(c) Draw the graph of the function $$ f(x)=2 x^{2}-x-6 $$ in the domain $$ -2 \leq x \leq 2 $$, where $$ x \in \mathbb{R} $$, on the co-ordinate grid opposite. Show your working out. ,

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To graph the quadratic function $$ f(x) = 2x^{2} - x - 6 $$ over the domain $$ -2 \leq x \leq 2 $$, follow these steps:

1. **Identify Key Features of the Graph**
   - **Vertex**: The highest or lowest point on the graph.
   - **Axis of Symmetry**: A vertical line that passes through the vertex.
   - **Y-intercept**: The point where the graph crosses the y-axis.
   - **X-intercepts (Roots)**: The points where the graph crosses the x-axis.

2. **Find the Vertex and Axis of Symmetry**
   
   For a quadratic function in the form $$ f(x) = ax^{2} + bx + c $$:

- **Axis of Symmetry**: $$ x = -\frac{b}{2a} $$
   - **Vertex**: $$ \left( -\frac{b}{2a}, f\left( -\frac{b}{2a} \right) \right) $$

Given $$ a = 2 $$, $$ b = -1 $$, and $$ c = -6 $$:

$$
   x = -\frac{-1}{2 \times 2} = \frac{1}{4}
   $$
   
   Plugging $$ x = \frac{1}{4} $$ back into $$ f(x) $$:

$$
   f\left( \frac{1}{4} \right) = 2\left( \frac{1}{4} \right)^2 - \frac{1}{4} - 6 = 2\left( \frac{1}{16} \right) - \frac{1}{4} - 6 = \frac{1}{8} - \frac{2}{8} - \frac{48}{8} = -\frac{49}{8} = -6.125
   $$
   
   So, the **vertex** is at $$ \left( \frac{1}{4}, -6.125 \right) $$ and the **axis of symmetry** is $$ x = \frac{1}{4} $$.

3. **Determine the Y-intercept**
   
   The y-intercept occurs when $$ x = 0 $$:

$$
   f(0) = 2(0)^2 - 0 - 6 = -6
   $$
   
   So, the y-intercept is at $$ (0, -6) $$.

4. **Find the X-intercepts (if any)**
   
   To find the x-intercepts, set $$ f(x) = 0 $$:

$$
   2x^{2} - x - 6 = 0
   $$
   
   Use the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:

$$
   x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times (-6)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4}
   $$
   
   Therefore:

$$
   x = \frac{1 + 7}{4} = 2 \quad \text{and} \quad x = \frac{1 - 7}{4} = -\frac{3}{2}
   $$
   
   So, the x-intercepts are at $$ (2, 0) $$ and $$ \left( -\frac{3}{2}, 0 \right) $$.

5. **Calculate Additional Points**

To ensure accuracy, calculate $$ f(x) $$ for several values within the domain $$ -2 \leq x \leq 2 $$.

| $$ x $$ | $$ f(x) = 2x^{2} - x - 6 $$ |
   |-------|----------------------------|
   | -2    | $$ 2(-2)^2 - (-2) - 6 = 8 + 2 - 6 = 4 $$ |
   | -1    | $$ 2(-1)^2 - (-1) - 6 = 2 + 1 - 6 = -3 $$ |
   | -0.5  | $$ 2(-0.5)^2 - (-0.5) - 6 = 0.5 + 0.5 - 6 = -5 $$ |
   | 0     | $$ 2(0)^2 - 0 - 6 = -6 $$ |
   | 0.5   | $$ 2(0.5)^2 - 0.5 - 6 = 0.5 - 0.5 - 6 = -6 $$ |
   | 1     | $$ 2(1)^2 - 1 - 6 = 2 - 1 - 6 = -5 $$ |
   | 1.5   | $$ 2(1.5)^2 - 1.5 - 6 = 4.5 - 1.5 - 6 = -3 $$ |
   | 2     | $$ 2(2)^2 - 2 - 6 = 8 - 2 - 6 = 0 $$ |

6. **Plot the Points on the Coordinate Grid**

Here's a list of points to plot:

