Question

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Answer 1

Answer:

Part 1) $f(x)=\sqrt{x-1}$ -----> Graph C

Part 2) $g(x)=-\sqrt{x}$ ----> Graph D

Part 3) $h(x)=\sqrt{x}$ -----> Graph A

Part 4) $j(x)=-\sqrt{x-1}$ ---> Graph E

Step-by-step explanation:

Part 1) we have

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

Find the domain of the function

The radicand must be positive

so

$x-1\geq 0$

Solve for x

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

Find the range

For x=1

$f(1)=\sqrt{1-1}=0$

so

The range is the interval ----> [0,∞)

All real numbers greater than or equal to 0

therefore

The function represent Graph C

Part 2) we have

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

Find the domain of the function

The radicand must be positive

so

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

Find the range

For x=0

$f(0)=-\sqrt{0}=0$

so

The range is the interval ----> (-∞,0]

All real numbers less than or equal to 0

therefore

The function represent Graph D

Part 3) we have

$h(x)=\sqrt{x}$

Find the domain of the function

The radicand must be positive

so

$x\geq 0$

Solve for x

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

Find the range

For x=0

$f(0)=\sqrt{0}=0$

so

The range is the interval ----> [0,∞)

All real numbers greater than or equal to 0

therefore

The function represent Graph A

Part 4) we have

$j(x)=-\sqrt{x-1}$

Find the domain of the function

The radicand must be positive

so

$x-1\geq 0$

Solve for x

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

Find the range

For x=1

$f(1)=-\sqrt{1-1}=0$

so

The range is the interval ----> (-∞,0]

All real numbers less than or equal to 0

therefore

The function represent Graph E

Answer 2

Answer:

Graph A → y=√x.

Graph B → y=(√x) - 1.

Graph C → y=√(x-1).

Graph D → y= -√x.

Graph E → y= -√(x-1)

Step-by-step explanation:

The graph 'A' intercepts the y-axis at (0, 0). Therefore it belongs to the function y=√x.

The graph 'D' is exactly the same graph 'A' but reflected across the x-axis. Therefore, it belongs to the function y=-√x.

The function 'C' is exactly the same function y=√x but translated one unit to the right, therefore, the solution function is y=√(x-1)

The graph 'E' is exactly the same graph 'C' but reflected across the x-axis, therefore the function is: y= -√(x-1)

In the options you have two times the function y=√x. I assume that's a mistake. The graph 'B' corresponds to y = (√x) - 1