Final answer:
The vertex of the quadratic function f(x) = x² + 2x + 2 is at (-1, 1), the graph opens upward, the y-intercept is at (0, 2), and a sketch would show an upward-opening parabola.
Step-by-step explanation:
The given quadratic function is f(x) = x² + 2x + 2. To find the vertex of the graph of the quadratic function, we use the formula -b/2a for the x-coordinate of the vertex, where a and b are the coefficients from the quadratic equation ax² + bx + c. For the given function, a = 1 and b = 2, giving us the vertex's x-coordinate as x = -2/(2*1) = -1. To find the y-coordinate, we substitute x = -1 into the equation to get y = (-1)² + 2(-1) + 2 = 1 - 2 + 2 = 1. Therefore, the vertex is at (-1, 1).
Since the coefficient a is positive, the graph of the quadratic function opens upward. The y-intercept is found by setting x = 0 in the equation, which yields y = 0² + 2*0 + 2 = 2. Hence, the y-intercept is at (0, 2).
A sketch of the graph would show a parabola opening upwards, with its vertex at (-1, 1) and crossing the y-axis at (0, 2).