### 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: To solve this problem, I need to use the given condition that $f(2 + t) = f(2 - t)$ for all $t.$
I can start by plugging in some simple values of $t$ and see what happens.
For example, if $t = 0,$ then $f(2) = f(2),$ which is obvious and not very helpful.
But if $t = 1,$ then $f(3) = f(1),$ which is more interesting.
This means that the graph of $f(x)$ is symmetric about the line $x = 2,$ which is the axis of symmetry of the parabola.
In other words, the vertex of the parabola is at $(2, f(2)).$
I can also use $t = 2$ and get $f(4) = f(0),$ which means that the $y$-intercept of the parabola is equal to the value of $f(4).$
Now I have a better idea of what the parabola looks like.
It is symmetric about $x = 2,$ has a vertex at $(2, f(2)),$ and passes through $(0, f(4))$ and $(4, f(4)).$
I can also use the fact that the vertex of a parabola is at $(-b/2, f(-b/2))$ for any quadratic $f(x) = x^2 + bx + c.$
This means that $-b/2 = 2,$ or $b = -4.$
So the quadratic is $f(x) = x^2 - 4x + c.$
To find $c,$ I can use any point on the parabola, such as $(0, f(4)).$
Plugging in $x = 0,$ I get $f(0) = c = f(4).$
So the quadratic is $f(x) = x^2 - 4x + f(4).$
Now I can find the values of $f(1),$ $f(2),$ and $f(4)$ by plugging in $x = 1,$ $x = 2,$ and $x = 4.$
I get $f(1) = 1 - 4 + f(4),$ $f(2) = 4 - 8 + f(4),$ and $f(4) = f(4).$
To compare them in increasing order, I need to subtract $f(4)$ from each one.
I get $f(1) - f(4) = -3,$ $f(2) - f(4) = -4,$ and $f(4) - f(4) = 0.$
So $f(2) - f(4) < f(1) - f(4) < f(4) - f(4),$ or equivalently, $f(2) < f(1) < f(4).$