The tangent line to the graph of function 
\[f\] at the point 
\[(5,-7)\] passes through the point 
\[(1,-1)\].
Find 
\[f'(5)\].

We are given that the tangent line to the graph of the function \( f \) at the point \( (5, -7) \) passes through the point \( (1, -1) \). Our task is to find \( f'(5) \), which represents the slope of the tangent line at \( x = 5 \).

### Step 1: Slope of the tangent line
Since the tangent line passes through the points \( (5, -7) \) and \( (1, -1) \), we can use the formula for the slope of a line:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Substitute the points \( (x_1, y_1) = (5, -7) \) and \( (x_2, y_2) = (1, -1) \):

\[
m = \frac{-1 - (-7)}{1 - 5} = \frac{-1 + 7}{1 - 5} = \frac{6}{-4} = -\frac{3}{2}
\]

Thus, the slope of the tangent line is \( m = -\frac{3}{2} \).

### Step 2: Interpretation of the slope
Since the slope of the tangent line at \( x = 5 \) is given by the derivative \( f'(5) \), we conclude that:

\[
f'(5) = -\frac{3}{2}
\]

Therefore, \( f'(5) = -\frac{3}{2} \).

Let me know if you'd like more details or have any questions.

Here are 5 related questions:
1. How can we find the equation of the tangent line using this slope?
2. What is the geometric interpretation of the derivative at a point?
3. How would you compute \( f'(x) \) for a general function \( f(x) \)?
4. What happens if a function's tangent at a point is horizontal?
5. How does knowing the slope help in approximating values near \( x = 5 \)?

**Tip:** The derivative at a point gives the slope of the tangent line, which can also help approximate function values near that point using linearization.

The tangent line to the graph of function f at the point (5,-7) passes through the point (1,-1). Find f'(5).

Learn how to find the derivative f'(5) by calculating the slope of the tangent line that passes through the points (5,-7) and (1,-1). This calculus problem covers key concepts such as the slope of a line and the interpretation of the derivative as the slope of the tangent line.

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