Calculate the range of $f$. Calculate the range of the function $f$ with $f(x) = x^2 - 2x$, $x\in\Bbb{R}$. My book has solved solutions but I don't get what is done:
$$f(x) = x^2 - 2x + (1^2) - (1^2)= (x-1)^2 -1$$
edit: sorry for wasting all of the people who've answered time, I didnt realize that I had to complete the square....  Wish I could delete it but I cant
 A: From the form $f(x)=(x-1)^2-1$ you can see immediately two things:


*

*Since the term $(x-1)^2$ is a square it is always non-negative, or in simpler words $\ge 0$. So, its lowest possible values is $0$ which can be indeed attained when $x=1$. So, $f(x)=\text{ positive or zero } -1 \ge 0-1=-1$. So $f(x)$ takes values greater than $-1$. So this the lower bound for the range of $f$.

*Is there also an upper bound for the values of $f$, or can $f(x)$ take all values that are greater than $-1$? To answer this, you should ask if you can increase the first term $(x-1)^2$ (that depends on $x$) as much as you want. The second term $-1$ is now irrelevant. Indeed $(x-1)^2$ gets bigger constantly as $x$ increases and in fact without an end. 


This allows you to conclude that $$\text{Range}(f)=[-1, +\infty)$$ and that is why they brought $f$ to this form.
A: For any $y\in\Bbb{R}$, $y^2\ge0$. So $y^2-1\ge-1$.
Substitute $y=x-1$. What does it imply? 
A: Hint:
i suppose that you are convinced that $f(x)=x^2-2x=(x-1)^2-1$. Now write this as: $f(x)=-1+(x-1)^2$ and note that the square is always not negative, and it is null for $x=1$.
So what you can say about $f(x)$?
