Identify the vertex, axis of symmetry, \( y \) - intercept, \( x \)-intercepts, and opening of the parabola, then sketch the graph. \( y=-2 x^{2}+8 x-5 \)

We start with the quadratic function   y = –2x² + 8x – 5 and identify its key features one by one. 1. Vertex:  • The x-coordinate of the vertex is given by –b/(2a). Here a = –2 and b = 8, so   x = –8/(2(–2)) = –8/(–4) = 2.  • To find the y-coordinate, plug x = 2 back into the equation:   y = –2(2)² + 8(2) – 5 = –2(4) + 16 – 5 = –8 + 16 – 5 = 3.  • Thus, the vertex is at (2, 3). 2. Axis of Symmetry:  • The axis of symmetry is the vertical line through the vertex, so   x = 2. 3. y‑Intercept:  • To find the y‑intercept, set x = 0:   y = –2(0)² + 8(0) – 5 = –5.  • Therefore, the y‑intercept is (0, –5). 4. x‑Intercepts:  • To find the x‑intercepts, set y = 0:   –2x² + 8x – 5 = 0.  • Multiply both sides by –1 (to simplify):   2x² – 8x + 5 = 0.  • Use the quadratic formula: x = [8 ± √(64 – 40)]/(2·2) = [8 ± √24]/4.  • Simplify √24: √24 = 2√6, so   x = (8 ± 2√6)/4 = (4 ± √6)/2.  • Thus, the x‑intercepts are at ((4 – √6)/2, 0) and ((4 + √6)/2, 0). 5. Opening:  • The coefficient a = –2 is less than 0, so the parabola opens downward. Graph Sketch: Here’s a rough sketch showing the vertex, intercepts, and axis of symmetry: y ↑ 4 | ● (2,3) Vertex 3 | | 2 | / \ 1 | / \ 0 | -------●---------●-------→ x –1 | / \ –2 | / \ –3 | / \ –4 | / \ –5 | ● (0,–5) (x‑intercepts around 0.78 and 3.22) | +--------------------------------- Notes: • The axis of symmetry is the vertical line x = 2. • The sketch shows the parabola opening downward. • The x‑intercepts (approximately x ≈ 0.78 and x ≈ 3.22) and y‑intercept (0, –5) are roughly plotted. This completes the identification of the key features and the sketch of the parabola.