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To find the difference quotient for each function, we use the formula:

\[
\frac{f(x + h) - f(x)}{h}
\]

We'll apply this formula to each function one by one.

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Problem 1: \( f(b) = -7b - 7 \)

Step 1: Find \( f(b + h) \)

\[
f(b + h) = -7(b + h) - 7 = -7b - 7h - 7
\]

Step 2: Subtract \( f(b) \) from \( f(b + h) \)

\[
f(b + h) - f(b) = (-7b - 7h - 7) - (-7b - 7) = -7h
\]

Step 3: Divide by \( h \)

\[
\frac{-7h}{h} = -7
\]

✔ Final Answer for Problem 1: -7

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Problem 2: \( f(k) = 4k - 5 \)

Step 1: Find \( f(k + h) \)

\[
f(k + h) = 4(k + h) - 5 = 4k + 4h - 5
\]

Step 2: Subtract \( f(k) \)

\[
f(k + h) - f(k) = (4k + 4h - 5) - (4k - 5) = 4h
\]

Step 3: Divide by \( h \)

\[
\frac{4h}{h} = 4
\]

✔ Final Answer for Problem 2: 4

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Problem 3: \( f(y) = y + 3 \)

Step 1: Find \( f(y + h) \)

\[
f(y + h) = (y + h) + 3 = y + h + 3
\]

Step 2: Subtract \( f(y) \)

\[
f(y + h) - f(y) = (y + h + 3) - (y + 3) = h
\]

Step 3: Divide by \( h \)

\[
\frac{h}{h} = 1
\]

✔ Final Answer for Problem 3: 1

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Problem 4: \( f(c) = -2c - 9 \)

Step 1: Find \( f(c + h) \)

\[
f(c + h) = -2(c + h) - 9 = -2c - 2h - 9
\]

Step 2: Subtract \( f(c) \)

\[
f(c + h) - f(c) = (-2c - 2h - 9) - (-2c - 9) = -2h
\]

Step 3: Divide by \( h \)

\[
\frac{-2h}{h} = -2
\]

✔ Final Answer for Problem 4: -2

---

Problem 5: \( f(c) = 6c^2 + 8c - 6 \)

Step 1: Find \( f(c + h) \)

\[
f(c + h) = 6(c + h)^2 + 8(c + h) - 6
\]
\[
= 6(c^2 + 2ch + h^2) + 8c + 8h - 6
\]
\[
= 6c^2 + 12ch + 6h^2 + 8c + 8h - 6
\]

Step 2: Subtract \( f(c) \)

\[
f(c + h) - f(c) = [6c^2 + 12ch + 6h^2 + 8c + 8h - 6] - [6c^2 + 8c - 6]
\]
\[
= 12ch + 6h^2 + 8h
\]

Step 3: Divide by \( h \)

\[
\frac{12ch + 6h^2 + 8h}{h} = 12c + 6h + 8
\]

✔ Final Answer for Problem 5: 12c + 6h + 8

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Problem 6: \( f(x) = 4x^2 + 6x + 6 \)

Step 1: Find \( f(x + h) \)

\[
f(x + h) = 4(x + h)^2 + 6(x + h) + 6
\]
\[
= 4(x^2 + 2xh + h^2) + 6x + 6h + 6
\]
\[
= 4x^2 + 8xh + 4h^2 + 6x + 6h + 6
\]

Step 2: Subtract \( f(x) \)

\[
f(x + h) - f(x) = [4x^2 + 8xh + 4h^2 + 6x + 6h + 6] - [4x^2 + 6x + 6]
\]
\[
= 8xh + 4h^2 + 6h
\]

Step 3: Divide by \( h \)

\[
\frac{8xh + 4h^2 + 6h}{h} = 8x + 4h + 6
\]

✔ Final Answer for Problem 6: 8x + 4h + 6

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Problem 7: \( f(x) = -10x^2 - 4x - 3 \)

