Explanation:
Okay, so I need to figure out what 999 in base 10 is in base six. Hmm, let me remember how to convert numbers from base 10 to another base. I think the standard method is to divide the number by the base and keep track of the remainders. Let me try to recall the exact steps.

First, I know that base six means each digit represents a power of six. So, the rightmost digit is 6^0 (which is 1), then 6^1 (which is 6), then 6^2 (which is 36), then 6^3 (which is 216), and so on. So, to convert 999 into base six, I need to find out how many times each power of six fits into 999, starting from the highest power possible.

But maybe the division method is more straightforward. Let me try that. The process is to divide the number by 6, record the remainder, then divide the quotient by 6 again, and repeat until the quotient is zero. Then, the base six number is the remainders read in reverse order. Let me try that.

So, starting with 999 divided by 6. Let me calculate that. 6 times 166 is 996, right? Because 6*160 is 960, and 6*6 is 36, so 960+36=996. Then 999 minus 996 is 3. So the first remainder is 3. So that's the least significant digit (rightmost digit) in base six.

Now, take the quotient, which is 166, and divide that by 6. Let me do 166 divided by 6. 6*27 is 162, so 166-162=4. So the remainder is 4. So the next digit is 4.

Now take the quotient 27 and divide by 6. 6*4 is 24, so 27-24=3. Remainder is 3. Next digit is 3.

Take the quotient 4 and divide by 6. Well, 4 divided by 6 is 0 with a remainder of 4. So the next digit is 4.

Now the quotient is 0, so we stop. Then, reading the remainders from last to first, we have 4, 3, 4, 3. Wait, let me check again.

Wait, let me write down each step properly:

Step 1: 999 ÷ 6 = 166 with remainder 3.

Step 2: 166 ÷ 6 = 27 with remainder 4.

Step 3: 27 ÷ 6 = 4 with remainder 3.

Step 4: 4 ÷ 6 = 0 with remainder 4.

So the remainders, in order from last to first, are 4, 3, 4, 3. Therefore, the base six number is 4343? Let me check if that's correct.

Wait, let me verify by converting 4343 base six back to decimal. Let's see:

Starting from the right, the digits are 4, 3, 4, 3. Wait, actually, in base six, the rightmost digit is 6^0, then 6^1, 6^2, 6^3. So the number 4343 in base six would be:

4*6^3 + 3*6^2 + 4*6^1 + 3*6^0.

Calculating each term:

6^0 = 1, so 3*1 = 3.

6^1 = 6, so 4*6 = 24.

6^2 = 36, so 3*36 = 108.

6^3 = 216, so 4*216 = 864.

Adding them up: 864 + 108 = 972; 972 + 24 = 996; 996 + 3 = 999. Perfect! That matches the original number. So 999 in base 10 is 4343 in base 6. Therefore, the answer is 4343 base six.

But let me just make sure I didn't make any mistakes in the division steps. Let me go through them again.

First division: 999 divided by 6. 6*160 is 960, 999-960=39. Then 39 divided by 6 is 6*6=36, so total quotient is 160+6=166, remainder 3. Correct.

Second division: 166 divided by 6. 6*20=120, 166-120=46. 46 divided by 6 is 7*6=42, remainder 4. So total quotient is 20+7=27, remainder 4. Correct.

Third division: 27 divided by 6. 6*4=24, remainder 3. Quotient 4, remainder 3. Correct.

Fourth division: 4 divided by 6 is 0 with remainder 4. Correct.

So the remainders are 3, 4, 3, 4. Wait, but when we read them in reverse order, it's 4, 3, 4, 3. So 4343. Yes, that's correct. So that seems right.

Alternatively, maybe I can try another method to cross-verify. Let me try expanding the number in terms of powers of six.

What's the highest power of 6 less than or equal to 999?

Compute 6^0 = 1

6^1 = 6

6^2 = 36

6^3 = 216

6^4 = 1296

Okay, 6^4 is 1296 which is larger than 999, so the highest power is 6^3 = 216.

So how many times does 216 go into 999?

Let me calculate 999 divided by 216. 216*4 = 864, 216*5=1080 which is too big. So 4 times. So the coefficient for 6^3 is 4. Then subtract 4*216=864 from 999: 999 - 864 = 135.

Now move to the next lower power, which is 6^2 = 36. How many times does 36 go into 135?

36*3=108, 36*4=144 which is too big. So coefficient is 3. Subtract 108 from 135: 135 - 108 = 27.

Next power is 6^1 = 6. How many times does 6 go into 27? 6*4=24, 6*5=30 which is too big. So coefficient is 4. Subtract 24 from 27: 3.

Next power is 6^0 = 1. How many times does 1 go into 3? 3 times. So coefficient is 3.

Therefore, the coefficients are 4 (for 6^3), 3 (for 6^2), 4 (for 6^1), 3 (for 6^0). So the number is 4343 in base six. Same result as before. So that confirms it.

Therefore, I think the answer is 4343 base six. Let me just check once more with another approach.

Alternatively, I can use the fact that 6^3 is 216, so 4*216 is 864, as before. Then 999 - 864 = 135. Then 6^2 is 36, 3*36 is 108, 135 - 108 = 27. Then 6^1 is 6, 4*6 is 24, 27 - 24 = 3. Then 3*1 is 3. So yes, same numbers. So 4, 3, 4, 3. So 4343.

Alternatively, maybe I can think of 999 divided by 6, then divide by 6 again, etc., but that's essentially the same as the first method.

Alternatively, maybe I can use the fact that 999 is 1000 - 1. Let me see if that helps. 1000 in base 10 is... Well, maybe not directly helpful. Let me think.

But 1000 in base 10 is 4*6^3 + 3*6^2 + 4*6 + 4? Wait, no. Wait, 6^3 is 216, 4*216 is 864, 6^4 is 1296. So 1000 is between 6^4 and 6^3. Maybe not helpful.

