Identify the turning point of the function [tex]f(x) = x^2 - 2x + 8[/tex] by writing its equation in vertex form.
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Answer (2)
f ( x ) = x 2 − 2 x + 8 T o co n v er t t h e s t an d a r d f or m y = a x 2 + b x + c o f a f u n c t i o n in t o v er t e x f or m y = a ( x − h ) 2 + k Here t h e p o in t ( h , k ) i s c a ll e d a s v er t e x h = 2 a − b , k = c − 4 a b 2
a = 1 , b = − 2 , c = 8 h = 2 a − b = 2 − ( − 2 ) = 2 2 = 1 k = c − 4 a b 2 = 8 − 4 ( − 2 ) 2 = 8 − 4 ( − 2 ) 2 = 8 − 4 4 = 8 − 1 = 7 y = ( x − 1 ) 2 + 7 V er t e x = ( h , k ) = ( 1 , 7 ) T h e \vertex o f t hi s g r a p h w i ll b e m o v e d o n e u ni t t o t h e r i g h t an d se v e n u ni t s u p f ro m ( 0 , 0 ) , t h e v er t e x o f i t s p a re n t y = x 2 .
The function f ( x ) = x 2 − 2 x + 8 has its turning point at the vertex, which is found to be ( 1 , 7 ) . This vertex form can be expressed as f ( x ) = ( x − 1 ) 2 + 7 . Therefore, the turning point of the function is at the point (1, 7).
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