1. Find the axis of symmetry of the quadratic equation y = -x^2 - 4x +1; Give your answer with x =
2. Find the vertex of the quadratic equation y = 2x^2 +8. Give your answer as an ordered pair (x, y)
3. Identify the zeros of the quadratic equation y = -(x -2)^2 + 4. Give your answer in series notation { }
1. To find the axis of symmetry of the quadratic equation y = -x^2 - 4x + 1, we need to find the x-coordinate of the vertex. The x-coordinate of the vertex can be found using the formula x = -b/2a. In this case, a = -1 and b = -4. Therefore, x = -(-4)/(2*(-1)) = 2. So, the axis of symmetry is x = 2.
2. To find the vertex of the quadratic equation y = 2x^2 + 8, we need to find the coordinates of the vertex which can be found using the formula (x, y) = (-b/2a, f(-b/2a)). In this case, a = 2 and b = 0. Therefore, (x, y) = (0, f(0)) = (0, 8). So, the vertex is (0, 8).
3. To identify the zeros of the quadratic equation y = -(x - 2)^2 + 4, we set y to 0 and solve for x. -(x - 2)^2 + 4 = 0. This can be simplified to (x - 2)^2 = 4. Taking the square root of both sides, we get x - 2 = ±2. Therefore, x = 2 ± 2. So, the zeros are {0, 4}.