SOLUTION: Find two points on the curve (C)······ x^2 y+y^4 +6=4+2x, where x = 1. At each of those points , find an equation of the tangent line to C.

Algebra ->  Functions -> SOLUTION: Find two points on the curve (C)······ x^2 y+y^4 +6=4+2x, where x = 1. At each of those points , find an equation of the tangent line to C.      Log On


   

Question 1023927: Find two points on the curve
(C)······ x^2 y+y^4 +6=4+2x,
where x = 1. At each of those points , find an equation of the tangent line to C.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
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%281%29%5E2y%2By%5E4%2B6=4%2B2%281%29
y%2By%5E4=0
%28y%5E3%2B1%29y=0
Two solutions:
y=0
and
y%5E3%2B1=0
%28y%2B1%29%28y%5E2-y%2B1%29=0
y=-1
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To find the slope of the tangent line, find the derivative at that point.
Differentiate implicitly,
x%5E2dy%2By%282xdx%29%2B4y%5E3dy=2dx
%28x%5E2%2B4y%5E3%29dy=%282-2xy%29dx
dy%2Fdx=%282-2xy%29%2F%28x%5E2%2B4y%5E3%29
When x=1, y=0,
m=dy%2Fdx=%282-0%29%2F%281%2B0%29=2
So then the tangent line is,
y-0=2%28x-1%29
y=2x-2
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When x=1, y=-1,
m=dy%2Fdx=%282-2%281%29%28-1%29%29%2F%281-4%29=-4%2F3
So then the tangent line is,
y%2B1=%28-4%2F3%29%28x-1%29
y=-%284%2F3%29x%2B4%2F3-3%2F3
y=-%284%2F3%29x%2B1%2F3
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