Question

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

Show your work.

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 -((-2)^2)/(4)= 8 -(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.`