Question

Find the points of any horizontal tangent lines: (x^2+xy+y^2) = 6. Am I doing something wrong?

Answer 1

\[x^2+xy +y^2 = 6 \]
\[2x+(x+y)y'+2yy'=0\]
\[y'(x+y+2y) = -2x\]
\[y'=\frac{ -2x }{ x+3y }\]

Answer 2

I'm thinking set y' = 0. Am I correct?

Answer 3

yes

Answer 4

check the derivative of your middle term: [xy]' = ???

Answer 5

x+y ? product rule...?

Answer 6

ohhhh i seeeeeeeee

Answer 7

yes... product rule:
\(\large [xy]'=x'y+y'x=y+y'x \)

Answer 8

\[2x+xy'+y+2yy' = 0\]
\[y'(x+2y) = -2x-y\]
\[y' = \frac{ -2x-y }{ x+2y }\]
so then \[-2x = y \] ?

Answer 9

yep...
now you need to find WHERE on the graph this occurs: y = -2x
so substitute this into the original equation and solve...

Answer 10

\(\large x^2+xy+y^2=6\), \(\large y=-2x \)
\(\large x^2+x(-2x)+(-2x)^2=6\)

????

Answer 11

ohhh i remember this now. sorry i learned this quite a while ago, and forgot some parts :'(

Answer 12

ok... cya....

Answer 13

thanks!