problem 6 is part of my study guide. I did not understand well how to solve it (1) A horizontal tangent will have slope m =8.Problem 6. Consider the curve defined by the equation 8ry - x2 - y? – 15 = 0.%3D(1) Find all of the points (x,y) on the curve where the tangent line to the curve ishorizontal.(2) Find all of the points (x, y) on the curve where the tangent line to the curve is vertical.Note: this type of curve is called a hyperbola.AdxSolution 6. To answer both of these questions, we will first find an expression for dy6.<>8.15(^M2.= 0Y 8xy – x² - y16-dydy2) - 2л - 2уdx8(y+xd.xdy(8x-2y) = 2x - 8ydxondy2x8yX-4y4x-y-d.x8x-2yA horizontal tangent will have slope m =dydx0; this occurs when x = 4y. Weplug this relationship into the curve equation to get the points on the curve thatsatisfy this,

Given curve 8xy-x2-y2-1516=0 For the slope of tangent to the curve, taking derivative with respect…

Problem 6. Consider the curve defined by the equation 8ry – x? - y² – = 0. (1) Find all of the points (x, y) on the curve where the tangent line to the curve is horizontal. (2) Find all of the points (x, y) on the curve where the tangent line to the curve is vertical. Note: this tyne of curve is called a huwnerbola

Problem 6. Consider the curve defined by the equation 8ry – x? - y² – = 0. (1) Find all of the points (x, y) on the curve where the tangent line to the curve is horizontal. (2) Find all of the points (x, y) on the curve where the tangent line to the curve is vertical. Note: this tyne of curve is called a huwnerbola

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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