Question

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

Answer 1

(1,7)

Working;
Start by finding the derivative of the function f(x);
$f^{'} (x)$=2x-2
Turning points occur when the derivative function=0;
Thus
2x-2=0
x=1
when x=1 then
f(1)=$1^{2} -2(1)+8$
=7
Turning point then is (1,7)