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I see that you've uploaded an image containing a graph and a problem related to the function \( f(x) = \frac{1}{x - 3} \). 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:
\[
m_{\text{secant}} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
\]
where \( f(x) \) is the function and \( (x_1, f(x_1)) \) and \( (x_2, f(x_2)) \) are two points on the curve.

In your case:
- The points are \( (-2, f(-2)) \) and another point on the curve is \( (x_2, f(x_2)) \).

The slope of the secant line will depend on the values you input for \( x_2 \) and \( f(x_2) \). 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 \( x = -2 \), we need to compute the derivative of \( f(x) \) and evaluate it at \( x = -2 \).

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

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

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

### Summary of Results:
- The slope of the tangent line at \( x = -2 \) is \( -\frac{1}{25} \).
- 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 \( f(x) \) 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.

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)).

Explore the calculation of the secant line and tangent line slopes for the function f(x) = 1 / (x - 3). This detailed guide explains the use of derivative rules and secant line formulas, suitable for calculus students in Grades 11-12 or first-year college.

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