**Quote:**

Does *bf*(*x*) not mean

*b*(*x*²/*b*² + 2*x* +4) = *bx*²/*b*² + 2*b*/*x* + 4*b*?

Apparently, you have overlooked the words "for each" and the space in between "

*b*" and "

*f*(

*x*)" in the question statement. In this particular question we do NOT deal with

*bf*(

*x*), but only with

*f*(

*x*), while the parameter

*b* can possess any value, except 0.

This is one of the toughest questions indeed. It's main idea is to use the well-known formula (

*a* +

*c*)² =

*a*² + 2

*ac* +

*c*² backwards. But you must be careful with identifying what

*a* and

*c* stand for in this particular question. You should start with

*a* (

*a* stands for

*x*/

*b*) and then identify

*c* from the term with the variable (2x). It stands for 2

*ac*.

The hint for using this well-known formula comes from the look of the expression of the function itself. If you get a very good understanding of how this formula is applied in this particular case and why this quadratic function possesses the least value when that square is 0, you'll improve your timing and eliminate a bunch of possible mistakes when dealing with easier cases and other adjoining types of questions.

P.S. If you studied properties of the quadratic equation and can use the formula for finding the required value of the variable, go for it. If you didn't - the provided explanation shows how to solve this question using just the well-known formula ((

*a* +

*c*)² =

*a*² + 2

*ac* +

*c*²), and the property of the square (≥ 0).