The range of the function f(x)=-sqrt{5-6x-x^2 | vedclass

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(B) Given $f(x) = -\sqrt{5-6x-x^2}$.First,we find the domain by solving $5-6x-x^2 \geq 0$.$x^2+6x-5 \leq 0$.Completing the square: $(x+3)^2 - 14 \leq

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$[-\sqrt{14}, 0]$

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(B) Given $f(x) = -\sqrt{5-6x-x^2}$.First,we find the domain by solving $5-6x-x^2 \geq 0$.$x^2+6x-5 \leq 0$.Completing the square: $(x+3)^2 - 14 \leq 0 \Rightarrow (x+3)^2 \leq 14$.So,$-\sqrt{14} \leq x+3 \leq \sqrt{14}$,which means $x \in [-3-\sqrt{14}, -3+\sqrt{14}]$.Now,let $y = f(x) = -\sqrt{14-(x+3)^2}$.Since $(x+3)^2 \geq 0$,the maximum value of $14-(x+3)^2$ is $14$ (at $x=-3$).Thus,the maximum value of $\sqrt{14-(x+3)^2}$ is $\sqrt{14}$.Since $y = -\sqrt{14-(x+3)^2}$,the minimum value is $-\sqrt{14}$ (at $x=-3$) and the maximum value is $0$ (when $14-(x+3)^2 = 0$).Therefore,the range is $[-\sqrt{14}, 0]$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

The range of the function $f(x)=-\sqrt{5-6x-x^2}$ is

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