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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 (Type an equation.)

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 (Type an equation.)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Calculus Problem: Tangent Line Equation** ### Problem Statement Find an equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at \( (2, 8) \). --- ### Task The equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at \( (2, 8) \) is: \[ \text{(Type an equation.)} \] --- To solve this problem, follow these steps: 1. **Find the function's derivative, \( f'(x) \)**: - Use the product rule: \( (u \cdot v)' = u' \cdot v + u \cdot v' \) - Let \( u = 5x - 9 \) and \( v = x + 6 \) - Calculate \( u' \) and \( v' \) 2. **Evaluate \( f'(x) \) at \( x = 2 \)** to find the slope of the tangent line at that point. 3. **Use the point-slope form of the line equation**: - Given the point \( (2, 8) \) and the slope found in step 2, - Apply the point-slope formula: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point, and \( m \) is the slope. This process will yield the equation of the tangent line at the given point.
Transcribed Image Text:**Calculus Problem: Tangent Line Equation** ### Problem Statement Find an equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at \( (2, 8) \). --- ### Task The equation of the line tangent to the graph of \( f(x) = (5x - 9)(x + 6) \) at \( (2, 8) \) is: \[ \text{(Type an equation.)} \] --- To solve this problem, follow these steps: 1. **Find the function's derivative, \( f'(x) \)**: - Use the product rule: \( (u \cdot v)' = u' \cdot v + u \cdot v' \) - Let \( u = 5x - 9 \) and \( v = x + 6 \) - Calculate \( u' \) and \( v' \) 2. **Evaluate \( f'(x) \) at \( x = 2 \)** to find the slope of the tangent line at that point. 3. **Use the point-slope form of the line equation**: - Given the point \( (2, 8) \) and the slope found in step 2, - Apply the point-slope formula: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point, and \( m \) is the slope. This process will yield the equation of the tangent line at the given point.
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