1. The problem is to analyze the function $y = 2x^2 - 4x + 1$ and find its key features such as intercepts and extrema. 2. The general form of a quadratic function is $y = ax^2 + bx + c$. Here, $a=2$, $b=-4$, and $c=1$. 3. To find the vertex (extremum), use the formula for the x-coordinate of the vertex: $$x = -\frac{b}{2a} = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1.$$ 4. Substitute $x=1$ back into the function to find the y-coordinate of the vertex: $$y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1.$$ 5. So, the vertex is at $(1, -1)$, which is a minimum point since $a=2 > 0$. 6. To find the y-intercept, set $x=0$: $$y = 2(0)^2 - 4(0) + 1 = 1.$$ So the y-intercept is at $(0,1)$. 7. To find the x-intercepts, set $y=0$ and solve for $x$: $$0 = 2x^2 - 4x + 1.$$ 8. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 2 \times 1}}{2 \times 2} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4}.$$ 9. Simplify the square root: $$\sqrt{8} = 2\sqrt{2}.$$ 10. So, $$x = \frac{4 \pm 2\sqrt{2}}{4} = \frac{4}{4} \pm \frac{2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}.$$ 11. Therefore, the x-intercepts are at $$\left(1 - \frac{\sqrt{2}}{2}, 0\right)$$ and $$\left(1 + \frac{\sqrt{2}}{2}, 0\right).$$ Final answer: - Vertex (minimum): $(1, -1)$ - Y-intercept: $(0, 1)$ - X-intercepts: $\left(1 - \frac{\sqrt{2}}{2}, 0\right)$ and $\left(1 + \frac{\sqrt{2}}{2}, 0\right)$