Step 1: Find \( f(x + h) \)

\[
f(x + h) = -10(x + h)^2 - 4(x + h) - 3
\]
\[
= -10(x^2 + 2xh + h^2) - 4x - 4h - 3
\]
\[
= -10x^2 - 20xh - 10h^2 - 4x - 4h - 3
\]

Step 2: Subtract \( f(x) \)

\[
f(x + h) - f(x) = [-10x^2 - 20xh - 10h^2 - 4x - 4h - 3] - [-10x^2 - 4x - 3]
\]
\[
= -20xh - 10h^2 - 4h
\]

Step 3: Divide by \( h \)

\[
\frac{-20xh - 10h^2 - 4h}{h} = -20x - 10h - 4
\]

✔ Final Answer for Problem 7: -20x - 10h - 4

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Problem 8: \( f(m) = -\frac{3}{-2m + 1} \)

Note: The problem says to leave the denominator in factored form. Also, note that the difference quotient is written as:

\[
\frac{f(m + h) - f(m)}{(m + h) - m} = \frac{f(m + h) - f(m)}{h}
\]

So we proceed normally.

Step 1: Find \( f(m + h) \)

\[
f(m + h) = -\frac{3}{-2(m + h) + 1} = -\frac{3}{-2m - 2h + 1}
\]

Step 2: Compute \( f(m + h) - f(m) \)

\[
f(m + h) - f(m) = -\frac{3}{-2m - 2h + 1} - \left( -\frac{3}{-2m + 1} \right)
= -\frac{3}{-2m - 2h + 1} + \frac{3}{-2m + 1}
\]

Factor out the 3:

\[
= 3 \left( \frac{1}{-2m + 1} - \frac{1}{-2m - 2h + 1} \right)
\]

Get common denominator:

Let’s write it as:

\[
= 3 \cdot \frac{ (-2m - 2h + 1) - (-2m + 1) }{ (-2m + 1)(-2m - 2h + 1) }
\]

Simplify numerator:

\[
(-2m - 2h + 1) - (-2m + 1) = -2m - 2h + 1 + 2m - 1 = -2h
\]

So:

\[
f(m + h) - f(m) = 3 \cdot \frac{ -2h }{ (-2m + 1)(-2m - 2h + 1) } = \frac{ -6h }{ (-2m + 1)(-2m - 2h + 1) }
\]

Step 3: Divide by \( h \)

\[
\frac{f(m + h) - f(m)}{h} = \frac{ -6h }{ h \cdot (-2m + 1)(-2m - 2h + 1) } = \frac{ -6 }{ (-2m + 1)(-2m - 2h + 1) }
\]

We can factor out a negative from the second denominator term if needed, but the problem says to leave denominator in factored form — so we leave it as is.

Alternatively, we can write:

\[
= \frac{ -6 }{ (-2m + 1)(-2(m + h) + 1) }
\]

But since the original function was written with \( -2m + 1 \), and the instruction is to leave denominator in factored form, we’ll keep it as:

\[
\frac{ -6 }{ (-2m + 1)(-2m - 2h + 1) }
\]

We can also factor out a -1 from the second factor to make it look nicer:

\[
-2m - 2h + 1 = -(2m + 2h - 1)
\]

So:

\[
\frac{ -6 }{ (-2m + 1) \cdot [-(2m + 2h - 1)] } = \frac{ -6 }{ - ( -2m + 1 )(2m + 2h - 1) } = \frac{6}{( -2m + 1 )(2m + 2h - 1)}
\]

Wait — let's double-check signs.

Original expression after dividing by h:

\[
\frac{ -6 }{ (-2m + 1)(-2m - 2h + 1) }
\]

Note that \( -2m - 2h + 1 = - (2m + 2h - 1) \), so:

\[
= \frac{ -6 }{ (-2m + 1) \cdot [ - (2m + 2h - 1) ] } = \frac{ -6 }{ - ( -2m + 1 )(2m + 2h - 1) } = \frac{6}{( -2m + 1 )(2m + 2h - 1)}
\]

This looks cleaner. But the problem didn’t specify to simplify signs — just to leave denominator in factored form. Either form is acceptable, but let’s go with the simplified sign version.

