Final answer:
The function k(t) = t² - 4t + 5 is a quadratic function with a vertex at (2,1), and since the 'a' term is positive, the parabola opens upwards making this point the minimum value. Therefore, the range of the function is all real numbers greater than or equal to 1, corresponding to option c) (1, ∞).
Step-by-step explanation:
The function k(t) = t² - 4t + 5 is a quadratic function, and its range can be determined by finding its vertex. The vertex form of a quadratic function is y = a(x-h)² + k, where (h,k) is the vertex of the parabola. To find the vertex of the function k(t), we can complete the square or use the fact that the x-coordinate of the vertex is given by -b/(2a).
The equation can be rearranged in the form of at² + bt + c. Here, a = 1, b = -4, and c = 5. Applying the formula for the x-coordinate of the vertex, we get t = -(-4)/(2· 1) = 2. To find the y-coordinate of the vertex, we substitute t = 2 into the original function k(t), giving us k(2) = 2² - 4·2 + 5 = 4 - 8 + 5 = 1.
Since a = 1 is positive, the parabola opens upwards, meaning that the value at the vertex is the minimum value of the function. Therefore, the function k(t) has a range of (1, ∞), which corresponds to option c) (1, ∞) from the given choices. This is the correct option as the range includes all real numbers greater than or equal to 1.