math related question?
find the coordinates of the minimum point of f(x)=x^2-2x+5
-------------------------------------------------------

answers:
Como say: f ' (x) = 2x - 2 = 0 for turning point
x = 1 for turning point
Turning point (1, 5)
-
MyRank say: f(x) = x² - 2x + 5 …..(1)

f’(x) = 2x - 2 + 0

f’(x) = 0

2x - 2 = 0

2(x-1) = 0

x = 1 substitute in 1

y = f(x) = 4

(x, y) = (1, 4) minimum point.
-
khalil say: f'(x) = 0
2x - 2 = 0
x = 1
f(1) = 4
Min(1, 4)
-
Φ² = Φ+1 say: The turning point of f(x) = ax² + bx + c is ( -b/(2a), f(-b/(2a)) )

Here this is ( 1, f(1) ), which is ( 1, 4 ).
a is positive, so this is the minimum point.
-
Puzzling say: Convert that to vertex form:

f(x) = (x² - 2x) + 5
f(x) = (x² - 2x + 1) + 4 <-- completing the square
f(x) = (x - 1)² + 4

So the vertex (a minimum) is at (1,4)
-
jgusa say: ygrofkaf
-