Question

1. reasoning consider a function of the form $f(x)=a\sqrt{x - h}+k$. describe when the function is increasing or decreasing when (a) $a > 0$ and (b) $a < 0$.

1. college prep the graph of which function is shown?

(a) $y = \sqrt{x - 12}$
(b) $y = \sqrt{x}-12$
(c) $y = \sqrt{x + 12}$
(d) $y = \sqrt{-x + 12}$

in exercises 33-36, write a rule for $g$ described by the transformations of the graph of $f$.

1. let the graph of $g$ be a reflection in the x-axis of the graph of $f(x)=\sqrt{x - 5}$.
2. let the graph of $g$ be a translation 4 units down of the graph of $f(x)=\sqrt{x}+7$.
3. let $g$ be a vertical stretch by a factor of 2, followed by a translation 2 units up of the graph of $f(x)=\sqrt{x}+3$.
4. let $g$ be a horizontal shrink by a factor of $\frac{2}{3}$, followed by a translation 4 units left of the graph of $f(x)=\sqrt{6x}$.

in exercises 37-44, describe the transformation from the graph of $f$ to the graph of $g$. (see example 4.)

1. $f(x)=\sqrt{x}+2$, $g(x)=f(x - h)$
2. $f(x)=\sqrt{x + 1}-3$, $g(x)=f(ax)$

39.

$x$	$-25$	$-16$	$-9$	$-4$	$-1$
$g(x)=a\cdot f(x)$	$22$	$18$	$14$	$10$	$6$

40.

$x$	$0$	$1$	$4$	$9$	$16$
$g(x)=f(x)+k$	$12$	$11$	$10$	$9$	$8$

1. $f(x)=\sqrt{x}-1$, $g$
2. $f(x)=\sqrt{x - 1}-2$, $g$

Explanation:

Problem 31

Step1: Analyze base function behavior

The base square root function $y=\sqrt{x-h}$ is increasing for $x>h$, as its derivative $\frac{1}{2\sqrt{x-h}}>0$ for $x>h$.

Step2: Apply vertical scale ($a>0$)

When $a>0$, the function $f(x)=a\sqrt{x-h}+k$ preserves the increasing trend of the base function, since scaling by a positive factor does not reverse slope sign. So $f(x)$ is increasing for $x>h$.

Step3: Apply vertical scale ($a<0$)

When $a<0$, the function $f(x)=a\sqrt{x-h}+k$ reflects the base function over the x-axis, reversing the slope sign. So $f(x)$ is decreasing for $x>h$.

Step1: Identify domain of the graph

The graph starts at $x=12$ (domain $x\geq12$), so the square root term must be $\sqrt{x-12}$ (since $\sqrt{x-12}$ is defined for $x\geq12$).

Step2: Match to options

Only option A has the form $y=\sqrt{x-12}$, which matches the domain and the shape of the square root graph starting at $(12,0)$ and increasing.

Step1: Apply x-axis reflection rule

A reflection over the x-axis transforms $f(x)$ to $-f(x)$.

Step2: Substitute $f(x)=\sqrt{x-5}$

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

Answer:

(a) When $a>0$, $f(x)$ is increasing for all $x > h$.
(b) When $a<0$, $f(x)$ is decreasing for all $x > h$.

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Problem 32