Graph Quadratic Functions - Worksheet - Free Printable

To solve the problem of graphing the quadratic functions, we need to follow a systematic approach for each function. Here's how we can do it step by step:

General Steps for Graphing Quadratic Functions:

1. Identify the standard form: The general form of a quadratic function is \( y = ax^2 + bx + c \).
2. Find the vertex:
- The x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).
- Substitute \( x \) back into the equation to find the y-coordinate.
3. Determine the axis of symmetry: The axis of symmetry is the vertical line \( x = -\frac{b}{2a} \).
4. Find the y-intercept: This is the point where \( x = 0 \).
5. Find the x-intercepts (if any): Solve \( ax^2 + bx + c = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
6. Plot additional points: Choose a few x-values on either side of the vertex and calculate the corresponding y-values.
7. Sketch the parabola: Use the vertex, intercepts, and additional points to sketch the graph.

Now, let's apply these steps to each of the given quadratic functions.

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Problem 1: \( y = 2x^2 - 1 \)

1. Vertex:
- \( a = 2 \), \( b = 0 \), \( c = -1 \)
- \( x = -\frac{b}{2a} = -\frac{0}{2 \cdot 2} = 0 \)
- \( y = 2(0)^2 - 1 = -1 \)
- Vertex: \( (0, -1) \)

2. Axis of symmetry: \( x = 0 \)

3. y-intercept: When \( x = 0 \), \( y = -1 \). So, the y-intercept is \( (0, -1) \).

4. x-intercepts:
- Solve \( 2x^2 - 1 = 0 \)
- \( 2x^2 = 1 \)
- \( x^2 = \frac{1}{2} \)
- \( x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \)
- x-intercepts: \( \left( \frac{\sqrt{2}}{2}, 0 \right) \) and \( \left( -\frac{\sqrt{2}}{2}, 0 \right) \)

5. Additional points:
- When \( x = 1 \), \( y = 2(1)^2 - 1 = 1 \). Point: \( (1, 1) \)
- When \( x = -1 \), \( y = 2(-1)^2 - 1 = 1 \). Point: \( (-1, 1) \)

6. Graph: The parabola opens upwards (since \( a > 0 \)) with the vertex at \( (0, -1) \), y-intercept at \( (0, -1) \), and x-intercepts at \( \left( \frac{\sqrt{2}}{2}, 0 \right) \) and \( \left( -\frac{\sqrt{2}}{2}, 0 \right) \).

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Problem 2: \( y = -2x^2 - 4x + 2 \)

1. Vertex:
- \( a = -2 \), \( b = -4 \), \( c = 2 \)
- \( x = -\frac{b}{2a} = -\frac{-4}{2 \cdot -2} = -\frac{-4}{-4} = -1 \)
- \( y = -2(-1)^2 - 4(-1) + 2 = -2(1) + 4 + 2 = -2 + 4 + 2 = 4 \)
- Vertex: \( (-1, 4) \)

2. Axis of symmetry: \( x = -1 \)

3. y-intercept: When \( x = 0 \), \( y = -2(0)^2 - 4(0) + 2 = 2 \). So, the y-intercept is \( (0, 2) \).

4. x-intercepts:
- Solve \( -2x^2 - 4x + 2 = 0 \)
- Divide by -2: \( x^2 + 2x - 1 = 0 \)
- Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \)
- x-intercepts: \( (-1 + \sqrt{2}, 0) \) and \( (-1 - \sqrt{2}, 0) \)

5. Additional points:
- When \( x = 1 \), \( y = -2(1)^2 - 4(1) + 2 = -2 - 4 + 2 = -4 \). Point: \( (1, -4) \)
- When \( x = -2 \), \( y = -2(-2)^2 - 4(-2) + 2 = -2(4) + 8 + 2 = -8 + 8 + 2 = 2 \). Point: \( (-2, 2) \)

6. Graph: The parabola opens downwards (since \( a < 0 \)) with the vertex at \( (-1, 4) \), y-intercept at \( (0, 2) \), and x-intercepts at \( (-1 + \sqrt{2}, 0) \) and \( (-1 - \sqrt{2}, 0) \).

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Problem 3: \( y = -3x^2 - 6x - 1 \)

1. Vertex:
- \( a = -3 \), \( b = -6 \), \( c = -1 \)
- \( x = -\frac{b}{2a} = -\frac{-6}{2 \cdot -3} = -\frac{-6}{-6} = -1 \)
- \( y = -3(-1)^2 - 6(-1) - 1 = -3(1) + 6 - 1 = -3 + 6 - 1 = 2 \)
- Vertex: \( (-1, 2) \)

2. Axis of symmetry: \( x = -1 \)

3. y-intercept: When \( x = 0 \), \( y = -3(0)^2 - 6(0) - 1 = -1 \). So, the y-intercept is \( (0, -1) \).

