Identify the vertex, axis of symmetry, \( y \) - intercept, x-intercepts, and opening of the parabola, then sketch the graph. \( y=-2 x^{2}+4 x-1 \)

We start with the quadratic function   y = –2x² + 4x – 1 Step 1. Find the Vertex: • The x-coordinate of the vertex is given by –b/(2a). Here, a = –2 and b = 4 so:   x = –4 / (2·–2) = –4 / –4 = 1. • Substitute x = 1 into the equation to find y:   y = –2(1)² + 4(1) – 1 = –2 + 4 – 1 = 1. • Thus, the vertex is (1, 1). Step 2. Determine the Axis of Symmetry: • The axis of symmetry is the vertical line through the vertex:   x = 1. Step 3. Find the y-intercept: • The y-intercept is found by setting x = 0:   y = –2(0)² + 4(0) – 1 = –1. • So the y-intercept is at (0, –1). Step 4. Find the x-intercepts: • Set y = 0 and solve:   0 = –2x² + 4x – 1. • Multiply the equation by –1 to simplify:   0 = 2x² – 4x + 1. • Use the quadratic formula:   x = [4 ± √(4² – 4·2·1)] / (2·2)   x = [4 ± √(16 – 8)] / 4   x = [4 ± √8] / 4   Since √8 = 2√2, we have:   x = [4 ± 2√2] / 4 = [2 ± √2] / 2. • Thus, the x-intercepts are at ((2 + √2)/2, 0) and ((2 – √2)/2, 0). Step 5. Determine the Opening: • Because the coefficient a = –2 is negative, the parabola opens downwards. Sketching the Graph: 1. Plot the vertex at (1, 1) which is the highest point since the parabola opens downward. 2. Draw the axis of symmetry as a vertical line through x = 1. 3. Plot the y-intercept at (0, –1). 4. Plot the x-intercepts at ((2 + √2)/2, 0) and ((2 – √2)/2, 0) (approximate values: (1 + 0.707, 0) ≈ (1.707, 0) and (1 – 0.707, 0) ≈ (0.293, 0)). 5. Sketch a smooth, downward-opening curve passing through these points. Summary: • Vertex: (1, 1) • Axis of Symmetry: x = 1 • y-intercept: (0, –1) • x-intercepts: ((2 + √2)/2, 0) and ((2 – √2)/2, 0) • Opening: Downward This completes the analysis and graph of the parabola y = –2x² + 4x – 1.