### Instruction: The quadratic $f(x) = x^2 + bx + c$ satisfies $f(2 + t) = f(2 - t)$ for all real numbers $t.$  Enter $f(1),$ $f(2),$ and $f(4)$ in increasing order.  For example, if you think $f(4) < f(2) < f(1),$ then enter "f(4), f(2), f(1)".### Response: This problem involves a quadratic function and some kind of symmetry.
I wonder if I can use the fact that $f(2 + t) = f(2 - t)$ to find the values of $b$ and $c$.
Let me try plugging in some values of $t.$
If $t = 0,$ then $f(2) = f(2),$ which is not very helpful.
If $t = 1,$ then $f(3) = f(1),$ which is more interesting.
That means $9 + 3b + c = 1 + b + c,$ or $b = -4.$
Now I have $f(x) = x^2 - 4x + c.$
To find $c,$ I can use another value of $t.$
If $t = 2,$ then $f(4) = f(0),$ which means $16 - 16 + c = c,$ or $c = 0.$
So I have $f(x) = x^2 - 4x,$ which is a simple quadratic.
To find $f(1),$ $f(2),$ and $f(4),$ I just need to plug in those values of $x.$
I get $f(1) = 1 - 4 = -3,$ $f(2) = 4 - 8 = -4,$ and $f(4) = 16 - 16 = 0.$
In increasing order, these are $f(2),$ $f(1),$ and $f(4).$