Find, by completing the square, the greatest value of the function $f(x)=1-6x-x^2$

QUADRATIC FUNCTION

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$f(x)=1-6x-x^2$

$=10-(9+6x+x^2)$

$=10-(3+x)^2$

Since $(3+x)^2$ is the square of a real number it cannot be negative, it is zero when $x=-3$, otherwise it is positive. Consequently $10-(3+x)^2$ is always less than or equal to $10$

$\therefore$ the greatest value of the function is $10$ and this occurs when $x=-3$.

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Asked: 2 months ago