Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.2 Working with Derivatives - 3.2 Exercises - Page 143: 21


The equation is $$\boxed{y_n=-\frac{1}{3}x-\frac{2}{3}}$$

Work Step by Step

This is a straight line and as we know the tangent to the straight line at every point is that very same straight line so the equation of the tangent to $y=3x-4$ at $P(1,-1)$ is $y=3x-4$. The normal line to the given tangent $y=mx+n$ at the point $P(a,f(a))$ is a straight line $y_n=cx+d$ such that $y_n(a)=f(a)$ and $c\cdot m=-1$. For this problem $a=1$ and $m=3$ which gives $$3c=-1,\quad 3\cdot1-4=c\cdot1+d.$$ From the first equation we get $c=-1/3$. Putting this into the second equation we get $$3-4=-\frac{1}{3}+d\Rightarrow -1=-\frac{1}{3}+d$$ which gives $$d=-\frac{2}{3}.$$ Finally the equation of the normal line is $$\boxed{y_n=-\frac{1}{3}x-\frac{2}{3}}$$
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