Find $g(y-1)$ given $g(y)=y^2+2y+1$

Question

If $g(y)=y^2+2y+1$, then $g(y-1)= \dots$

$g(y-1)=(y-1)^2+2(y-1)+1$.

This is where I get stuck.

Answer 1

Life is better with $(a-b)^{2} = a^{2} - 2ab + b^{2}$

Answer 2

You have at least two options:

Either you could expand things out after your first step...

 g(y-1) = (y-1)^2 + 2(y-1) + 1
        = (y^2 - 2y + 1) + (2y - 2) + 1
        = y^2 -2y + 2y +1 -2 +1
        = y^2

Alternately you could note that g(y) can be factorised...

 g(y) = y^2 + 2y + 1
      = (y+1)(y+1)

so

 g(y-1) = ((y-1) + 1)((y-1) + 1)
        = (y)(y) = y^2 

Answer 3

Starting from where you're at: \begin{align} g(y-1)&=(y-1)^2+2(y-1)+1 \\ &= \underbrace{\left(y^2 - 2y + 1\right)}_{(y-1)^2} + \underbrace{2y -2}_{2(y-1)} + 1\\ &= y^2 -2y +2y +1 +1 -2\\ &= y^2 \end{align}

EDIT: Whenever we write $a^2$, that means $a\cdot a$. So, in this particular instance, we can think of $y-1$ as our "$a$."

\begin{align} (y-1)^2 &= (y-1)(y-1) \\ &= y(y-1) -1(y-1) \quad \text{(distribute first factor)}\\ &= (y\cdot y - y\cdot 1) + (-y+1) \quad \text{(distribute again)}\\ &= y^2 -y-y + 1\\ &= y^2 -2y + 1 \end{align}