Transformations' order

I get why you can move the curve for (a*(x-b))^2 where ever you want horizontally and then stretch as you want and it stays in the same place. Why does (ax-b)^2 behave any differently(and rather weirdly)?

I get how the parentheses force the order in the first one. I want to know the reasoning behind why the second one functions (pun intended) the way it does. Please explain in as detail and verbosely as you wish. Thanks.

• How does $(a(x-b))^2$ behave strangely relative to $(ax-b)^2$ when $(ax-b)^2=(ax-ab/a)^2=(a(x-b/a))^2$, which is $(a(x-b))^2$ shifted elsewhere? – wjm Apr 4 at 7:39
• Once you change the value of b for the second one and then try to change the value of 'a' it moves horizontally. (I made graphs with variables on desmos) – user134505 Apr 4 at 7:51
• Yes. The second expression moves horizontally and stretches itself as you vary $a$, and this horizontal movement and stretching is described exactly by the answer below. – wjm Apr 4 at 8:03

$$\left(a\left(x-\frac{b}{a}\right)\right)^2$$ if you want to put it into a form that is similar to the first one.
• @user134505 The "why" is a very subjective thing. If you ask me, my form shows clearly that the function behaves like the standard square function, shifted by $\frac{b}{a}$ and then streched by a factor of $a$. Why does it do that? Well, because it is a standard square function, shifted by $\frac{b}{a}$ and then streched by a factor of $a$. So, how else would it behave? – 5xum Apr 4 at 7:59
• @user134505 Exactly. Or, you could look at it this way: In the first case, you shift first, and then stretch. Therefore, the stretch also affects the shift. In the second case, you stretch only x, and then you shift. But now, when you apply the shift, $x$ is $a$-times bigger, so it "sees" any shift as being $a$-times smaller, i.e. it "sees" only a shift by $\frac{b}{a}$ instead of a shift by $b$. (If you suddenly became twice as big, everything around you would appear half the size) – 5xum Apr 4 at 8:15