What is the relationship between the minimal polynomial of $\alpha$ and the minimal polynomial of a linear function of $\alpha$?

Question

Let $k$ be a field, with algebraic closure $\overline{k}$, and let $\eta\in \overline{k}$. Let $f(x) \in k[x]$ be the irreducible polynomial of $\eta $ over $k$. Say I wanted to find the irreducible polynomial of $\eta -1 $, or more generally, the irreducible polynomial of $a\eta +b$ where $a,b \in k$? How is that irreducible polynomial related to $f(x)$?

Answer 1

Claim: Let $a \neq 0$. The irreducible polynomial of $a\eta + b$ is $f(\frac{1}{a}(x-b))$.

Proof: First we check that $a\eta+b$ is a root of $f(\frac{1}{a}(x-b))$. $$ f \left( \frac{1}{a} (a\eta+b-b) \right) = f \left( \frac{1}{a} (a \eta) \right) = f(\eta) = 0 $$ since $a \neq 0$. Also, $f( \frac{1}{a}(x-b))$ is irreducible because a linear substitution of an irreducible polynomial is irreducible. QED

I know I've seen the last statement as an argument in textbooks (e.g. Lang's Algebra), but can anyone offer any justification or intuition for why that is?