Alternatively, maybe I can use some other base conversion tricks, but I think the methods I've already used are sufficient and they both give me 4343. So I think that's correct.

But just to be thorough, let me check with another example. Let's say I convert 4343 base 6 to decimal again:

Starting from the right:

Digits: 4 (position 3), 3 (position 2), 4 (position 1), 3 (position 0). Wait, actually, positions are from right starting at 0. Wait, the number is 4 3 4 3. So from left to right, it's 4*6^3 + 3*6^2 + 4*6^1 + 3*6^0. Which is 4*216 + 3*36 + 4*6 + 3*1. Which is 864 + 108 + 24 + 3. 864 + 108 is 972, plus 24 is 996, plus 3 is 999. Perfect. So that's correct.

Therefore, I can be confident that 999 in base 10 is 4343 in base 6.

**Final Answer**
The base six equivalent of $999_{10}$ is \boxed{4343}.

To convert the decimal number $ 999_{10} $ into base six, we use the **division-remainder method**, which is a standard approach for converting numbers from base 10 to another base.

---

### Step-by-Step Conversion

We repeatedly divide the number by 6 and record the remainders. The remainders, read in reverse order, give the digits of the number in base six.

1. **Divide 999 by 6:**
   $$
   999 \div 6 = 166 \text{ remainder } 3
   $$
   - Remainder: 3

2. **Divide 166 by 6:**
   $$
   166 \div 6 = 27 \text{ remainder } 4
   $$
   - Remainder: 4

3. **Divide 27 by 6:**
   $$
   27 \div 6 = 4 \text{ remainder } 3
   $$
   - Remainder: 3

4. **Divide 4 by 6:**
   $$
   4 \div 6 = 0 \text{ remainder } 4
   $$
   - Remainder: 4

Now, reading the remainders from **last to first**, we get the base six number:
$$
4\ 3\ 4\ 3
$$

---

### Verification

To confirm, we can convert $ 4343_6 $ back to base 10:

$$
4 \times 6^3 + 3 \times 6^2 + 4 \times 6^1 + 3 \times 6^0
= 4 \times 216 + 3 \times 36 + 4 \times 6 + 3 \times 1
= 864 + 108 + 24 + 3 = 999
$$

This matches the original number, confirming the conversion is correct.

---

### Final Answer

$$
\boxed{4343}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what 999 in base 10 is in base six. Hmm, let me remember how to convert numbers from base 10 to another base. I think the standard method is to divide the number by the base and keep track of the remainders. Let me try to recall the exact steps.

First, I know that base six means each digit represents a power of six. So, the rightmost digit is 6^0 (which is 1), then 6^1 (which is 6), then 6^2 (which is 36), then 6^3 (which is 216), and so on. So, to convert 999 into base six, I need to find out how many times each power of six fits into 999, starting from the highest power possible.

But maybe the division method is more straightforward. Let me try that. The process is to divide the number by 6, record the remainder, then divide the quotient by 6 again, and repeat until the quotient is zero. Then, the base six number is the remainders read in reverse order. Let me try that.

So, starting with 999 divided by 6. Let me calculate that. 6 times 166 is 996, right? Because 6*160 is 960, and 6*6 is 36, so 960+36=996. Then 999 minus 996 is 3. So the first remainder is 3. So that's the least significant digit (rightmost digit) in base six.

Now, take the quotient, which is 166, and divide that by 6. Let me do 166 divided by 6. 6*27 is 162, so 166-162=4. So the remainder is 4. So the next digit is 4.

Now take the quotient 27 and divide by 6. 6*4 is 24, so 27-24=3. Remainder is 3. Next digit is 3.

Take the quotient 4 and divide by 6. Well, 4 divided by 6 is 0 with a remainder of 4. So the next digit is 4.

Now the quotient is 0, so we stop. Then, reading the remainders from last to first, we have 4, 3, 4, 3. Wait, let me check again.

Wait, let me write down each step properly:

Step 1: 999 ÷ 6 = 166 with remainder 3.

Step 2: 166 ÷ 6 = 27 with remainder 4.

Step 3: 27 ÷ 6 = 4 with remainder 3.

Step 4: 4 ÷ 6 = 0 with remainder 4.

So the remainders, in order from last to first, are 4, 3, 4, 3. Therefore, the base six number is 4343? Let me check if that's correct.

Wait, let me verify by converting 4343 base six back to decimal. Let's see:

Starting from the right, the digits are 4, 3, 4, 3. Wait, actually, in base six, the rightmost digit is 6^0, then 6^1, 6^2, 6^3. So the number 4343 in base six would be:

4*6^3 + 3*6^2 + 4*6^1 + 3*6^0.

Calculating each term:

6^0 = 1, so 3*1 = 3.

6^1 = 6, so 4*6 = 24.

6^2 = 36, so 3*36 = 108.

6^3 = 216, so 4*216 = 864.

Adding them up: 864 + 108 = 972; 972 + 24 = 996; 996 + 3 = 999. Perfect! That matches the original number. So 999 in base 10 is 4343 in base 6. Therefore, the answer is 4343 base six.

But let me just make sure I didn't make any mistakes in the division steps. Let me go through them again.

First division: 999 divided by 6. 6*160 is 960, 999-960=39. Then 39 divided by 6 is 6*6=36, so total quotient is 160+6=166, remainder 3. Correct.

Second division: 166 divided by 6. 6*20=120, 166-120=46. 46 divided by 6 is 7*6=42, remainder 4. So total quotient is 20+7=27, remainder 4. Correct.

Third division: 27 divided by 6. 6*4=24, remainder 3. Quotient 4, remainder 3. Correct.

Fourth division: 4 divided by 6 is 0 with remainder 4. Correct.

So the remainders are 3, 4, 3, 4. Wait, but when we read them in reverse order, it's 4, 3, 4, 3. So 4343. Yes, that's correct. So that seems right.

Alternatively, maybe I can try another method to cross-verify. Let me try expanding the number in terms of powers of six.