Actually, let’s check numerically to verify.

Let m = 0, h = 1

Then f(m) = -3/( -0 + 1 ) = -3

f(m+h) = f(1) = -3/( -2 + 1 ) = -3/(-1) = 3

Difference quotient = (3 - (-3)) / 1 = 6

Now plug into our answer:

\[
\frac{6}{( -0 + 1 )(0 + 2 - 1)} = \frac{6}{1 \cdot 1} = 6 ✔
\]

If we used the other form:

\[
\frac{ -6 }{ (1)(-1) } = \frac{-6}{-1} = 6 ✔
\]

Both work. But since the problem says “leave your denominator in factored form”, and doesn’t specify to simplify signs, either is fine. However, the form without negative signs in front is cleaner.

But note: the original function is written as \( f(m) = -\frac{3}{-2m+1} \), which is equivalent to \( \frac{3}{2m - 1} \). Maybe we should rewrite the function first?

Wait — let’s re-express the function:

\[
f(m) = -\frac{3}{-2m + 1} = \frac{3}{2m - 1}
\]

Because multiplying numerator and denominator by -1:

\[
-\frac{3}{-2m + 1} = \frac{3}{2m - 1}
\]

That might be simpler! Let’s redo Problem 8 using this equivalent form.

Let \( f(m) = \frac{3}{2m - 1} \)

Then:

Step 1: \( f(m + h) = \frac{3}{2(m + h) - 1} = \frac{3}{2m + 2h - 1} \)

Step 2: \( f(m + h) - f(m) = \frac{3}{2m + 2h - 1} - \frac{3}{2m - 1} \)

Common denominator:

\[
= 3 \cdot \frac{ (2m - 1) - (2m + 2h - 1) }{ (2m + 2h - 1)(2m - 1) } = 3 \cdot \frac{ 2m - 1 - 2m - 2h + 1 }{ (2m + 2h - 1)(2m - 1) } = 3 \cdot \frac{ -2h }{ (2m + 2h - 1)(2m - 1) }
\]

Step 3: Divide by h:

\[
\frac{ -6h }{ h \cdot (2m + 2h - 1)(2m - 1) } = \frac{ -6 }{ (2m + 2h - 1)(2m - 1) }
\]

Now, this matches our earlier result if we consider:

Earlier we had \( \frac{6}{( -2m + 1 )(2m + 2h - 1)} \)

But \( -2m + 1 = -(2m - 1) \), so:

\[
\frac{6}{ [-(2m - 1)] (2m + 2h - 1) } = \frac{ -6 }{ (2m - 1)(2m + 2h - 1) }
\]

Same as above.

So final answer is:

\[
\frac{ -6 }{ (2m - 1)(2m + 2h - 1) }
\]

Or, if we want to match the original function’s form, we could write:

\[
\frac{ -6 }{ (-2m + 1)(-2m - 2h + 1) }
\]

But since the problem says “leave your denominator in factored form”, and doesn’t specify which form, I think the cleanest is:

\[
\frac{ -6 }{ (2m - 1)(2m + 2h - 1) }
\]

However, looking back at the original function: \( f(m) = -\frac{3}{-2m+1} \), which is the same as \( \frac{3}{2m - 1} \), so using \( 2m - 1 \) is fine.

But to be safe, let’s present the answer as:

\[
\frac{ -6 }{ (2m - 1)(2m + 2h - 1) }
\]

✔ Final Answer for Problem 8: \(\frac{ -6 }{ (2m - 1)(2m + 2h - 1) }\)

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Final Answers:

1. -7
2. 4
3. 1
4. -2
5. 12c + 6h + 8
6. 8x + 4h + 6
7. -20x - 10h - 4
8. \(\frac{ -6 }{ (2m - 1)(2m + 2h - 1) }\)

Final Answer:
1. -7
2. 4
3. 1
4. -2
5. 12c + 6h + 8
6. 8x + 4h + 6
7. -20x - 10h - 4
8. \(\frac{ -6 }{ (2m - 1)(2m + 2h - 1) }\)