Find the parabola with equation y = ax^2 + bx+c whose tangent line at (1,1) has equation y = 3x − 2, and satisfies y''(2) = 3.

Answer 1

f'(1)=3 and f''(2)=3

Y'=2ax+b

Y''=2a

So 2a=3 --> a=3/2

2(3/2)(1)+b=3

B=0

Ffinally per (1,1)

(3/2)(1)^2+c=1

So c= -1/2

Y=(3/2)x^2-1/2