Question

Using correct vocabulary, select the phrases describing how the following equationschanged the parent function

Answer 1

Let's transform the parent function step by step and name each.

The order in which we do the operations will get us to two possible answers.

The first order, we will start by doing the translation of "x" and then its reflection and then do the stretch and then translation of "y". So, starting from the traslation, we need to look for term without variables beside the "x" and alone in the function.

Inside the cubic root, we have a -2, we can get it by doing a horizontal translation of 2 units right:

`f_1(x)=f_0(x-2)=\sqrt[3]{x-2}`

Now, to get the negative sign onto "x", we need to do a y-axis reflection:

`f_2(x)=f_1(-x)=\sqrt[3]{-x-2}`

Now, we see that there is a coefficient in front of the square root, so we need to get it there by making a vertical compression of 4, also named, vertical stretch of 1/4:

`f_3(x)=(1)/(4)f_2(x)=(1)/(4)\sqrt[3]{-x-2}`

Outside, we have another "-2", which we can get by doing a vertical translation of 2 units down:

`f_3(x)=f_3(x)-2=(1)/(4)\sqrt[3]{-x-2}-2`

And we achieved the function. In this case, we have done:

1 - horizontal translation of 2 units right

2 - y-axis reflection

3 - vertical stretch of 1/4

4 - vertical translation of 2 units down

However, using these alternatives we can still get another answer.

We start by doing the y-axis reflection:

`f_1(x)=f_0(-x)=\sqrt[3]{-x}`

Then, we do a horizontal translation of 2 units left:

`f_2(x)=f_1(x+2)=\sqrt[3]{-(x+2)}=\sqrt[3]{-x-2}`

And now we complete with the same last two:

Vertical stretch of 1/4:

`f_3(x)=(1)/(4)f_2(x)=(1)/(4)\sqrt[3]{-x-2}`

And vertical translation of 2 units down:

`f_4(x)=f_3(x)-2=(1)/(4)\sqrt[3]{-x-2}-2`

Notice that we have got to the same function, but using a different answer:

1 - y-axis reflection

2 - horizontal translation of 2 units left

3 - vertical stretch of 1/4

4 - vertical translation of 2 units down