SOLUTION: Vertex at (2,1) and passes the point (0,-3). Find the quadratic function. f(x)=

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Vertex at (2,1) and passes the point (0,-3). Find the quadratic function. f(x)=      Log On


   

Question 360031: Vertex at (2,1) and passes the point (0,-3). Find the quadratic function. f(x)=
Answer by Jk22(389) About Me  (Show Source):
You can put this solution on YOUR website!
derivative of f should be 0 in 2 : f(x) = a(x-2)^2 + c

f(2) = c = 1
f(0) = 4a + c = -3 = 4a + 1 => a = -1

result : f(x) = -(x-2)^2 + 1 = -(x^2-4x+4) + 1 = -x^2 + 4x - 3


Verification : f(0) = -3, f'(x) = -2x+4 => f'(2) = 0 (maximum)