Final answer:
The correct graph of the equation y = -16t^2 + 64t + 16 is Option B, a parabola with the vertex at (2, 80), which opens downward, and passes through the points (0, 16) and approximately (4.25, 0).
Step-by-step explanation:
To determine the correct graph of the given quadratic equation y = -16t^2 + 64t + 16, we must identify its key features, such as the vertex, direction of opening, and x-intercepts. The standard form of a quadratic equation is at^2 + bt + c = 0, and in this case, the coefficients are a = -16, b = 64, and c = 16. The vertex can be found using the formula -b/(2a), which gives us -64/(2 * -16) = 2. The y-coordinate of the vertex is found by plugging this t-value back into the equation, resulting in y = -16(2)^2 + 64(2) + 16 = 80. Thus, the vertex is (2, 80). Because the coefficient a is negative, the parabola opens downward. To find the x-intercepts, we set y to 0 and solve for t; the quadratic formula could be used if needed, but in this case it is clear by inspection that the x-intercepts are at (0, 16), and where the expression inside the square root equals to zero, approximately at (4.25, 0). Thus, the correct graph would be Option B: A parabola with vertex (2, 80) that opens downward and passes through the points (0, 16) and approximately (4.25, 0).