Hey guys, its my first post here and I was wondering if you could help.
I need to show the point on the graph of:
x^2-3xy+y^2=1
has a horizontal tangent line.
I am assume that since I know when its horizontal = 0 for slope and y'=0
ps (when I am writing y' it = yprime)
So step one is find the derivative.
The teacher told us the derivative is 2x-3y-3xy'-2yy'=0
is this right ? can anyone explain it?
thanks to both of you, I still just don't understand it, the teacher really confused me by not showing the rational explanation that the book uses.
If anyone could do a step by step solution with text I would appreciate it.
I think if I could see the whole problem start to finish maybe I could grasp it better
ok, skeeter, I'm with so far far, can you continue?
thx guys
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ok, so y = 2/3x
now I take 2/3x and plug it into the original equation.
So,
x^2-3xy+y^2=1
becomes
x^2-3x(2/3x) + (2/3x)2 = 1
right??????? I think I am close . help me finish !!!!
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so next is
x^2 - 2x^2 +4/9x^2 = 1
then what ???????????
I'm offically stuck !!!!!