$\displaystyle x^2-8x+23=(x-p)^2+q$ for all values of x
Find p and q, does this mean I can just use x as anything such as 0?
if so using simultaneous equations I got $\displaystyle -8=p^2-p^2$
What now?
Completing the square is covered in post 3 of this topic: http://www.mathhelpforum.com/math-he...tic-curve.html
Comparing coefficients says that two equations are equal if and only if the coefficients on each side match.
For example given that ax^2+bx+c = x^2 - 12x+1 comparing coefficients tells you that a = 1, b=-12 and c = 1