Final answer:
To find a parametrization of the graph y=2x^3 + 4x - 1, we substitute the x-part with our parameter t and match it with the given y-value. Only the first parametrization satisfies the original equation, making (i) the correct parametrization and option (H) the correct answer.
Step-by-step explanation:
To parametrize the graph of y=2x^3 + 4x - 1, we are looking for a set of parametric equations that represent all the points on this curve. A parametrization (t, y(t)) should satisfy the original equation when we substitute the x-part with our parameter t.
For the first parametrization (i): (2t, 16t^3 + 8t - 1), substituting x with 2t gives us:
y = 2(2t)^3 + 4(2t) - 1 = 16t^3 + 8t - 1.
This matches the y-value of the parametrization, therefore it is a valid parametrization.
For the second parametrization (ii): (t + 1, 2t^3 + 6t^2 + 10t + 5), this does not satisfy the original equation for an arbitrary parameter t, as it results in a different cubic equation not equivalent to y = 2x^3 + 4x - 1.
For the third parametrization (iii): (2t, 4t^3 + 8t - 2), substituting x with 2t gives us: y = 2(2t)^3 + 4(2t) - 1 = 16t^3 + 8t - 1. However, this does not match the y-value of the parametrization which is 4t^3 + 8t - 2, therefore it is not valid either.
Thus, the correct answer is option (H) (i) only.
Complete Question:
Which of the following provides a parametrization of the graph of y=2x^3 + 4x − 1?
(i) (2t, 16t^3 + 8t − 1)
(ii) (t + 1, 2t^3 + 6t^2 + 10t + 5)
(iii) (2t, 4t^3 + 8t−2)
(A) (iii) only
(B) (ii) and (iii) only
(C) (i) and (ii) only
(D) all of them
(E) none of them
(F) (ii) only
(G) (i) and (iii) only
(H) (i) only