1. **Stating the problem:** We are given the quadratic function $$y = x^2 - 2x - 8$$ and asked to find: a. The coordinates of the vertex (titik puncak). b. The graph of the function. 2. **Formula for the vertex of a parabola:** For a quadratic function $$y = ax^2 + bx + c$$, the x-coordinate of the vertex is given by: $$x = -\frac{b}{2a}$$ The y-coordinate is found by substituting this x back into the function. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = -2$$, and $$c = -8$$. 4. **Calculate the x-coordinate of the vertex:** $$x = -\frac{-2}{2 \times 1} = \frac{2}{2} = 1$$ 5. **Calculate the y-coordinate of the vertex:** Substitute $$x=1$$ into the function: $$y = (1)^2 - 2(1) - 8 = 1 - 2 - 8 = -9$$ 6. **Vertex coordinates:** The vertex is at $$ (1, -9) $$. 7. **Graph description:** The parabola opens upwards (since $$a=1 > 0$$) with vertex at $$ (1, -9) $$. 8. **Summary:** - Vertex: $$ (1, -9) $$ - Equation: $$y = x^2 - 2x - 8$$ This completes the solution for the vertex and the graph description.