Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises - Page 146: 58


Horizontal tangents at $(1, 0)$ and $(1, -4)$. Vertical tangents at $(0, -2)$ and $(2, -2)$.

Work Step by Step

$\dfrac{d}{dx}(4x^2)+\dfrac{d}{dx}(y^2)-\dfrac{d}{dx}(8x)+\dfrac{d}{dx}(4y)+\dfrac{d}{dx}(4)=0\rightarrow$ $8x+\dfrac{dy}{dx}(2y)-8+\dfrac{dy}{dx}(4)=0\rightarrow$ $\dfrac{dy}{dx}=\dfrac{8x-8}{-2y-4}.$ Case I: Horizontal tangent $\rightarrow \dfrac{dy}{dx}=0\rightarrow8x-8=0\rightarrow x=1.$ Substituting in $x=1$ into the original equation gives: $4(1^2)+y^2-8(1)+4y+4=y(y+4)=0\rightarrow$ $y=-4$ or $y=0\rightarrow$ horizontal tangents at $(1, 0)$ and $(1, -4)$. Case II: Vertical tangents$\rightarrow\dfrac{dy}{dx}=$undefined$\rightarrow-2y-4=0\rightarrow y=-2.$ Substituting $y=-2$ into the original equation gives: $4x^2+(-2)^2-8x+4(-2)+4=4x(x-2)\rightarrow x=0$ or $x=2\rightarrow$ vertical tangent at $(0, -2)$ and $(2, -2)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.