problem 6 is part of my study guide. I did not understand well how to solve it **Problem 6.** Consider the curve defined by the equation \(8xy - x^2 - y^2 - \frac{15}{16} = 0\).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 type of curve is called a hyperbola.***Solution 6.** To answer both of these questions, we will first find an expression for \(\frac{dy}{dx}\).Using the product rule and implicit differentiation:\[8\left(\frac{d}{dx}(x)y + x\frac{d}{dx}(y)\right)\]This becomes:\[8(y + x\frac{dy}{dx})\]Substituting this into the original equation:\[8(y + x\frac{dy}{dx}) - 2x - 2y = 0\]Simplifying, we solve for \(\frac{dy}{dx}\):\[8x\frac{dy}{dx} = 2x - 8y\]\[\frac{dy}{dx} = \frac{x - 4y}{4x - y}\]1. A horizontal tangent will have slope \(\frac{dy}{dx} = 0\); this occurs when \(x - 4y = 0\). We can plug this relationship into the equation to get the points on the curve that satisfy this.2. A vertical tangent will have the denominator of \(\frac{dy}{dx}\) equal to zero; this occurs when \(4x - y = 0\).**Diagrams and Annotations:**- On the right side, there is a triangle drawn with arrows pointing downward, labeled as \(d/dx(C)\) and an expression \(x \cdot d/dx(^{}\)).- Certain expressions are highlighted to emphasize specific calculations: - \(y^2 - \frac{15}{16} = 0\) - \(x - 4y\) - \(4x - y\) boxed for emphasis.

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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