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mafiozo [28]
3 years ago
7

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

Mathematics
2 answers:
liq [111]3 years ago
5 0

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}

<em>Find the domain of the function</em>

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

<em>Find the range</em>

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}

<em>Find the domain of the function</em>

The radicand must be positive

so

x\geq 0

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

All real numbers greater than or equal to 0

<em>Find the range</em>

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}

<em>Find the domain of the function</em>

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

<em>Find the range</em>

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}

<em>Find the domain of the function</em>

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

<em>Find the range</em>

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

pashok25 [27]3 years ago
4 0

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

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