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Identify the turning point of the function f(x)=x^2-2x+8 by writing it's 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}[/tex]

 [tex] a=1, \ b=-2, \ c=8\\ \\ h=\frac{-b}{2a}=\frac{-(-2)}{2}=\frac{2}{2}=1\\ \\ k= c - \frac{b^2}{4a}=8 -\frac{(-2)^2}{4}=8 -\frac{(-2)^2}{4}= 8 -\frac{4}{4}=8-1=7\\ \\ y = (x - 1)^2 + 7 \\ \\Vertex = (h, k) = (1, 7)\\The \vertex \ of \ this \ graph \ will \ be \ moved \ one \ unit \ to \ the \ right\\ and \ seven \ units \ up \ from \ (0,0),\ the \ vertex \ of \ its \ parent \ y = x^2. [/tex]

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