What's the highest power of 6 less than or equal to 999?

Compute 6^0 = 1

6^1 = 6

6^2 = 36

6^3 = 216

6^4 = 1296

Okay, 6^4 is 1296 which is larger than 999, so the highest power is 6^3 = 216.

So how many times does 216 go into 999?

Let me calculate 999 divided by 216. 216*4 = 864, 216*5=1080 which is too big. So 4 times. So the coefficient for 6^3 is 4. Then subtract 4*216=864 from 999: 999 - 864 = 135.

Now move to the next lower power, which is 6^2 = 36. How many times does 36 go into 135?

36*3=108, 36*4=144 which is too big. So coefficient is 3. Subtract 108 from 135: 135 - 108 = 27.

Next power is 6^1 = 6. How many times does 6 go into 27? 6*4=24, 6*5=30 which is too big. So coefficient is 4. Subtract 24 from 27: 3.

Next power is 6^0 = 1. How many times does 1 go into 3? 3 times. So coefficient is 3.

Therefore, the coefficients are 4 (for 6^3), 3 (for 6^2), 4 (for 6^1), 3 (for 6^0). So the number is 4343 in base six. Same result as before. So that confirms it.

Therefore, I think the answer is 4343 base six. Let me just check once more with another approach.

Alternatively, I can use the fact that 6^3 is 216, so 4*216 is 864, as before. Then 999 - 864 = 135. Then 6^2 is 36, 3*36 is 108, 135 - 108 = 27. Then 6^1 is 6, 4*6 is 24, 27 - 24 = 3. Then 3*1 is 3. So yes, same numbers. So 4, 3, 4, 3. So 4343.

Alternatively, maybe I can think of 999 divided by 6, then divide by 6 again, etc., but that's essentially the same as the first method.

Alternatively, maybe I can use the fact that 999 is 1000 - 1. Let me see if that helps. 1000 in base 10 is... Well, maybe not directly helpful. Let me think.

But 1000 in base 10 is 4*6^3 + 3*6^2 + 4*6 + 4? Wait, no. Wait, 6^3 is 216, 4*216 is 864, 6^4 is 1296. So 1000 is between 6^4 and 6^3. Maybe not helpful.

Alternatively, maybe I can use some other base conversion tricks, but I think the methods I've already used are sufficient and they both give me 4343. So I think that's correct.

But just to be thorough, let me check with another example. Let's say I convert 4343 base 6 to decimal again:

Starting from the right:

Digits: 4 (position 3), 3 (position 2), 4 (position 1), 3 (position 0). Wait, actually, positions are from right starting at 0. Wait, the number is 4 3 4 3. So from left to right, it's 4*6^3 + 3*6^2 + 4*6^1 + 3*6^0. Which is 4*216 + 3*36 + 4*6 + 3*1. Which is 864 + 108 + 24 + 3. 864 + 108 is 972, plus 24 is 996, plus 3 is 999. Perfect. So that's correct.

Therefore, I can be confident that 999 in base 10 is 4343 in base 6.

**Final Answer**
The base six equivalent of $999_{10}$ is \boxed{4343}.

To convert the decimal number $ 999_{10} $ into base six, we use the **division-remainder method**, which is a standard approach for converting numbers from base 10 to another base.

---

### Step-by-Step Conversion

We repeatedly divide the number by 6 and record the remainders. The remainders, read in reverse order, give the digits of the number in base six.

1. **Divide 999 by 6:**
   $$
   999 \div 6 = 166 \text{ remainder } 3
   $$
   - Remainder: 3

2. **Divide 166 by 6:**
   $$
   166 \div 6 = 27 \text{ remainder } 4
   $$
   - Remainder: 4

3. **Divide 27 by 6:**
   $$
   27 \div 6 = 4 \text{ remainder } 3
   $$
   - Remainder: 3

4. **Divide 4 by 6:**
   $$
   4 \div 6 = 0 \text{ remainder } 4
   $$
   - Remainder: 4

Now, reading the remainders from **last to first**, we get the base six number:
$$
4\ 3\ 4\ 3
$$

---

### Verification

To confirm, we can convert $ 4343_6 $ back to base 10:

$$
4 \times 6^3 + 3 \times 6^2 + 4 \times 6^1 + 3 \times 6^0
= 4 \times 216 + 3 \times 36 + 4 \times 6 + 3 \times 1
= 864 + 108 + 24 + 3 = 999
$$

This matches the original number, confirming the conversion is correct.

---

### Final Answer

$$
\boxed{4343}
$$

Solution: We know that $6^{4}>999_{10}>6^{3}$. So, we can tell that $999_{10}$ in base six will have four digits. $6^{3}=216$, which can go into 999 four times at most, leaving $999-4\cdot216 = 135$ for the next three digits. $6^{2}=36$ goes into 135 three times at most, leaving us with $135-3\cdot36 = 27$. Then, $6^{1}=6$ goes into 27 four times at most, leaving $27-4\cdot6 = 3$ for the ones digit. All together, the base six equivalent of $999_{10}$ is $\boxed{4343_6}$.


Explanation:
Okay, so I need to figure out how to express a + b + c + d + e + f in terms of k, where the original parabola is y = ax² + bx + c with vertex (h, k), and after reflecting it over the line y = k, we get another parabola y = dx² + ex + f.

First, let me recall some properties of parabolas. The vertex form of a parabola is y = a(x - h)² + k, right? So if the original parabola has vertex (h, k), then its equation can be written as y = a(x - h)² + k. Comparing this with the standard form y = ax² + bx + c, I can expand the vertex form to find the coefficients b and c in terms of a and h and k.

Let me do that. Expanding y = a(x - h)² + k:

y = a(x² - 2hx + h²) + k
= ax² - 2ahx + ah² + k

So comparing with y = ax² + bx + c, we have:

b = -2ah
c = ah² + k

Okay, so that gives me expressions for b and c in terms of a, h, and k.

