How do you complete the square $f (x) = x^{2} + 6 x - 3$ $f(x)=x^2+6x-3$ ?

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Answer 1 · 2016-11-18T02:41:02

$f (x) = 1 (x^{2} + 6 x + m - m) - 3$ $f(x)= 1(x^2 + 6x + m - m) - 3$

$m = {(\frac{b}{2})}^{2} = {(\frac{6}{2})}^{2} = 9$ $m = (b/2)^2 = (6/2)^2 = 9$

$f (x) = 1 (x^{2} + 6 x + 9 - 9) - 3$ $f(x) = 1(x^2 + 6x + 9 - 9) - 3$

$f (x) = 1 {(x + 3)}^{2} - 9 - 3$ $f(x) = 1(x + 3)^2 - 9 - 3$

$f (x) = {(x + 3)}^{2} - 12$ $f(x) = (x + 3)^2 - 12$

Hopefully this helps!

How do you complete the square f(x)=x2+6x−3f(x)=x^2+6x-3?