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3. Consider the graph of f(x)=x²-9x.The tangent has a y-intercept of 16. Determine the exact are of the triangle formed in the second quadrant by the tangent line and the axes. Let A (a, a³-3a) represent the tangency point. since tangent line passes through point B(0,16), we have: Equation of tangent line:y=3x+16 -16 x-int: x= 3 16 a³-3a-16 a-o m, = f'(a)=3a²-9 a³-3a-16 = 3a²-9 a a³-3a-16=3a³-9a mAB -16=2a³ -8=a³ a=-2 →m, = 3(-2)* -9 m₁ = 3 128 3 units²

3. Consider the graph of f(x)=x²-9x.The tangent has a y-intercept of 16. Determine the exact are of the triangle formed in the second quadrant by the tangent line and the axes. Let A (a, a³-3a) represent the tangency point. since tangent line passes through point B(0,16), we have: Equation of tangent line:y=3x+16 -16 x-int: x= 3 16 a³-3a-16 a-o m, = f'(a)=3a²-9 a³-3a-16 = 3a²-9 a a³-3a-16=3a³-9a mAB -16=2a³ -8=a³ a=-2 →m, = 3(-2)* -9 m₁ = 3 128 3 units²

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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3. Consider the graph of f(x)=x²-9x.The tangent has a y-intercept of 16. Determine the exact are of the triangle formed in the second quadrant by the tangent line and the axes. Let A (a, a³-3a) represent the tangency point. since tangent line passes through point B(0,16), we have: Equation of tangent line:y=3x+16 -16 x-int: x= a³-3a-16 a-o m, = f'(a)=3a²-9 a³-3a-16 -= 3a²-9 a a³-3a-16=3a³-9a -16=2a³ -8=a³ a=-2 →m, = 3(-2)* -9 m₁ = 3 3 16 3 units² A₁ = (16 128 3
Transcribed Image Text:3. Consider the graph of f(x)=x²-9x.The tangent has a y-intercept of 16. Determine the exact are of the triangle formed in the second quadrant by the tangent line and the axes. Let A (a, a³-3a) represent the tangency point. since tangent line passes through point B(0,16), we have: Equation of tangent line:y=3x+16 -16 x-int: x= a³-3a-16 a-o m, = f'(a)=3a²-9 a³-3a-16 -= 3a²-9 a a³-3a-16=3a³-9a -16=2a³ -8=a³ a=-2 →m, = 3(-2)* -9 m₁ = 3 3 16 3 units² A₁ = (16 128 3
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