Now, the problem says that this parabola is reflected about the line y = k. I need to figure out what the equation of the reflected parabola is.

Reflecting a graph over a horizontal line y = k... Let me think. If I have a point (x, y) on the original parabola, then its reflection over the line y = k would be the point (x, 2k - y). Because reflecting over y = k means that the distance from y to k is the same as the distance from the reflected point to k, but on the other side. So if the original point is y, then the reflected point is 2k - y.

Therefore, if the original parabola is y = ax² + bx + c, then the reflected parabola would have the equation obtained by replacing y with 2k - y. Wait, let me check that.

Let me think again. Suppose (x, y) is on the original parabola. Then (x, 2k - y) is on the reflected parabola. So if I want to find the equation of the reflected parabola, I can take the original equation and replace y with 2k - y.

So starting with y = ax² + bx + c, replacing y with 2k - y gives:

2k - y = ax² + bx + c
=> -y = ax² + bx + c - 2k
=> y = -ax² - bx - c + 2k

Therefore, the equation of the reflected parabola is y = -ax² - bx + (2k - c).

Wait, so that would mean that the coefficients d, e, f are:

d = -a
e = -b
f = 2k - c

Therefore, the reflected parabola is y = -a x² - b x + (2k - c).

But let me check if this is correct. Let me take a simple example. Suppose the original parabola is y = x², which has vertex at (0, 0). Reflecting over y = 0 (the x-axis) would give y = -x². Let's see if the formula works here.

Original equation: y = x². So a = 1, b = 0, c = 0. Then according to the formula, the reflected equation would be y = -1 x² - 0 x + (2*0 - 0) = -x². Which is correct. So that seems to work.

Another example: take a parabola with vertex (h, k). Let me take h = 1, k = 2, and a = 1. Then original equation is y = (x - 1)^2 + 2. Expanding that: y = x² - 2x + 1 + 2 = x² - 2x + 3. So here, a = 1, b = -2, c = 3.

Reflecting over y = 2. According to the formula, the reflected equation should be y = -1 x² - (-2)x + (2*2 - 3) = -x² + 2x + (4 - 3) = -x² + 2x + 1.

Let me check if this is correct. The original parabola has vertex (1, 2). Reflecting over y = 2 would result in a parabola that is symmetric to the original with respect to y = 2. Since the original vertex is on the line y = 2, the reflected parabola should have the same vertex? Wait, no. Wait, reflecting over y = k, which is the vertex's y-coordinate. So if the original vertex is (h, k), then reflecting over y = k would leave the vertex unchanged? Wait, that seems contradictory. Wait, if you reflect a point over a line that passes through it, then the point remains the same. So the vertex (h, k) is on the line y = k, so reflecting it over y = k would leave it unchanged. Therefore, the reflected parabola should have the same vertex (h, k).

But according to the equation I derived, the reflected parabola is y = -x² + 2x + 1. Let me check its vertex.

The vertex of y = -x² + 2x + 1 can be found by completing the square or using vertex formula. The x-coordinate of the vertex is at -b/(2a) = -2/(2*(-1)) = -2 / (-2) = 1. Then y-coordinate is y(1) = -1 + 2 + 1 = 2. So the vertex is (1, 2), which is the same as the original. So that checks out.

So the reflected parabola has the same vertex, which makes sense. So the formula seems to work here.

Therefore, my conclusion is that the reflected parabola has equation y = -a x² - b x + (2k - c). Therefore, coefficients are d = -a, e = -b, f = 2k - c.

Therefore, the original coefficients are a, b, c, and the reflected ones are d = -a, e = -b, f = 2k - c.

Therefore, the sum a + b + c + d + e + f is equal to:

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

Let me compute that term by term:

a - a = 0
b - b = 0
c - c = 0
Then we have 2k left.

Wait, so the total sum is 0 + 0 + 0 + 2k = 2k?

Wait, that seems surprisingly simple. So regardless of a, b, c, h, etc., the sum a + b + c + d + e + f equals 2k?

But let me check with my previous example.

Original parabola: y = x² - 2x + 3. So a = 1, b = -2, c = 3. Then d = -1, e = 2, f = 2*2 - 3 = 4 - 3 = 1.

So sum a + b + c + d + e + f = 1 + (-2) + 3 + (-1) + 2 + 1 = (1 - 2 + 3) + (-1 + 2 + 1) = (2) + (2) = 4.

Which is 2k, since k = 2. 2*2 = 4. Correct.

Another example: take original parabola y = (x - h)^2 + k. Let me take h = 0, k = 5. Then original equation is y = x² + 5. So a = 1, b = 0, c = 5. Then reflected parabola is y = -x² + 0x + (2*5 - 5) = -x² + 0x + 5. So d = -1, e = 0, f = 5. Then sum a + b + c + d + e + f = 1 + 0 + 5 + (-1) + 0 + 5 = (1 -1) + (0 + 0) + (5 + 5) = 0 + 0 + 10 = 10, which is 2k = 10. Correct.

Another test case: take a parabola with vertex (h, k) where h is not zero. Let me take h = 2, k = 3, a = -1. Original equation: y = -1(x - 2)^2 + 3. Expanding that: y = - (x² - 4x + 4) + 3 = -x² + 4x - 4 + 3 = -x² + 4x - 1. So here, a = -1, b = 4, c = -1. Then reflecting over y = 3. According to formula, d = -a = 1, e = -b = -4, f = 2k - c = 6 - (-1) = 7. Therefore, the reflected equation is y = x² - 4x + 7.

Calculating the sum a + b + c + d + e + f: (-1) + 4 + (-1) + 1 + (-4) + 7 = (-1 -1) + (4 -4) + (1 + 7) = (-2) + 0 + 8 = 6. Which is 2k = 6. Correct.

So seems like regardless of the original coefficients, the sum is always 2k. Therefore, the answer is 2k.

But let me think again if there's a way to see this without going through all the examples.

