For the given​ function, find ​(a) the equation of the secant line through the points where x has the given values and ​(b) the equation of the tangent line when x has the first value.

To solve this problem, let's follow these steps for a given function \( f(x) \):

### Part (a): Equation of the Secant Line

1. **Secant Line Concept**: The secant line through two points on the graph of a function is the line that passes through both points.
2. **Points on the Graph**: If the function is \( f(x) \) and the given \( x \)-values are \( x_1 \) and \( x_2 \), the points on the graph will be \( (x_1, f(x_1)) \) and \( (x_2, f(x_2)) \).
3. **Slope of the Secant Line**: The slope \( m \) of the secant line through these points is given by:
   \[
   m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.
   \]
4. **Equation of the Secant Line**: Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line, we substitute one of the points (e.g., \( (x_1, f(x_1)) \)) and the calculated slope to get the equation of the secant line.

### Part (b): Equation of the Tangent Line

1. **Tangent Line Concept**: The tangent line to the graph of a function at a point is the line that touches the graph at that point without crossing it.
2. **Slope of the Tangent Line**: The slope of the tangent line at a point \( x_1 \) is given by the derivative \( f'(x) \) evaluated at \( x = x_1 \).
3. **Equation of the Tangent Line**: Using the point-slope form of a line, the equation of the tangent line at \( x = x_1 \) is:
   \[
   y - f(x_1) = f'(x_1)(x - x_1).
   \]

To proceed further, I need the specific function \( f(x) \) and the given \( x \)-values to provide concrete equations for the secant and tangent lines. Could you please provide these details?

### Would You Like to Provide the Function and Values?

Once provided, I can solve the problem step-by-step. 

### Related Questions:
1. How do we find the derivative of a function at a given point?
2. What is the significance of the secant line in calculus?
3. How does the tangent line represent instantaneous rates of change?
4. How do the secant and tangent lines differ geometrically?
5. Can we find the normal line equation to a curve at a point?

**Tip:** Understanding the geometric meaning of secant and tangent lines helps in visualizing concepts in calculus more effectively.

For the given function, find (a) the equation of the secant line through the points where x has the given values and (b) the equation of the tangent line when x has the first value.

Learn how to find the equation of the secant line through two points and the tangent line at a given point for a specific function. This solution includes key calculus concepts such as the secant line, tangent line, and derivatives. Ideal for students in Grades 11-12.

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