### 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: I notice that the given condition implies that the graph of $f(x)$ is symmetric about the line $x = 2.$
That means the vertex of the parabola must lie on that line, so $f(2)$ is the minimum value of $f(x).$
I also notice that $f(2) = f(2 + t) = f(2 - t)$ implies that $f(2) = 4 + 2b + c.$
I can use this equation to eliminate $c$ from $f(x) = x^2 + bx + c$ and write $f(x) = x^2 + bx + f(2) - 4 - 2b.$
Now I can plug in $x = 1$ and $x = 4$ and compare their values to $f(2).$
I get $f(1) = 1 + b + f(2) - 4 - 2b = f(2) - 3 - b$ and $f(4) = 16 + 4b + f(2) - 4 - 2b = f(2) + 12 + 2b.$
Since $f(2)$ is the minimum, I know that $f(4) > f(2) > f(1).$
Therefore, the correct answer is "f(1), f(2), f(4)".