4. x-intercepts:
- Solve \( -3x^2 - 6x - 1 = 0 \)
- Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-3)(-1)}}{2(-3)} = \frac{6 \pm \sqrt{36 - 12}}{-6} = \frac{6 \pm \sqrt{24}}{-6} = \frac{6 \pm 2\sqrt{6}}{-6} = -1 \mp \frac{\sqrt{6}}{3} \)
- x-intercepts: \( \left( -1 + \frac{\sqrt{6}}{3}, 0 \right) \) and \( \left( -1 - \frac{\sqrt{6}}{3}, 0 \right) \)

5. Additional points:
- When \( x = 1 \), \( y = -3(1)^2 - 6(1) - 1 = -3 - 6 - 1 = -10 \). Point: \( (1, -10) \)
- When \( x = -2 \), \( y = -3(-2)^2 - 6(-2) - 1 = -3(4) + 12 - 1 = -12 + 12 - 1 = -1 \). Point: \( (-2, -1) \)

6. Graph: The parabola opens downwards (since \( a < 0 \)) with the vertex at \( (-1, 2) \), y-intercept at \( (0, -1) \), and x-intercepts at \( \left( -1 + \frac{\sqrt{6}}{3}, 0 \right) \) and \( \left( -1 - \frac{\sqrt{6}}{3}, 0 \right) \).

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Problem 4: \( y = x^2 + 6x \)

1. Vertex:
- \( a = 1 \), \( b = 6 \), \( c = 0 \)
- \( x = -\frac{b}{2a} = -\frac{6}{2 \cdot 1} = -3 \)
- \( y = (-3)^2 + 6(-3) = 9 - 18 = -9 \)
- Vertex: \( (-3, -9) \)

2. Axis of symmetry: \( x = -3 \)

3. y-intercept: When \( x = 0 \), \( y = 0^2 + 6(0) = 0 \). So, the y-intercept is \( (0, 0) \).

4. x-intercepts:
- Solve \( x^2 + 6x = 0 \)
- Factor: \( x(x + 6) = 0 \)
- \( x = 0 \) or \( x = -6 \)
- x-intercepts: \( (0, 0) \) and \( (-6, 0) \)

5. Additional points:
- When \( x = 1 \), \( y = 1^2 + 6(1) = 1 + 6 = 7 \). Point: \( (1, 7) \)
- When \( x = -4 \), \( y = (-4)^2 + 6(-4) = 16 - 24 = -8 \). Point: \( (-4, -8) \)

6. Graph: The parabola opens upwards (since \( a > 0 \)) with the vertex at \( (-3, -9) \), y-intercept at \( (0, 0) \), and x-intercepts at \( (0, 0) \) and \( (-6, 0) \).

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Problem 5: \( y = -x^2 - 6x - 4 \)

1. Vertex:
- \( a = -1 \), \( b = -6 \), \( c = -4 \)
- \( x = -\frac{b}{2a} = -\frac{-6}{2 \cdot -1} = -\frac{-6}{-2} = -3 \)
- \( y = -(-3)^2 - 6(-3) - 4 = -9 + 18 - 4 = 5 \)
- Vertex: \( (-3, 5) \)

2. Axis of symmetry: \( x = -3 \)

3. y-intercept: When \( x = 0 \), \( y = -(0)^2 - 6(0) - 4 = -4 \). So, the y-intercept is \( (0, -4) \).

4. x-intercepts:
- Solve \( -x^2 - 6x - 4 = 0 \)
- Multiply by -1: \( x^2 + 6x + 4 = 0 \)
- Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( x = \frac{-6 \pm \sqrt{6^2 - 4(1)(4)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 16}}{2} = \frac{-6 \pm \sqrt{20}}{2} = \frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5} \)
- x-intercepts: \( (-3 + \sqrt{5}, 0) \) and \( (-3 - \sqrt{5}, 0) \)

5. Additional points:
- When \( x = 1 \), \( y = -(1)^2 - 6(1) - 4 = -1 - 6 - 4 = -11 \). Point: \( (1, -11) \)
- When \( x = -5 \), \( y = -(-5)^2 - 6(-5) - 4 = -25 + 30 - 4 = 1 \). Point: \( (-5, 1) \)

6. Graph: The parabola opens downwards (since \( a < 0 \)) with the vertex at \( (-3, 5) \), y-intercept at \( (0, -4) \), and x-intercepts at \( (-3 + \sqrt{5}, 0) \) and \( (-3 - \sqrt{5}, 0) \).

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Problem 6: \( y = -(x - 5)^2 - 3 \)

1. Vertex:
- The equation is in vertex form \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.
- Here, \( h = 5 \) and \( k = -3 \)
- Vertex: \( (5, -3) \)

2. Axis of symmetry: \( x = 5 \)

3. y-intercept: When \( x = 0 \), \( y = -(0 - 5)^2 - 3 = -(25) - 3 = -25 - 3 = -28 \). So, the y-intercept is \( (0, -28) \).

4. x-intercepts:
- Solve \( -(x - 5)^2 - 3 = 0 \)
- \( -(x - 5)^2 = 3 \)
- \( (x - 5)^2 = -3 \)
- No real solutions (since a square cannot be negative)
- No x-intercepts

5. Additional points:
- When \( x = 4 \), \( y = -(4 - 5)^2 - 3 = -(-1)^2 - 3 = -1 - 3 = -4 \). Point: \( (4, -4) \)
- When \( x = 6 \), \( y = -(6 - 5)^2 - 3 = -(1)^2 - 3 = -1 - 3 = -4 \). Point: \( (6, -4) \)

6. Graph: The parabola opens downwards (since \( a < 0 \)) with the vertex at \( (5, -3) \), y-intercept at \( (0, -28) \), and no x-intercepts.

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Final Answer:

\[
\boxed{
\begin{array}{c}
\text{Graph each quadratic function as described above.}
\end{array}
}
\]