Let $f:\mathbb R→\mathbb R^3$ be defined by $f(x)=\langle 8x,6x,−8x\rangle$. Is f a linear transformation? I am not sure on what process I should have when dealing with this problem?

I don't know what f(y) is equal to..

To fill in the blanks, the answers must be be written as a formula that returns a vector, which can be done using angle brackets.

For example:&lt;8,6,-8> which is a result with 3 dimensions.
If I put a result that doesn't have 3 dimensions, then I will get an error.

In the "choose" drop-down menu for part a) and part b), it has 2 options that can be selected. The 2 options are "Yes, they are equal" and "No, they are not equal"

For part c), the two options are "f is a linear transformation" and "f is not a linear transformation"
When we write $f(x) = \langle 8x,6x,-8x \rangle$, it means that $f$ assigns to each real number $x$ the 3-dimensional vector $\langle 8x,6x,-8x \rangle$. Thus, for example, $f(1) = \langle 8,-6,-8 \rangle$ and $f(0)$ is the zero vector.

So, $f(y)$ is just the vector $\langle 8y,6y,-8y \rangle$ and $f(x+y)$ is...

It is easy to check that, in fact, $f(x+y) = f(x) + f(y)$.
Please find the solution, let me know if you have questions.