### 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: The given condition means that the function $f$ is symmetric about the line $x = 2,$ since $f(x) = f(4 - x)$ for any $x.$
To see why, let $x = 2 + t$ and $y = 2 - t,$ then $f(x) = f(y)$ by assumption, and also $x + y = 4.$
Therefore, $f(x) = f(4 - x)$ for any $x.$
A quadratic function that is symmetric about a vertical line has the form $f(x) = a(x - h)^2 + k,$ where $h$ is the line of symmetry and $k$ is the vertex value.
In this case, $h = 2,$ so we can write $f(x) = a(x - 2)^2 + k.$
To find $a$ and $k,$ we can use the fact that $f(x)$ is a quadratic with coefficient $1$ for the $x^2$ term.
This means that $a = 1.$
To find $k,$ we can plug in any value of $x$ and use the given formula for $f(x).$
For example, if we plug in $x = 0,$ we get $f(0) = 1(0 - 2)^2 + k = 4 + k.$
But we also know that $f(0) = c,$ the constant term of the original quadratic.
Therefore, $c = 4 + k,$ or $k = c - 4.$
So, we have $f(x) = (x - 2)^2 + c - 4.$
Now, we can evaluate $f(1),$ $f(2),$ and $f(4)$ using this formula.
$f(1) = (1 - 2)^2 + c - 4 = c - 3.$
$f(2) = (2 - 2)^2 + c - 4 = c - 4.$
$f(4) = (4 - 2)^2 + c - 4 = 4 + c - 4 = c.$
Therefore, the values of $f(1),$ $f(2),$ and $f(4)$ depend on the value of $c,$ which we do not know.
However, we can still order them by comparing their differences.
We have $f(1) - f(2) = c - 3 - (c - 4) = 1,$ so $f(1) > f(2).$
We also have $f(4) - f(2) = c - (c - 4) = 4,$ so $f(4) > f(2).$
Finally, we have $f(4) - f(1) = c - (c - 3) = 3,$ so $f(4) > f(1).$
Therefore, the correct order is $f(2),$ $f(1),$ $f(4).$