1. **State the problem:** We need to sketch the graph of the quadratic function $$f(x) = 2x^2 - 4x - 2$$. 2. **Recall the formula for the vertex of a parabola:** The vertex of a parabola given by $$f(x) = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. 3. **Calculate the vertex:** Given $$a=2$$, $$b=-4$$, and $$c=-2$$, $$x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1$$. 4. **Find the y-coordinate of the vertex:** Substitute $$x=1$$ into the function: $$f(1) = 2(1)^2 - 4(1) - 2 = 2 - 4 - 2 = -4$$. So, the vertex is at $$(1, -4)$$. 5. **Determine the axis of symmetry:** The axis of symmetry is the vertical line $$x=1$$. 6. **Find the y-intercept:** Set $$x=0$$: $$f(0) = 2(0)^2 - 4(0) - 2 = -2$$. So, the y-intercept is $$(0, -2)$$. 7. **Find the x-intercepts (roots):** Solve $$2x^2 - 4x - 2 = 0$$. Divide both sides by 2: $$x^2 - 2x - 1 = 0$$. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$. So, the x-intercepts are $$1 + \sqrt{2}$$ and $$1 - \sqrt{2}$$. 8. **Sketch the graph:** - The parabola opens upwards since $$a=2 > 0$$. - Vertex at $$(1, -4)$$. - Axis of symmetry at $$x=1$$. - Y-intercept at $$(0, -2)$$. - X-intercepts at approximately $$(1 + 1.414, 0) = (2.414, 0)$$ and $$(1 - 1.414, 0) = (-0.414, 0)$$. This information allows plotting the parabola accurately on the grid from -6 to 6 on both axes. **Final answer:** The graph is a parabola opening upwards with vertex at $$(1, -4)$$, axis of symmetry $$x=1$$, y-intercept at $$(0, -2)$$, and x-intercepts at $$1 \pm \sqrt{2}$$.