### 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 condition $f(2 + t) = f(2 - t)$ implies that the graph of $f$ is symmetric about the vertical line $x = 2.$
That means that the vertex of the parabola must lie on that line, so I can write $f(x) = a(x - 2)^2 + k$ for some constants $a$ and $k.$
Expanding this form, I get $f(x) = ax^2 - 4ax + 4a + k.$
Comparing this with the given form, I see that $a = 1,$ $b = -4a = -4,$ and $c = 4a + k = 4 + k.$
Now I can find $f(1),$ $f(2),$ and $f(4)$ by plugging in the values of $x.$
$f(1) = 1 - 4 + 4 + k = 1 + k.$
$f(2) = 0 + k.$
$f(4) = 16 - 16 + 4 + k = 4 + k.$
Since $k$ is the same for all three values, the order is determined by the coefficients of $k.$
The smallest coefficient is $0,$ then $1,$ then $4,$ so the order is $f(2),$ $f(1),$ $f(4).$