Math Problem Statement

Find the slope of the secant line and the tangent line at specific points for the function f(x) = 1/(x - 3). The graph shows the function and a point at (-2, f(-2)).

Solution

I see that you've uploaded an image containing a graph and a problem related to the function f(x)=1x3f(x) = \frac{1}{x - 3}. The image asks about finding the slope of the secant line and the tangent line at specific points.

Let’s break down the steps for solving this problem.

1. Secant Line Slope

The slope of a secant line passing through two points on a graph is given by the formula: msecant=f(x2)f(x1)x2x1m_{\text{secant}} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} where f(x)f(x) is the function and (x1,f(x1))(x_1, f(x_1)) and (x2,f(x2))(x_2, f(x_2)) are two points on the curve.

In your case:

  • The points are (2,f(2))(-2, f(-2)) and another point on the curve is (x2,f(x2))(x_2, f(x_2)).

The slope of the secant line will depend on the values you input for x2x_2 and f(x2)f(x_2). Once you plug in these values, the formula can give the secant line’s slope.

2. Tangent Line Slope

To find the slope of the tangent line at x=2x = -2, we need to compute the derivative of f(x)f(x) and evaluate it at x=2x = -2.

The function is f(x)=1x3f(x) = \frac{1}{x - 3}.

The derivative of f(x)f(x) using the power rule and chain rule is: f(x)=1(x3)2f'(x) = -\frac{1}{(x - 3)^2} Now, to find the slope of the tangent line at x=2x = -2, we substitute x=2x = -2 into the derivative: f(2)=1(23)2=125f'(-2) = -\frac{1}{(-2 - 3)^2} = -\frac{1}{25}

Therefore, the slope of the tangent line at x=2x = -2 is 125-\frac{1}{25}.

Summary of Results:

  • The slope of the tangent line at x=2x = -2 is 125-\frac{1}{25}.
  • For the secant line slope, you'll need to provide the second point to calculate it.

Would you like to clarify any part of this, or do you need help with the secant line calculation?

Here are 5 related questions:

  1. How is the derivative of f(x)f(x) found using the chain rule?
  2. What does the secant line represent geometrically?
  3. How does the slope of the tangent line compare to the slope of the secant line?
  4. How can the tangent line equation be written once we know the slope?
  5. How do vertical asymptotes affect the behavior of the function?

Tip: Remember that the slope of the tangent line at any point on a curve gives the instantaneous rate of change at that point.

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Math Problem Analysis

Mathematical Concepts

Calculus
Secant Line
Tangent Line
Derivatives
Functions

Formulas

Slope of the secant line: m_secant = (f(x2) - f(x1)) / (x2 - x1)
Derivative of f(x) = 1 / (x - 3): f'(x) = -1 / (x - 3)^2

Theorems

Secant line formula
Derivative rule (power rule and chain rule)

Suitable Grade Level

Grades 11-12 / First-year College Calculus