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

Lilith
[tex] 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=\frac{-b}{2a} , \ \ \ \ k= c - \frac{b^2}{4a}\\ \\ a=1 ,\ b=-2 , \ c = 8 [/tex]

[tex]h=\frac{-b}{2a}=\frac{-(-2)}{2 }=\frac{2}{2}=1 \\\\ k=c - \frac{b^2}{4a} =8 - \frac{ (-2)^2}{4 } = 8-1 = 7 \\ \\ y= (x-1)^2 + 7 \\ \\ So \ the \ turning \ point \ is \ (1,7)[/tex]


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