Original equation: y = ax² + bx + c.

Reflected equation: y = -a x² - b x + (2k - c).

Therefore, adding all coefficients:

Original coefficients: a, b, c.

Reflected coefficients: -a, -b, 2k - c.

Sum: a + b + c + (-a) + (-b) + (2k - c) = (a - a) + (b - b) + (c - c) + 2k = 0 + 0 + 0 + 2k = 2k.

Yes, that's straightforward. So regardless of the values of a, b, c, h, etc., the sum is always 2k.

Therefore, the answer is 2k.

But let me check if there's any possible mistake here. For example, is the reflection over y = k really resulting in the equation y = -a x² - b x + (2k - c)?

Another way to think about reflection: if you have a parabola and you reflect it over a horizontal line, the shape is inverted? Wait, not necessarily. Wait, reflecting over y = k would invert the parabola with respect to that line.

But in terms of equations, if you have a function y = f(x), then reflecting over y = k would give y = 2k - f(x). So yes, that's the same as replacing y with 2k - y in the original equation.

So if original equation is y = f(x), then the reflected equation is y = 2k - f(x).

Therefore, if original equation is y = ax² + bx + c, then reflected equation is y = 2k - ax² - bx - c. Which is y = -ax² - bx + (2k - c). Which is what I had before.

Therefore, the coefficients are correct.

Therefore, the sum is 2k.

Therefore, the answer is 2k.

But let me check if the problem says "the parabola with equation y = ax² + bx + c and vertex (h, k)" so maybe there is some relation between h and k? But in the problem, they just say vertex (h, k), so h is arbitrary. However, in our calculation, we didn't need h. The sum ended up being 2k regardless of h.

But let me think again. Suppose we have vertex (h, k). Then the original equation is y = a(x - h)^2 + k. Then, when we reflect over y = k, we get y = -a(x - h)^2 + k. Wait, is that correct? Wait, if you reflect over y = k, then the vertex remains the same, but the parabola opens in the opposite direction.

Wait, so if original parabola is y = a(x - h)^2 + k, then reflecting over y = k would invert the coefficient a?

Wait, let me think. If you have a point (x, y) on the original parabola, then the reflection over y = k is (x, 2k - y). So if the original parabola is y = a(x - h)^2 + k, then the reflected parabola is 2k - y = a(x - h)^2 + k. Therefore, solving for y: y = 2k - a(x - h)^2 - k = -a(x - h)^2 + k.

So the reflected parabola is y = -a(x - h)^2 + k.

But expanding this, we get y = -a(x² - 2hx + h²) + k = -a x² + 2a h x - a h² + k.

Comparing with the previous expression for the reflected equation, which was y = -a x² - b x + (2k - c).

But in the original equation, we had y = a(x - h)^2 + k = a x² - 2a h x + a h² + k. Therefore, original coefficients: a, b = -2a h, c = a h² + k.

So the reflected equation, when expanded, is y = -a x² + 2a h x - a h² + k.

Comparing with the previous expression from the reflection formula: y = -a x² - b x + (2k - c).

Let me check if these are the same.

From the reflection formula:

- b x = -(-2a h) x = 2a h x. So that matches the coefficient of x in the expanded reflected equation.

Then the constant term: 2k - c. Original c is a h² + k. Therefore, 2k - c = 2k - (a h² + k) = k - a h².

But in the expanded reflected equation, the constant term is -a h² + k. Which is same as k - a h². So yes, matches.

Therefore, both methods agree.

Therefore, the reflected equation is indeed y = -a x² - b x + (2k - c).

Therefore, the sum a + b + c + d + e + f is 2k.

Therefore, the answer is 2k.

But let me check once more with another example. Let me take a parabola with vertex (h, k) where h is not zero, and a is negative.

Let me take h = 3, k = 4, a = -2. Original equation: y = -2(x - 3)^2 + 4. Expanding: y = -2(x² - 6x + 9) + 4 = -2x² + 12x - 18 + 4 = -2x² + 12x - 14. So here, a = -2, b = 12, c = -14.

Reflecting over y = 4. According to formula, reflected equation is y = -a x² - b x + (2k - c) = -(-2)x² - 12x + (8 - (-14)) = 2x² - 12x + 22.

Let me check if this is correct. The original parabola has vertex (3, 4). Reflecting over y = 4 should give a parabola with the same vertex, opening upwards (since original opens downward).

Let me check the reflected equation: y = 2x² - 12x + 22. Let me find its vertex.

x-coordinate is -b/(2a) = 12/(4) = 3. Then y-coordinate is y(3) = 2*(9) - 12*3 + 22 = 18 - 36 + 22 = 4. Correct. So vertex is (3, 4).

Now, let's compute the sum a + b + c + d + e + f. Original coefficients: a = -2, b = 12, c = -14. Reflected coefficients: d = 2, e = -12, f = 22.

Sum: (-2) + 12 + (-14) + 2 + (-12) + 22 = (-2 -14) + (12 -12) + (2 + 22) = (-16) + 0 + 24 = 8. Which is 2k = 2*4 = 8. Correct.

Therefore, this seems to hold.

Therefore, regardless of the values of a, b, c, h, the sum a + b + c + d + e + f equals 2k.

Therefore, the answer is 2k.

But let me think again if there's a possibility that the problem is expecting something else. For example, maybe the question is in terms of the original coefficients, but since the problem says "Express a + b + c + d + e + f in terms of k," and we have shown that it's 2k regardless of other variables, then 2k is the answer.

Alternatively, maybe there's a different approach. Let me think.

Another way: evaluate the original parabola at x = 1, and the reflected parabola at x = 1. Then sum all coefficients is the value of the polynomial at x = 1.

Wait, because for any polynomial y = ax² + bx + c, the value at x = 1 is a(1)^2 + b(1) + c = a + b + c. Similarly, for the reflected parabola, the value at x = 1 is d + e + f. Therefore, the total sum a + b + c + d + e + f is equal to [value of original at x=1] + [value of reflected at x=1].

