+701 hello everybody,
hope you've had an awesome holiday and happy new year!
i do not understand how this problem turns into\(f(x)=9(x-1/3)+2\)
here is the original thing:
\(f(x)=9x^2-6x+3\)
i got an answer that is totally off:
\(f(x)=9(x-1/3)^2+6\)
i don't understand where i went wrong, and every time i attempt to do the problem, i keep getting different results each time.
can someone please explain how and why this equals to \(f(x)=9(x-1/3)^2+2\)
thank you! <3
+37153 f(x) =9x^2-6x+3 'complete the square' to do this , leading coeff has to be 1
divide by 9 (we'll put it back in shortly)
f(x)/9 = x^2 - 2/3 x +1/3 now complete the square
x^2 - 2/3 x + 1/9 -1/9 +1/3
f(x) / 9 = (x-1/3)^2 - 1/9 + 3/9
f(x) / 9 = (x- 1/3)^2 + 2/9 Now remember that division by 9? Put the 9 back in by multiplying by 9
f(x) = 9 (x-1/3)^2 + 2 END !
