The graph shows the function $f(x)$, and you're being asked to estimate the difference quotient, $\frac{f(2+h) - f(2)}{h}$, for $h = 1$ and $h = -1$. This expression approximates the derivative of the function at $x = 2$.

Step-by-Step Process:

1. When $h = 1$:

   • We need to find $f(2+h) = f(3)$ and $f(2)$.
   • From the graph, we estimate:
     • $f(3) \approx 3.0$ (where the blue curve at $x = 3$).
     • $f(2) \approx 1.5$ (from the blue curve at $x = 2$).

   Now, substitute into the difference quotient formula: $\frac{f(2+1) - f(2)}{1} = \frac{f(3) - f(2)}{1} = \frac{3.0 - 1.5}{1} = 1.5$

2. When $h = -1$:

   • We need to find $f(2+h) = f(1)$ and $f(2)$.
   • From the graph, we estimate:
     • $f(1) \approx 2.0$ (from the blue curve at $x = 1$).
     • $f(2) \approx 1.5$ (from the blue curve at $x = 2$).

   Now, substitute into the difference quotient formula: $\frac{f(2+(-1)) - f(2)}{-1} = \frac{f(1) - f(2)}{-1} = \frac{2.0 - 1.5}{-1} = \frac{0.5}{-1} = -0.5$

Summary of Results:

• For $h = 1$, the difference quotient is $1.5$.
• For $h = -1$, the difference quotient is $-0.5$.

Would you like further details on the steps or have any questions?

Here are 5 follow-up questions for further practice:

1. What is the general formula for the difference quotient, and how does it relate to the derivative?
2. How would you calculate the derivative of the function $f(x)$ at $x = 2$ using these estimates?
3. Can you derive an approximation for $f'(2)$ using symmetric difference (using both $h = 1$ and $h = -1$)?
4. How would the difference quotient change if we used smaller values of $h$?
5. How can you improve the accuracy of your estimate for the derivative from a graph?

Tip: The difference quotient formula is a crucial tool for estimating the slope of the tangent line to the graph at a point and is the foundation of the concept of derivatives in calculus.