We are given the function $f(x) = 5x^2 - 6x + 5$, and the problem asks us to simplify the expression for the difference quotient:

$\frac{f(5+h) - f(5)}{h}$

We are also asked to express the result as $ah + b$, and find the values of $a$ and $b$.

Step-by-Step Solution:

1. Substitute $f(5+h)$ into the function: $f(5+h) = 5(5+h)^2 - 6(5+h) + 5$ Expand each part: $= 5((5+h)^2) - 6(5+h) + 5$ $= 5(25 + 10h + h^2) - 6(5 + h) + 5$ $= 5(25) + 5(10h) + 5(h^2) - 6(5) - 6h + 5$ $= 125 + 50h + 5h^2 - 30 - 6h + 5$ $= 100 + 50h + 5h^2 - 6h$ $= 100 + 44h + 5h^2$

2. Calculate $f(5)$: $f(5) = 5(5^2) - 6(5) + 5$ $= 5(25) - 30 + 5$ $= 125 - 30 + 5 = 100$

3. Form the difference $f(5+h) - f(5)$: $f(5+h) - f(5) = (100 + 44h + 5h^2) - 100$ $= 44h + 5h^2$

4. Simplify the difference quotient: $\frac{f(5+h) - f(5)}{h} = \frac{44h + 5h^2}{h}$ Factor out $h$: $= h(44 + 5h) / h$ Cancel $h$: $= 44 + 5h$

Thus, the simplified form is $44 + 5h$. In the form $ah + b$, we have:

$a = 5, \quad b = 44$

Final Answer:

$a = 5, \quad b = 44$

Would you like further details or have any questions?

Related Questions:

1. How do you find the derivative of a quadratic function like $f(x) = 5x^2 - 6x + 5$?
2. What is the meaning of the difference quotient in calculus?
3. How can you determine the slope of a tangent line using the limit definition of the derivative?
4. How does expanding binomials, like $(5+h)^2$, help in calculus problems?
5. What happens to the difference quotient as $h$ approaches 0?

Tip: Always simplify terms before applying the difference quotient formula—it helps prevent mistakes in algebraic manipulation.