$y'=-{1\over 3}x^{-2}$
$y'=-{1\over 3x^2}$
$y'=-{1\over 3\left({1\over 6}\right)^2}=-{36\over 3}=-12$
When $x=1/6$, $y=1/(3(1/6))=2$
$y-2=-12(x-1/6)$
$y-2=-12x+2$
$12x+y-4=0$
JH :-)
This blog goes back a few years! while I was teaching in various schools in NSW, Australia. It contains a wide range of mathematics explanations, notes, and resources covering secondary-level mathematics topics. The easiest ways to navigate are by using the search bar or browsing through the labels/tags. If you find my content helpful and would like to support my work, a coffee donation is greatly appreciated — thank you!
No comments:
Post a Comment