Okay, let's see. I need to find the sum of the coefficients a, b, c, d, e, f for two parabolas related by reflection over the line y = k. The original parabola has the equation y = ax² + bx + c and vertex (h, k). When it's reflected over y = k, it becomes y = dx² + ex + f. Then, I need to express a + b + c + d + e + f in terms of k. Hmm, interesting.

First, let me recall that the vertex form of a parabola is y = a(x - h)² + k, where (h, k) is the vertex. So, the original parabola can be written as y = a(x - h)² + k. If I expand this, it would be y = a(x² - 2hx + h²) + k = ax² - 2ahx + ah² + k. Comparing this with the standard form y = ax² + bx + c, I can see that:

b = -2ah and c = ah² + k.

So, the coefficients of the original parabola are related to a, h, and k. Now, when the parabola is reflected over the line y = k, what happens? Reflecting a point (x, y) over the line y = k would result in the point (x, 2k - y). So, the reflection of the original parabola would have points (x, 2k - y), where (x, y) is on the original parabola.

Let me substitute y from the original equation into the reflection formula. So, the reflected y-coordinate would be 2k - y = 2k - [ax² + bx + c]. Therefore, the equation of the reflected parabola is y = 2k - ax² - bx - c. But the problem states that this reflected parabola has the equation y = dx² + ex + f. So, equating these two expressions:

dx² + ex + f = -ax² - bx + (2k - c).

Therefore, by comparing coefficients:

d = -a,

e = -b,

f = 2k - c.

So, now I can express d, e, and f in terms of a, b, c, and k. Since we need to find a + b + c + d + e + f, let's substitute the expressions for d, e, f:

a + b + c + (-a) + (-b) + (2k - c) = a + b + c - a - b + 2k - c.

Simplifying this:

a - a + b - b + c - c + 2k = 0 + 0 + 0 + 2k = 2k.

Wait, that's interesting. All the terms cancel out except for 2k. So, the sum a + b + c + d + e + f is equal to 2k. Is that possible?

Let me check my steps again to make sure I didn't make a mistake. Starting with the original parabola in vertex form: y = a(x - h)² + k. Expanding gives y = ax² - 2ahx + ah² + k, so coefficients are b = -2ah, c = ah² + k. Then reflecting over y = k: the reflection of any point (x, y) is (x, 2k - y). So substituting into the original equation, we have 2k - y = ax² + bx + c. Therefore, y = 2k - ax² - bx - c. So the reflected parabola is y = -ax² - bx + (2k - c). Comparing with y = dx² + ex + f, so d = -a, e = -b, f = 2k - c. Then summing a + b + c + d + e + f: a + b + c + (-a) + (-b) + (2k - c) = 2k. That seems correct.

But let me test with a specific example to verify. Suppose the original parabola is y = x², which has vertex at (0, 0). Wait, but in this case, k = 0. Then reflecting over y = 0 (the x-axis) would give y = -x². So the original coefficients are a = 1, b = 0, c = 0. The reflected parabola is y = -x², so d = -1, e = 0, f = 0. Then a + b + c + d + e + f = 1 + 0 + 0 + (-1) + 0 + 0 = 0. But 2k = 2*0 = 0. So that works.

Another example: original parabola is y = 2x² - 4x + 5. Let's find its vertex. The x-coordinate of the vertex is -b/(2a) = 4/(2*2) = 1. The y-coordinate is 2*(1)^2 -4*(1) +5 = 2 -4 +5 = 3. So vertex is (1, 3). So k = 3. Reflecting over y = 3. The original equation is y = 2x² -4x +5. Reflecting over y = 3 would invert the parabola. Let's compute the reflection:

For any point (x, y) on the original parabola, the reflected point is (x, 6 - y). So substitute y = 2x² -4x +5 into 6 - y: 6 - (2x² -4x +5) = -2x² +4x +1. So the reflected parabola is y = -2x² +4x +1. So coefficients d = -2, e = 4, f = 1. Original coefficients: a=2, b=-4, c=5. Sum a + b + c + d + e + f = 2 + (-4) +5 + (-2) +4 +1 = 2 -4 +5 -2 +4 +1 = (2-4) + (5-2) + (4+1) = (-2) + (3) + (5) = 6. But 2k = 2*3 =6. So that works too.

Therefore, the answer should be 2k. But let me check another example to be thorough. Let's take a parabola that's already symmetric over y = k. Suppose original parabola is y = k. Wait, that's a horizontal line, not a parabola. Hmm, maybe take a vertical line. Wait, all parabolas are functions, so they can't be horizontal lines. Let me think of a different example. Suppose original parabola is y = -3x² + 6x - 2. Let's find its vertex. The vertex x-coordinate is -b/(2a) = -6/(2*(-3)) = -6/-6 =1. The y-coordinate is -3*(1)^2 +6*(1) -2 = -3 +6 -2=1. So vertex is (1,1). So k=1. Reflecting over y=1. The reflected parabola would be y = 2*1 - (-3x² +6x -2) = 2 +3x² -6x +2 = 3x² -6x +4. So original coefficients: a=-3, b=6, c=-2. Reflected coefficients: d=3, e=-6, f=4. Sum: -3 +6 -2 +3 -6 +4 = (-3+6) + (-2+3) + (-6+4) = 3 +1 -2 =2. Which is 2k=2*1=2. Correct again.

So, all my test cases confirm that the sum is 2k. Therefore, the answer is 2k. So, expressed in terms of k, the sum is 2k. Therefore, the answer is \boxed{2k}.
Therefore, the sum of the coefficients $a+b+c+d+e+f$ is $\boxed{2k}$.
</think>