How to Simplify f(3a-4) for the Function f(x)=-4x+5?

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In summary, to find and simplify $f(3a-4)$, we substitute $3a-4$ for $x$ in the definition of $f(x)$, which is $-4x+5$. This results in $-12a+16+5=-12a+21$ after simplification. Substitution is similar to throwing a value into a portal, where $x$ is the entry and may appear multiple or zero times as the exit.
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Let f(x)=-4x+5. Find and simplify f(3a-4)

I have the answer, but I don't know how to get there.
Missed it on a test.
Answer= = -12a+21
 
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Monk said:
Let f(x)=-4x+5. Find and simplify f(3a-4)
When you are asked to find $f(3a-4)$, it means that you have to substitute $3a-4$ for every occurrence of $x$ in the definition of $f(x)$. The definition (i.e., the right-hand side) of $f(x)$ is $-4x+5$. It has a single occurrence of $x$. This occurrence must be replaced by $3a-4$. Thus, the result of substitution is $-4(3a-4)+5$. The remaining step is simplification. Here you have to "multiply through", i.e., apply distributivity of multiplication over addition. The expression becomes $-4\cdot(3a)-4\cdot(-4)+5$, which simplifies to $-12a+16+5=-12a+21$.

I have not played modern games, but if I am right, substitution is similar to engaging a portal. The definition like $f(x)=-4x+5$ is a portal with entry $x$ on the left-hand side and exit $x$ on the right-hand side. Applying $f(x)$ to $3a-4$ is similar to throwing $3a-4$ into the portal, so it appears unchanged at the exit. Note that even though $x$ appears once on the left-hand side of the definition of $f$, it may appear several or zero times at the right-hand side, i.e., the portal may have several or zero exits.
 

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