Left 1unit, Reflect over x-axis, up 2 units

Left 1unit, Reflect over y-axis, down 2 units

Right 1 unit, Reflect over x-axis, up 2 units

Right 1unit, Reflect over y-axis, up 2 units

$y=\sqrt{x+3}-1$

$y=\sqrt{x-3}-1$

$y=\sqrt{x-3}+1$

$y=-3\sqrt{x+1}$

$y=\sqrt{x+5}-1$

$y=-5\sqrt{x}-1$

$y=\sqrt{x-5}-1$

$y=\sqrt{x+5}+1$

$y=\sqrt{x-1}-2$

$y=-\sqrt{x-1}-2$

$y=-\sqrt{x+1}-2$

$y=\sqrt{x+1}-2$

The domain is $\left[5,\ \infty\right)$  

The domain is $\left[-5,\ \infty\right)$  

The range is  $\left[3,\ \infty\right)$  

The range is  $\left(-\infty,\ 3\right]$  

This function is decreasing, not increasing.

$\left(-\infty,\ \infty\right)$

$\left[1,\ \infty\right)$

$\left[-2,\ \infty\right)$

$\left(1,\ \infty\right)$

(−∞,∞)

[0,∞)

(−∞,0]

All cube root functions are different.

Down 2

Up 2

Left 2

Reflection and down 2

Translate Right 2, Down 4

Translate Left 2, Down 4

Translate Left 2, Up 4

Translate Right 2, Up 4

$\left(2,\ \infty\right)$

$\left(3,\ \infty\right)$

$\left[0,\ 4\right]$

$\left(-\infty,\ \infty\right)$

f(x) = − ∛(x + 3) − 1

f(x) = − ∛(x - 3) − 1

f(x) = ∛(x + 3) − 1

f(x) = ∛(x - 1) + 3

x = -3 y = 1

x = 3 y = -1

x = 1 y = -3

x = -1 y = 3

$f\left(x\right)\ =\ \frac{1}{x-3}+1$

$f\left(x\right)\ =\ \frac{1}{x+3}+1$

The domain is all real numbers except -3

The range is all real numbers except 1

This function is reflected over the x-axis.

$y=\frac{1}{x-1}-2$

$y=\frac{1}{x+1}+2$

$y=\frac{1}{x+1}-2$

$y=\frac{1}{x-2}+1$

$y=\frac{1}{x+2}+1$

$y=\frac{1}{x+2}-1$

$y=\frac{1}{x-2}+1$

$y=\frac{1}{x-2}-1$

It is located 4 units higher.

It is located 4 units to the right.

It is located 4 units to the left.

It is 4 times taller.

 $\left(-\infty,-2\right)\cup\left(-2,\infty\right)$  

 $\left(-\infty,\ 2\right)\cup\left(2,\infty\right)$  

 $\left(-\infty,-3\right)\cup\left(-3,\infty\right)$  

 $\left(-\infty,\ 3\right)\cup\left(3,\infty\right)$  

$f\left(x\right)=\frac{1}{x-1}+2$

$f\left(x\right)=\frac{1}{x+2}-1$

$f\left(x\right)=\frac{1}{x+1}+2$

$f\left(x\right)=\frac{1}{x-2}+1$

(1, ∞)

None

(2, ∞)

(-∞, ∞)

The interval of decrease is  $\left(0,\ \infty\right)$  

The domain is  $\left[0,\ \infty\right)$  

The range is  $\left(-\infty,\ -2\right]$  

The range is  $\left[-2,\ \infty\right]$  

The interval of decrease is  $\left(-2,\ \infty\right)$  

 $\left[2,\infty\right)$  

 $\left(-\infty\ -2\right)$  

 $\left[-2,\ \infty\right)$  

 $\left[-4,\ \infty\right)$  

 $\left(-\infty,\ -6\right)\cup\left(-6,\ \infty\right)$  

 $\left(-\infty,\ 6\right)\cup\left(6,\ \infty\right)$  

 $\left(-\infty,\ 3\right)\cup\left(3,\ \infty\right)$  

 $\left(-\infty,\ \infty\right)$  

$f\left(x\right)\ =\ \sqrt{x-2}-1$

$f\left(x\right)\ =\ \sqrt{x+2}-1$

$f\left(x\right)\ =\ \sqrt{x-1}-2$

$f\left(x\right)\ =\ -\sqrt{x-2}-1$

f(x) = ∛(x - 1)

f(x) = - ∛(x + 1)

f(x) = ∛(x) + 1

f(x) = -∛(x) +1

$f\left(x\right)\ =\ \frac{1}{x+4}+1$

$f\left(x\right)\ =\ -\frac{1}{x-4}+1$

$f\left(x\right)\ =\ \frac{1}{x-4}+1$

$f\left(x\right)=-\frac{1}{x+4}+1$

$f\left(x\right)\ =\ \sqrt{x+5}-5$

$f\left(x\right)\ =\ -\sqrt{x+5}-5$

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

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

x = -2

x = 2

y = -6

y = 6