x = 5 → -(1)² + 5 = -1 + 5 = 4
- x = 6 → -(2)² + 5 = -4 + 5 = 1

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

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Problem 6: f(x) = x²

Simplest quadratic!

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

- x = -2 → (-2)² = 4
- x = -1 → (-1)² = 1
- x = 0 → 0² = 0
- x = 1 → 1² = 1
- x = 2 → 2² = 4

Table for #6:
| x | -2 | -1 | 0 | 1 | 2 |
|---|----|----|---|---|---|
|f(x)| 4 | 1 | 0 | 1 | 4 |

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Final Answer:

Here are all completed tables:

1) f(x) = -x² + 4x - 2
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
|f(x)| -2 | 1 | 2 | 1 | -2 |

2) f(x) = x² + 8x + 13
| x | -6 | -5 | -4 | -3 | -2 |
|----|----|----|----|----|----|
|f(x)| 1 | -2 | -3 | -2 | 1 |

3) f(x) = x² - 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) = x²
| x | -2 | -1 | 0 | 1 | 2 |
|---|----|----|---|---|---|
|f(x)| 4 | 1 | 0 | 1 | 4 |

Use these (x, f(x)) pairs to plot points on each grid and draw smooth parabolas through them.