Expert: Ahmed Salami - 11/24/2011

Question
a]Express f(x)=x^2-6x+14 in the form f(x)=(x-h)^2+k,where h and k are to be determined.

b] Hence, or otherwise, write down the coordinates of the vertex of the parabola with equation y-x^2-6x+14.

Answer
Hi Kevin,
For a quadratic form x² + kx, we complete the square by adding (k/2)² and then factorising as (x - k/2)².
So, for f(x) = x² - 6x + 14
f(x) = x² - 6x + 3² - 3² + 14
= (x² - 6x + 9) + (14 - 9)
= (x - 3)² + 5

If y = (x - 3)² + 5, we can already see that y cannot be smaller than 5 because (x - 3)² is always positive and has a minimum value of 0 at x = 3. The coordinates of the vertex (minimum value in this case) is therefore (3, 5).

Regards