Find an equation of the line tangent to the graph of f(x) = (5x-9)(x+6) at (2,8). The equation of the line tangent to the graph of f(x) = (5x-9)(x+6) at (2,8) is y=41x-74 (Type an equation.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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ISBN:9780079039897
Author:Carter
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Chapter3: Linear And Nonlinear Functions
Section: Chapter Questions
Problem 26MCQ
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### Tangent Line Problem

**Objective**: Find the equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at the point \((2, 8)\).

---

### Solution

**Given**:
- Function, \( f(x) = (5x - 9)(x + 6) \)
- Point, \((2, 8)\)

**Solution**:
The equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at \((2,8)\) is:
\[ y = 41x - 74 \]

*Type an equation.*

---

### Explanation of the Process

To find the equation of a tangent line, follow these general steps:

1. **Find the derivative \( f'(x) \)**: This will give you the slope of the tangent line at any point \( x \).

2. **Evaluate \( f'(x) \) at \( x = 2 \)**: Substitute \( x = 2 \) into the derivative to find the slope of the tangent line at the given point.

3. **Use the point-slope form of a line**: With the slope from step 2 and the given point, use the point-slope form to find the equation of the tangent line.

In this specific example, after differentiation and substitution, the slope at \( x = 2 \) was found to be 41, leading directly to the equation \( y = 41x - 74 \).

*Derivation details could be further provided in a complete lesson to guide through each differentiation and substitution step.*
Transcribed Image Text:### Tangent Line Problem **Objective**: Find the equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at the point \((2, 8)\). --- ### Solution **Given**: - Function, \( f(x) = (5x - 9)(x + 6) \) - Point, \((2, 8)\) **Solution**: The equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at \((2,8)\) is: \[ y = 41x - 74 \] *Type an equation.* --- ### Explanation of the Process To find the equation of a tangent line, follow these general steps: 1. **Find the derivative \( f'(x) \)**: This will give you the slope of the tangent line at any point \( x \). 2. **Evaluate \( f'(x) \) at \( x = 2 \)**: Substitute \( x = 2 \) into the derivative to find the slope of the tangent line at the given point. 3. **Use the point-slope form of a line**: With the slope from step 2 and the given point, use the point-slope form to find the equation of the tangent line. In this specific example, after differentiation and substitution, the slope at \( x = 2 \) was found to be 41, leading directly to the equation \( y = 41x - 74 \). *Derivation details could be further provided in a complete lesson to guide through each differentiation and substitution step.*
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