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 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: Upward-opening parabola (positive $ x^2 $ coefficient). Vertex at $ x = -\frac{8}{2(1)} = -4 $. Vertex is $ (-4, -3) $. Symmetric around $ x = -4 $.

Plot points and sketch.

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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: Upward-opening parabola. Vertex at $ x = -\frac{-2}{2(1)} = 1 $. Vertex is $ (1, -3) $. Symmetric about $ x = 1 $.

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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: Downward-opening parabola. Vertex at $ x = -\frac{-2}{2(-1)} = -1 $. Vertex is $ (-1, 1) $. Roots at $ x = -2 $ and $ x = 0 $.

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

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

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: Vertex at $ (4, 5) $, symmetric, opens down.

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

Simple standard parabola.

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: Upward-opening parabola with vertex at origin $ (0, 0) $, symmetric about $ y $-axis.

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✔ Summary of Steps for Each Problem:

For every problem:
1. Plug in each $ x $ value into the function.
2. Compute $ f(x) $.
3. Fill the table.
4. Plot the points on the grid.
5. Draw a smooth parabolic curve through them.
6. Identify key features: vertex, direction, symmetry.

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📌 Tips for Graphing Quadratics:

- Vertex: Use $ x = -\frac{b}{2a} $ for standard form $ f(x) = ax^2 + bx + c $
- Direction: If $ a > 0 $, opens up; if $ a < 0 $, opens down
- Symmetry: Parabolas are symmetric about the vertical line through the vertex
- Y-intercept: Plug in $ x = 0 $
- X-intercepts: Solve $ f(x) = 0 $ (if needed)

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Now you can complete the tables and sketch each graph based on these values. Let me know if you'd like me to generate the actual graphs or help with any specific one!