- $$ (-2, 4) $$
   - $$ \left( -\frac{3}{2}, 0 \right) $$
   - $$ (-1, -3) $$
   - $$ (-0.5, -5) $$
   - $$ \left( \frac{1}{4}, -6.125 \right) $$ (Vertex)
   - $$ (0, -6) $$
   - $$ (0.5, -6) $$
   - $$ (1, -5) $$
   - $$ (1.5, -3) $$
   - $$ (2, 0) $$

7. **Sketch the Graph**

Using the points plotted:

- **Plot the Vertex** at $$ \left( \frac{1}{4}, -6.125 \right) $$.
   - **Plot the Intercepts** at $$ (0, -6) $$, $$ \left( -\frac{3}{2}, 0 \right) $$, and $$ (2, 0) $$.
   - **Plot Additional Points** to define the curvature of the parabola.
   - **Draw the Parabola**: Since the coefficient of $$ x^{2} $$ is positive (2), the parabola opens upwards.

Below is an approximate ASCII representation of the graph within the domain $$ -2 \leq x \leq 2 $$. Note that this is a simplified visualization.

```
    y
    6 |                     
    5 |                     
    4 |        * (-2,4)      
    3 |                     
    2 |                     
    1 |                     
    0 |* (-1.5,0)         * (2,0)
   -1 |                     
   -2 |                     
   -3 |    * (-1,-3)      * (1.5,-3)
   -4 |                     
   -5 |  * (-0.5,-5)    * (1,-5)
   -6 |        * (0,-6)   * (0.5,-6)
   -7 |                     
       -------------------------------
          -2  -1   0   1   2      x
   ```

**Notes:**
   - The vertex at $$ \left( \frac{1}{4}, -6.125 \right) $$ lies slightly below $$ (0.5, -6) $$ and $$ (0, -6) $$.
   - The symmetry axis $$ x = \frac{1}{4} $$ helps in ensuring the parabola is symmetric on both sides.
   - Connect the plotted points with a smooth, curved line to form the parabola.

8. **Final Graph Interpretation**

The graph of $$ f(x) = 2x^{2} - x - 6 $$ is a parabola opening upwards with:
   
   - **Vertex** at $$ \left( \frac{1}{4}, -6.125 \right) $$
   - **Y-intercept** at $$ (0, -6) $$
   - **X-intercepts** at $$ \left( -\frac{3}{2}, 0 \right) $$ and $$ (2, 0) $$
   
   This graph accurately represents the function within the specified domain $$ -2 \leq x \leq 2 $$.
To graph the function $$ f(x) = 2x^{2} - x - 6 $$ between $$ x = -2 $$ and $$ x = 2 $$, follow these steps: 1. **Find the Vertex and Axis of Symmetry**: - Vertex at $$ \left( \frac{1}{4}, -6.125 \right) $$ - Axis of symmetry at $$ x = \frac{1}{4} $$ 2. **Determine Intercepts**: - Y-intercept at $$ (0, -6) $$ - X-intercepts at $$ \left( -\frac{3}{2}, 0 \right) $$ and $$ (2, 0) $$ 3. **Plot Additional Points**: - $$ (-2, 4) $$ - $$ (-1, -3) $$ - $$ (-0.5, -5) $$ - $$ (0.5, -6) $$ - $$ (1, -5) $$ - $$ (1.5, -3) $$ 4. **Sketch the Parabola**: - Draw a smooth, upward-opening curve through the plotted points, ensuring symmetry around the axis $$ x = \frac{1}{4} $$. The graph shows a parabola with the vertex at $$ \left( \frac{1}{4}, -6.125 \right) $$, y-intercept at $$ (0, -6) $$, and x-intercepts at $$ \left( -\frac{3}{2}, 0 \right) $$ and $$ (2, 0) $$ within the domain $$ -2 \leq x \leq 2 $$.

(c) Draw the graph of the function \( f(x)=2 x^{2}-x-6 \) in the domain \( -2 \leq x \leq 2 \), where \( x \in \mathbb{R} \), on the co-ordinate grid opposite. Show your working out. ,

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