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Find the co-ordinates of the minimum of the parabola?
"By completeing the square, find the co-ordinates of the minimum of the parabola y=x^2 -10x +22."
I've just revised how to complete the square, although I learnt it a different way this time. I worked it down to:
(X - 5)^2 -3
How do I find out the co-ordinates?
Thanks
1 Answer
- The Integral ∫Lv 710 years agoFavorite Answer
y = x^2 -10x + 22 ===> completing the square
y = x^2 -10x + (-10/2)^2 - (-10/2)^2 + 22
y = x^2 -10x + (-5)^2 - (-5)^2 + 22
y = x^2 -10x + 25 - 25 + 22
y = x^2 -10x + 25 - 3
y = (x - 5)^2 - 3 <===== vertex form of the parabola
the vertex is (5 , -3) <===== just take the opposite sign of whats inside the parentheses and same as sign of -3.
since our leading coefficient of the parabola is positive which is +x^2, we should know that the vertex is minimum.
