Question

Find an equation for the graphed function below by using transformations of the grade h of a toolkit function.

Answer 1

`y^{}=\sqrt[]{x-1}\text{ +2}`

Step-by-step explanation

Step 1

the original function is

`y=\sqrt[\square]{x}`

we can see that the original function was shifted

a) one unit to the rigth

b) 2 units up

so

Step 2

A) shifted one unit to the right:

To shift, move, or translate horizontally, replace y = f(x) with y = f(x + c) (left by c) or y = f(x - c) (right by c).so

C=1, to the rigth, so f(x-1)

replace

`\begin{gathered} y=\sqrt[]{x} \\ y^(\prime)=\sqrt[]{x-1} \end{gathered}`

Step 3

B)now, shifted 2 units up.

To move a function up, you add outside the function: f (x) + b is f (x) moved up b units. Moving the function down works the same way; f (x) – b is f (x) moved down b units.

hence,

`\begin{gathered} y^(\prime)=\sqrt[]{x-1} \\ b=2,so \\ y^(\prime)^(\prime)=\sqrt[]{x-1}\text{ +2} \end{gathered}`

therefore, the answer is

`y^{}=\sqrt[]{x-1}\text{ +2}`

I hope this helps you