slope of a secant line Given f1x2 = x2,
 (A) Find the slope of the secant line for a = 1 and h = 2 and 1, respectively. 
Graph y = f1x2 and the two secant lines.
 (B) Find and simplify the slope of the secant line for a = 1 and h any nonzero 
number.
 (C) Find the limit of the expression in part (B).
 (D) Discuss possible interpretations of the limit in part (C)

Given the function \( f(x) = x^2 \), let's go through each part.

### Part (A)
The slope of the secant line between two points on \( f(x) \) is given by:
\[
\text{slope} = \frac{f(a + h) - f(a)}{h}
\]

For \( a = 1 \):
1. When \( h = 2 \):
   \[
   f(1 + 2) = f(3) = 3^2 = 9, \quad f(1) = 1^2 = 1
   \]
   \[
   \text{slope} = \frac{f(3) - f(1)}{2} = \frac{9 - 1}{2} = \frac{8}{2} = 4
   \]

2. When \( h = 1 \):
   \[
   f(1 + 1) = f(2) = 2^2 = 4, \quad f(1) = 1^2 = 1
   \]
   \[
   \text{slope} = \frac{f(2) - f(1)}{1} = \frac{4 - 1}{1} = 3
   \]

So, the slopes of the secant lines for \( h = 2 \) and \( h = 1 \) are 4 and 3, respectively.

To graph \( y = f(x) = x^2 \) and the two secant lines, plot the curve \( y = x^2 \) and draw lines with these slopes passing through the point \( (1,1) \).

### Part (B)
Now, let’s find the slope of the secant line for \( a = 1 \) and any nonzero \( h \).

1. Substitute \( a = 1 \) into the general formula:
   \[
   \text{slope} = \frac{f(1 + h) - f(1)}{h}
   \]
2. Substitute \( f(1 + h) = (1 + h)^2 \) and \( f(1) = 1^2 = 1 \):
   \[
   \text{slope} = \frac{(1 + h)^2 - 1}{h}
   \]
3. Expand \( (1 + h)^2 = 1 + 2h + h^2 \):
   \[
   \text{slope} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = \frac{h(2 + h)}{h} = 2 + h
   \]

So, the slope of the secant line for \( a = 1 \) and any nonzero \( h \) simplifies to \( 2 + h \).

### Part (C)
To find the limit of the expression \( 2 + h \) as \( h \) approaches 0:
\[
\lim_{h \to 0} (2 + h) = 2
\]

### Part (D)
The limit found in part (C) represents the slope of the **tangent line** to the curve \( y = x^2 \) at the point \( x = 1 \). Geometrically, as \( h \) approaches 0, the secant line approaches the tangent line, so the slope of the secant line approaches the slope of the tangent line. This slope of 2 is the derivative \( f'(1) \), which represents the instantaneous rate of change of \( f(x) = x^2 \) at \( x = 1 \).

Would you like to go over any specific part in more detail, or have any questions about the concepts here?

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### Related Questions
1. How would the slope of the secant line change if \( f(x) = x^3 \)?
2. What is the general formula for the slope of a secant line for a function \( f(x) \)?
3. How does the concept of a secant line help in understanding derivatives?
4. What happens if we choose other values for \( a \) in this example?
5. How would you calculate the tangent line equation at \( x = 1 \)?

### Tip:
The slope of a secant line gives an average rate of change between two points, while the slope of a tangent line shows the instantaneous rate of change at a single point.

Given f(x) = x^2, find the slope of the secant line for a = 1 and h = 2 and 1, respectively. Graph y = f(x) and the two secant lines. Find and simplify the slope of the secant line for a = 1 and h any nonzero number. Find the limit of the expression in part (B). Discuss possible interpretations of the limit in part (C).

Explore how to calculate the slope of secant lines for the function f(x) = x^2 at specific points, find limits, and discuss interpretations related to derivatives. This guide is suitable for high school calculus students.

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