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NARA [144]
3 years ago
11

Consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0
<span>The derivative of your function is:  
1 + (1*dy/dx + y) - 4y(dy/dx) = 0
2 + dy/dx + y - 4y(dy/dx) = 0
(2 + y) + dy/dx (1 - 4y) = 0
(dy/dx) = -(2 + y)/(1 - 4y)
 
and at point (2,1),
dy/dx = -(2 + 1)/(1 - 4*1)
          = -3 / -3
          = 1
 
Slope of tangent line = 1</span>
viktelen [127]3 years ago
5 0
The derivative is actually (-y-1)/(4y+x)
then you plug in the points (2,1) ==>
(-1-1)/(4*1+2) which equals -1/3
then you just plug in your info into a point-slope formula and get:
y-1=-1/3(x-2) which can then be reduced
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