So maybe we can compute this.

Original parabola: y = ax² + bx + c. At x = 1, y = a + b + c.

Reflected parabola: y = -a x² - b x + (2k - c). At x = 1, y = -a - b + 2k - c.

Therefore, total sum is (a + b + c) + (-a - b + 2k - c) = (a - a) + (b - b) + (c - c) + 2k = 0 + 0 + 0 + 2k = 2k.

Same result.

Therefore, regardless of the method, the answer is 2k.

Therefore, I think the answer is 2k.

**Final Answer**
\boxed{2k}

To solve the problem, we are given a parabola with equation:

$$
y = ax^2 + bx + c
$$

and vertex $(h, k)$. This parabola is then reflected over the horizontal line $y = k$, resulting in a new parabola with equation:

$$
y = dx^2 + ex + f
$$

We are asked to express the sum $a + b + c + d + e + f$ in terms of $k$.

---

### Step 1: Express the Original Parabola in Vertex Form

The vertex form of a parabola is:

$$
y = a(x - h)^2 + k
$$

Expanding this:

$$
y = a(x^2 - 2hx + h^2) + k = ax^2 - 2ahx + ah^2 + k
$$

Comparing with the standard form $y = ax^2 + bx + c$, we identify:

- $b = -2ah$
- $c = ah^2 + k$

---

### Step 2: Reflect the Parabola Over the Line $y = k$

To reflect a point $(x, y)$ over the line $y = k$, the reflected point is $(x, 2k - y)$. So, if the original parabola is $y = ax^2 + bx + c$, the reflected parabola is obtained by replacing $y$ with $2k - y$:

$$
2k - y = ax^2 + bx + c
\Rightarrow y = -ax^2 - bx + (2k - c)
$$

Thus, the reflected parabola has coefficients:

- $d = -a$
- $e = -b$
- $f = 2k - c$

---

### Step 3: Compute the Sum $a + b + c + d + e + f$

Now, we compute the sum of all coefficients:

$$
a + b + c + d + e + f = a + b + c + (-a) + (-b) + (2k - c)
$$

Grouping like terms:

- $a - a = 0$
- $b - b = 0$
- $c - c = 0$
- Remaining term: $2k$

So the total sum is:

$$
a + b + c + d + e + f = 2k
$$

---

### Final Answer

$$
\boxed{2k}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how to express a + b + c + d + e + f in terms of k, where the original parabola is y = ax² + bx + c with vertex (h, k), and after reflecting it over the line y = k, we get another parabola y = dx² + ex + f.

First, let me recall some properties of parabolas. The vertex form of a parabola is y = a(x - h)² + k, right? So if the original parabola has vertex (h, k), then its equation can be written as y = a(x - h)² + k. Comparing this with the standard form y = ax² + bx + c, I can expand the vertex form to find the coefficients b and c in terms of a and h and k.

Let me do that. Expanding y = a(x - h)² + k:

y = a(x² - 2hx + h²) + k
= ax² - 2ahx + ah² + k

So comparing with y = ax² + bx + c, we have:

b = -2ah
c = ah² + k

Okay, so that gives me expressions for b and c in terms of a, h, and k.

Now, the problem says that this parabola is reflected about the line y = k. I need to figure out what the equation of the reflected parabola is.

Reflecting a graph over a horizontal line y = k... Let me think. If I have a point (x, y) on the original parabola, then its reflection over the line y = k would be the point (x, 2k - y). Because reflecting over y = k means that the distance from y to k is the same as the distance from the reflected point to k, but on the other side. So if the original point is y, then the reflected point is 2k - y.

Therefore, if the original parabola is y = ax² + bx + c, then the reflected parabola would have the equation obtained by replacing y with 2k - y. Wait, let me check that.

Let me think again. Suppose (x, y) is on the original parabola. Then (x, 2k - y) is on the reflected parabola. So if I want to find the equation of the reflected parabola, I can take the original equation and replace y with 2k - y.

So starting with y = ax² + bx + c, replacing y with 2k - y gives:

2k - y = ax² + bx + c
=> -y = ax² + bx + c - 2k
=> y = -ax² - bx - c + 2k

Therefore, the equation of the reflected parabola is y = -ax² - bx + (2k - c).

Wait, so that would mean that the coefficients d, e, f are:

d = -a
e = -b
f = 2k - c

Therefore, the reflected parabola is y = -a x² - b x + (2k - c).

But let me check if this is correct. Let me take a simple example. Suppose the original parabola is y = x², which has vertex at (0, 0). Reflecting over y = 0 (the x-axis) would give y = -x². Let's see if the formula works here.

Original equation: y = x². So a = 1, b = 0, c = 0. Then according to the formula, the reflected equation would be y = -1 x² - 0 x + (2*0 - 0) = -x². Which is correct. So that seems to work.

Another example: take a parabola with vertex (h, k). Let me take h = 1, k = 2, and a = 1. Then original equation is y = (x - 1)^2 + 2. Expanding that: y = x² - 2x + 1 + 2 = x² - 2x + 3. So here, a = 1, b = -2, c = 3.

Reflecting over y = 2. According to the formula, the reflected equation should be y = -1 x² - (-2)x + (2*2 - 3) = -x² + 2x + (4 - 3) = -x² + 2x + 1.

Let me check if this is correct. The original parabola has vertex (1, 2). Reflecting over y = 2 would result in a parabola that is symmetric to the original with respect to y = 2. Since the original vertex is on the line y = 2, the reflected parabola should have the same vertex? Wait, no. Wait, reflecting over y = k, which is the vertex's y-coordinate. So if the original vertex is (h, k), then reflecting over y = k would leave the vertex unchanged? Wait, that seems contradictory. Wait, if you reflect a point over a line that passes through it, then the point remains the same. So the vertex (h, k) is on the line y = k, so reflecting it over y = k would leave it unchanged. Therefore, the reflected parabola should have the same vertex (h, k).

