Graph the equation \( y=x^{2}-2 x-8 \) on the accompanying including the roots and the vertex.

We want to graph the quadratic   y = x² – 2x – 8. Step 1. Find the Roots: To find the x-intercepts, set y = 0:   x² – 2x – 8 = 0. Factor the quadratic:   (x – 4)(x + 2) = 0. So, the roots are:   x – 4 = 0 ⟹ x = 4,   x + 2 = 0 ⟹ x = –2. These give the points (4, 0) and (–2, 0). Step 2. Find the Vertex: For a quadratic in the form y = ax² + bx + c, the x-coordinate of the vertex is given by   x = –b/(2a). Here, a = 1 and b = –2:   x = –(–2)/(2·1) = 2/2 = 1. Now substitute x = 1 into the equation to get y:   y = (1)² – 2(1) – 8 = 1 – 2 – 8 = –9. Thus, the vertex is at (1, –9). Step 3. Find the y-intercept: Set x = 0:   y = 0² – 2·0 – 8 = –8. So, the y-intercept is (0, –8). Step 4. Graphing: 1. Plot the vertex at (1, –9). This is the lowest point on the parabola (since a > 0, the parabola opens upward). 2. Plot the roots (–2, 0) and (4, 0). 3. Plot the y-intercept (0, –8). 4. Draw a smooth curve through these points forming a “U” shape. The parabola is symmetric about the vertical line x = 1. Here is a rough sketch of the graph: y ↑ 5 | 4 | 3 | 2 | 1 | y=0 —+————•————•————•————→ x | (–2,0) (4,0) -1 | -2 | -3 | -4 | -5 | -6 | -7 | -8 | • (0,–8) -9 | • (1,–9) -10 | | +-------------------------------- -2 -1 0 1 2 3 4 Remember, the parabola is symmetric about the vertical line line x = 1. You can plot additional points on both sides of the vertex if needed for accuracy. That is the graph of y = x² – 2x – 8, complete with its roots (–2, 0) and (4, 0) and vertex (1, –9).