SOLUTION: I need help. I need to state the range of the function f(x)= (x-2)^2 + 1 I have solved it this far..... (x-2) (x-2) + 1 x^2-2x-2x+4 +1 x^2-4x+5 and then I equaled it to zer

Algebra ->  Graphs -> SOLUTION: I need help. I need to state the range of the function f(x)= (x-2)^2 + 1 I have solved it this far..... (x-2) (x-2) + 1 x^2-2x-2x+4 +1 x^2-4x+5 and then I equaled it to zer      Log On


   

Question 48991: I need help. I need to state the range of the function f(x)= (x-2)^2 + 1
I have solved it this far.....
(x-2) (x-2) + 1
x^2-2x-2x+4 +1
x^2-4x+5 and then I equaled it to zero.
X^2-4x + 5 = 0
Now, I am stuck. Am I doing it right? If I am finding the range I am trying to find Y, right? So I need to solve x so I can find y. But what if I can not go any further?
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 - 4x + 5 is a parabola with a vertex of (-b/2a,f(x))
vertex:(2,1) and the parabola opens up
Range: y+%3E=+1
graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%28x-2%29%5E2%2B1%29