But according to the equation I derived, the reflected parabola is y = -x² + 2x + 1. Let me check its vertex.

The vertex of y = -x² + 2x + 1 can be found by completing the square or using vertex formula. The x-coordinate of the vertex is at -b/(2a) = -2/(2*(-1)) = -2 / (-2) = 1. Then y-coordinate is y(1) = -1 + 2 + 1 = 2. So the vertex is (1, 2), which is the same as the original. So that checks out.

So the reflected parabola has the same vertex, which makes sense. So the formula seems to work here.

Therefore, my conclusion is that the reflected parabola has equation y = -a x² - b x + (2k - c). Therefore, coefficients are d = -a, e = -b, f = 2k - c.

Therefore, the original coefficients are a, b, c, and the reflected ones are d = -a, e = -b, f = 2k - c.

Therefore, the sum a + b + c + d + e + f is equal to:

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

Let me compute that term by term:

a - a = 0
b - b = 0
c - c = 0
Then we have 2k left.

Wait, so the total sum is 0 + 0 + 0 + 2k = 2k?

Wait, that seems surprisingly simple. So regardless of a, b, c, h, etc., the sum a + b + c + d + e + f equals 2k?

But let me check with my previous example.

Original parabola: y = x² - 2x + 3. So a = 1, b = -2, c = 3. Then d = -1, e = 2, f = 2*2 - 3 = 4 - 3 = 1.

So sum a + b + c + d + e + f = 1 + (-2) + 3 + (-1) + 2 + 1 = (1 - 2 + 3) + (-1 + 2 + 1) = (2) + (2) = 4.

Which is 2k, since k = 2. 2*2 = 4. Correct.

Another example: take original parabola y = (x - h)^2 + k. Let me take h = 0, k = 5. Then original equation is y = x² + 5. So a = 1, b = 0, c = 5. Then reflected parabola is y = -x² + 0x + (2*5 - 5) = -x² + 0x + 5. So d = -1, e = 0, f = 5. Then sum a + b + c + d + e + f = 1 + 0 + 5 + (-1) + 0 + 5 = (1 -1) + (0 + 0) + (5 + 5) = 0 + 0 + 10 = 10, which is 2k = 10. Correct.

Another test case: take a parabola with vertex (h, k) where h is not zero. Let me take h = 2, k = 3, a = -1. Original equation: y = -1(x - 2)^2 + 3. Expanding that: y = - (x² - 4x + 4) + 3 = -x² + 4x - 4 + 3 = -x² + 4x - 1. So here, a = -1, b = 4, c = -1. Then reflecting over y = 3. According to formula, d = -a = 1, e = -b = -4, f = 2k - c = 6 - (-1) = 7. Therefore, the reflected equation is y = x² - 4x + 7.

Calculating the sum a + b + c + d + e + f: (-1) + 4 + (-1) + 1 + (-4) + 7 = (-1 -1) + (4 -4) + (1 + 7) = (-2) + 0 + 8 = 6. Which is 2k = 6. Correct.

So seems like regardless of the original coefficients, the sum is always 2k. Therefore, the answer is 2k.

But let me think again if there's a way to see this without going through all the examples.

Original equation: y = ax² + bx + c.

Reflected equation: y = -a x² - b x + (2k - c).

Therefore, adding all coefficients:

Original coefficients: a, b, c.

Reflected coefficients: -a, -b, 2k - c.

Sum: a + b + c + (-a) + (-b) + (2k - c) = (a - a) + (b - b) + (c - c) + 2k = 0 + 0 + 0 + 2k = 2k.

Yes, that's straightforward. So regardless of the values of a, b, c, h, etc., the sum is always 2k.

Therefore, the answer is 2k.

But let me check if there's any possible mistake here. For example, is the reflection over y = k really resulting in the equation y = -a x² - b x + (2k - c)?

Another way to think about reflection: if you have a parabola and you reflect it over a horizontal line, the shape is inverted? Wait, not necessarily. Wait, reflecting over y = k would invert the parabola with respect to that line.

But in terms of equations, if you have a function y = f(x), then reflecting over y = k would give y = 2k - f(x). So yes, that's the same as replacing y with 2k - y in the original equation.

So if original equation is y = f(x), then the reflected equation is y = 2k - f(x).

Therefore, if original equation is y = ax² + bx + c, then reflected equation is y = 2k - ax² - bx - c. Which is y = -ax² - bx + (2k - c). Which is what I had before.

Therefore, the coefficients are correct.

Therefore, the sum is 2k.

Therefore, the answer is 2k.

But let me check if the problem says "the parabola with equation y = ax² + bx + c and vertex (h, k)" so maybe there is some relation between h and k? But in the problem, they just say vertex (h, k), so h is arbitrary. However, in our calculation, we didn't need h. The sum ended up being 2k regardless of h.

But let me think again. Suppose we have vertex (h, k). Then the original equation is y = a(x - h)^2 + k. Then, when we reflect over y = k, we get y = -a(x - h)^2 + k. Wait, is that correct? Wait, if you reflect over y = k, then the vertex remains the same, but the parabola opens in the opposite direction.

Wait, so if original parabola is y = a(x - h)^2 + k, then reflecting over y = k would invert the coefficient a?

Wait, let me think. If you have a point (x, y) on the original parabola, then the reflection over y = k is (x, 2k - y). So if the original parabola is y = a(x - h)^2 + k, then the reflected parabola is 2k - y = a(x - h)^2 + k. Therefore, solving for y: y = 2k - a(x - h)^2 - k = -a(x - h)^2 + k.

So the reflected parabola is y = -a(x - h)^2 + k.

But expanding this, we get y = -a(x² - 2hx + h²) + k = -a x² + 2a h x - a h² + k.

Comparing with the previous expression for the reflected equation, which was y = -a x² - b x + (2k - c).

