SOLUTION: Find the exact vertex of the parabola algebraically... f(x) = (x - 1)^2 + (x + 3) Could you please provide a detailed explantion of how you achieved the solution. Thank

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the exact vertex of the parabola algebraically... f(x) = (x - 1)^2 + (x + 3) Could you please provide a detailed explantion of how you achieved the solution. Thank      Log On


   

Question 192073: Find the exact vertex of the parabola algebraically...

f(x) = (x - 1)^2 + (x + 3)

Could you please provide a detailed explantion of how you achieved the solution. Thanks!
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The "vertex form" is
y= a(x-h)2+k
where
(h,k) is the vertex
.
The idea is to make your equation look like that:
f(x) = (x - 1)^2 + (x + 3)
f(x) = (x - 1)(x - 1) + (x + 3)
f(x) = x^2 - 2x + 1 + x + 3
f(x) = x^2 - x + 4
Group x-terms:
f(x) = (x^2 - x) + 4
Complete the square:
f(x) = (x^2 - x + __ ) + 4
f(x) = (x^2 - x + 1/4 ) + 4 -1/4
f(x) = (x - 1/2)^2 + 16/4 -1/4
f(x) = (x - 1/2)^2 + 15/4
.
vertex = (1/2, 15/4)