Question

find the equation of the square root function described contextually as follows: - its domain is ( x leq 0 ) and its range is ( y leq -5 ). - it passes through the point (-9, -23). - ( b = -1 ) use the general formula ( f(x) = asqrt{b(x - h)} + k ). the corresponding square root function is: ( f(x) = squaresqrt{square(square)}square )

Explanation:

Step1: Identify given values

Given $b=-1$, domain $x\leq0$, range $y\leq-5$, point $(-9,-23)$.

Step2: Find $h$ from domain

For $\sqrt{b(x-h)}=\sqrt{-(x-h)}$, radicand $\geq0$:
$-(x-h)\geq0 \implies x-h\leq0 \implies x\leq h$.
Since $x\leq0$, $h=0$.

Step3: Find $k$ from range

The range of $a\sqrt{-(x-0)}+k$ is $y\leq k$ (since $a$ will be negative to match range). Given $y\leq-5$, so $k=-5$.

Step4: Substitute point to find $a$

Substitute $x=-9$, $f(x)=-23$, $h=0$, $k=-5$, $b=-1$ into $f(x)=a\sqrt{b(x-h)}+k$:
$-23 = a\sqrt{-1(-9-0)} + (-5)$
Simplify radicand: $\sqrt{9}=3$
$-23 = 3a -5$
Solve for $a$:
$3a = -23 + 5 = -18$
$a = \frac{-18}{3} = -6$

Step5: Build the function

Substitute $a=-6$, $b=-1$, $h=0$, $k=-5$ into the general formula:
$f(x) = -6\sqrt{-1(x-0)} -5$
Simplify the expression inside the square root: $-1(x-0)=-x$
$f(x) = -6\sqrt{-x} -5$

Answer:

$f(x) = -6\sqrt{-x} -5$