Explanation Given Information: The provided equation of the curve is, x 2 + y 2 + 6 x − 3 y + 7 = 0 and point is ( − 1 , 2 ) . Formula used: The slope of tangent to a curve y = f ( x ) at point ( x , y ) is given by the derivative of the curve at that point. The equation of a line passing through points ( x 1 , y 1 ) and slope m is given by, y − y 1 = m ( x − x 1 ) Calculation: Consider the provided equation of curve, x 2 + y 2 + 6 x − 3 y + 7 = 0 Now, find the derivative d y d x from x 2 + y 2 + 6 x − 3 y + 7 = 0 by taking the derivative term by term on both sides of the equation as, d ( x 2 ) d x + d d x ( y 2 ) + d d x ( 6 x ) − d d x ( 3 y ) + d d x ( 7 ) = 0 2 x + 2 y y ′ + 6 − 3 y ′ = 0 Solve for d y d x as, 2 x + 2 y y ′ + 6 − 3 y ′ = 0 y &

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

12th Edition

ISBN: 9781337630542

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 11.3, Problem 28E

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To calculate: The equation of the tangent line to the curve x2+y2+6x−3y+7=0 at (−1,2).

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Chapter 11.3, Problem 27E

Chapter 11.3, Problem 29E

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Chapter 11 Solutions

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

Ch. 11.1 - 1. Ch. 11.1 - 2. If Ch. 11.1 - Prob. 3CP Ch. 11.1 - 4. Find . Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Find the derivatives of the functions in Problems...

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The equation of the tangent line to the curve x 2 + y 2 + 6 x − 3 y + 7 = 0 at ( − 1 , 2 ) .

Solution Summary: The author calculates the equation of the tangent line to the curve, underset_y=-4x-2, and the slope of a line passing through points.

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To calculate: The equation of the tangent line to the curve x2+y2+6x−3y+7=0 at (−1,2).

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Chapter 11 Solutions