Question

Find all the polynomials f (t ) of degree ≤ 2 [of the form f (t) = a + bt + ct2] whose graphs run through the points (1, 3) and (2, 6), such that f ′(1) = 1 [where f ′(t) denotes the derivative?

Answer 1

Answer:

f(t) = 4 - 3t + 2t^2

Step-by-step explanation:

Given:

f(t) = a + bt + ct^2

Differentiate w.r.t. "t"

f'(t) = b + 2ct

1) f(1) = 3

a + b +c = 3...........1)

2) f(2) = 6

a + 2b + 4c = 6.........2)

3) f'(1) = 1

b + 2c = 1 ...............3)

Solving above three simultaneous equations:

a = 4

b = -3

c = 2

Therefore equation becomes:

f(t) = 4 + (-3)t + (2)t^2