Question

Identify the turning point of the function f(x)=x^2-2x+8 by writing its equation in vertex form.

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Answer 1

`f(x)=x^2-2x+8 \\ \\ To \ convert \ the \ standard \ form \ y = ax^2 + bx + c \\ \\ of \ a \ function \ into \ vertex \ form \\ \\ y = a(x - h)^2 + k \\ \\ Here \ the \ point \ (h, k) \ is \ called \ as \ vertex \\ \\ h=(-b)/(2a) , \ \ \ \ k= c - (b^2)/(4a)\\ \\ a=1 ,\ b=-2 , \ c = 8`


`h=(-b)/(2a)=(-(-2))/(2 )=(2)/(2)=1 \\\\ k=c - (b^2)/(4a) =8 - ( (-2)^2)/(4 ) = 8-1 = 7 \\ \\ y= (x-1)^2 + 7 \\ \\ So \ the \ turning \ point \ is \ (1,7)`