But in the original equation, we had y = a(x - h)^2 + k = a x² - 2a h x + a h² + k. Therefore, original coefficients: a, b = -2a h, c = a h² + k.

So the reflected equation, when expanded, is y = -a x² + 2a h x - a h² + k.

Comparing with the previous expression from the reflection formula: y = -a x² - b x + (2k - c).

Let me check if these are the same.

From the reflection formula:

- b x = -(-2a h) x = 2a h x. So that matches the coefficient of x in the expanded reflected equation.

Then the constant term: 2k - c. Original c is a h² + k. Therefore, 2k - c = 2k - (a h² + k) = k - a h².

But in the expanded reflected equation, the constant term is -a h² + k. Which is same as k - a h². So yes, matches.

Therefore, both methods agree.

Therefore, the reflected equation is indeed y = -a x² - b x + (2k - c).

Therefore, the sum a + b + c + d + e + f is 2k.

Therefore, the answer is 2k.

But let me check once more with another example. Let me take a parabola with vertex (h, k) where h is not zero, and a is negative.

Let me take h = 3, k = 4, a = -2. Original equation: y = -2(x - 3)^2 + 4. Expanding: y = -2(x² - 6x + 9) + 4 = -2x² + 12x - 18 + 4 = -2x² + 12x - 14. So here, a = -2, b = 12, c = -14.

Reflecting over y = 4. According to formula, reflected equation is y = -a x² - b x + (2k - c) = -(-2)x² - 12x + (8 - (-14)) = 2x² - 12x + 22.

Let me check if this is correct. The original parabola has vertex (3, 4). Reflecting over y = 4 should give a parabola with the same vertex, opening upwards (since original opens downward).

Let me check the reflected equation: y = 2x² - 12x + 22. Let me find its vertex.

x-coordinate is -b/(2a) = 12/(4) = 3. Then y-coordinate is y(3) = 2*(9) - 12*3 + 22 = 18 - 36 + 22 = 4. Correct. So vertex is (3, 4).

Now, let's compute the sum a + b + c + d + e + f. Original coefficients: a = -2, b = 12, c = -14. Reflected coefficients: d = 2, e = -12, f = 22.

Sum: (-2) + 12 + (-14) + 2 + (-12) + 22 = (-2 -14) + (12 -12) + (2 + 22) = (-16) + 0 + 24 = 8. Which is 2k = 2*4 = 8. Correct.

Therefore, this seems to hold.

Therefore, regardless of the values of a, b, c, h, the sum a + b + c + d + e + f equals 2k.

Therefore, the answer is 2k.

But let me think again if there's a possibility that the problem is expecting something else. For example, maybe the question is in terms of the original coefficients, but since the problem says "Express a + b + c + d + e + f in terms of k," and we have shown that it's 2k regardless of other variables, then 2k is the answer.

Alternatively, maybe there's a different approach. Let me think.

Another way: evaluate the original parabola at x = 1, and the reflected parabola at x = 1. Then sum all coefficients is the value of the polynomial at x = 1.

Wait, because for any polynomial y = ax² + bx + c, the value at x = 1 is a(1)^2 + b(1) + c = a + b + c. Similarly, for the reflected parabola, the value at x = 1 is d + e + f. Therefore, the total sum a + b + c + d + e + f is equal to [value of original at x=1] + [value of reflected at x=1].

So maybe we can compute this.

Original parabola: y = ax² + bx + c. At x = 1, y = a + b + c.

Reflected parabola: y = -a x² - b x + (2k - c). At x = 1, y = -a - b + 2k - c.

Therefore, total sum is (a + b + c) + (-a - b + 2k - c) = (a - a) + (b - b) + (c - c) + 2k = 0 + 0 + 0 + 2k = 2k.

Same result.

Therefore, regardless of the method, the answer is 2k.

Therefore, I think the answer is 2k.

**Final Answer**
\boxed{2k}

To solve the problem, we are given a parabola with equation:

$$
y = ax^2 + bx + c
$$

and vertex $(h, k)$. This parabola is then reflected over the horizontal line $y = k$, resulting in a new parabola with equation:

$$
y = dx^2 + ex + f
$$

We are asked to express the sum $a + b + c + d + e + f$ in terms of $k$.

---

### Step 1: Express the Original Parabola in Vertex Form

The vertex form of a parabola is:

$$
y = a(x - h)^2 + k
$$

Expanding this:

$$
y = a(x^2 - 2hx + h^2) + k = ax^2 - 2ahx + ah^2 + k
$$

Comparing with the standard form $y = ax^2 + bx + c$, we identify:

- $b = -2ah$
- $c = ah^2 + k$

---

### Step 2: Reflect the Parabola Over the Line $y = k$

To reflect a point $(x, y)$ over the line $y = k$, the reflected point is $(x, 2k - y)$. So, if the original parabola is $y = ax^2 + bx + c$, the reflected parabola is obtained by replacing $y$ with $2k - y$:

$$
2k - y = ax^2 + bx + c
\Rightarrow y = -ax^2 - bx + (2k - c)
$$

Thus, the reflected parabola has coefficients:

- $d = -a$
- $e = -b$
- $f = 2k - c$

---

### Step 3: Compute the Sum $a + b + c + d + e + f$

Now, we compute the sum of all coefficients:

$$
a + b + c + d + e + f = a + b + c + (-a) + (-b) + (2k - c)
$$

Grouping like terms:

- $a - a = 0$
- $b - b = 0$
- $c - c = 0$
- Remaining term: $2k$

So the total sum is:

$$
a + b + c + d + e + f = 2k
$$

---

### Final Answer

$$
\boxed{2k}
$$

Solution:
The equation of the original parabola can be written as
\[y = a(x - h)^2 + k.\]The equation of the reflected parabola is then
\[y = -a(x - h)^2 + k.\]Thus,
\[ax^2 + bx + c + dx^2 + ex + f = 2k.\]Setting $x = 1,$ we get $a + b + c + d + e + f = \boxed{2k}.$