Answer:
![W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}](https://tex.z-dn.net/?f=W%3D%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3A%20a%2Cb%5Cin%5Cmathbb%7BR%7D%20%5C%7D)
Observe that if the vector
is in W then it satisfies:
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{c}a+2b\\b\\-3a\end{array}\right]=a\left[\begin{array}{c}1\\0\\-3\end{array}\right]+b\left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%2B2b%5C%5Cb%5C%5C-3a%5Cend%7Barray%7D%5Cright%5D%3Da%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
This means that each vector in W can be expressed as a linear combination of the vectors ![\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C0%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%2C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C1%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Also we can see that those vectors are linear independent. Then the set
is a basis for W and the dimension of W is 2.
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20a_n%3D2-5%28n-1%29%5Cimplies%20a_n%3D%5Cstackrel%7B%5Cstackrel%7Ba_1%7D%7B%5Cdownarrow%20%7D%7D%7B2%7D%2B%28n-1%29%28%5Cstackrel%7B%5Cstackrel%7Bd%7D%7B%5Cdownarrow%20%7D%7D%7B-5%7D%29)
so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

just a quick note on notation:
![\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_n%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bthe%20term%20before%20it%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_%7Bn-1%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_5%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B5-1%7D%5Cimplies%20a_4%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_%7B12%7D%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B12-1%7D%5Cimplies%20a_%7B11%7D%7D)
f(-4) = 3(-4)^2 - 2(-4)
f(-4) = 3(16) + 8
f(-4) = 48 + 8
f(-4) = 56
I think the answer is 56 :)
Please give BRAINIEST :)