Question 1023927
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{{{(1)^2y+y^4+6=4+2(1)}}}
{{{y+y^4=0}}}
{{{(y^3+1)y=0}}}
Two solutions:
{{{y=0}}} 
and
{{{y^3+1=0}}}
{{{(y+1)(y^2-y+1)=0}}}
{{{y=-1}}}
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To find the slope of the tangent line, find the derivative at that point.
Differentiate implicitly,
{{{x^2dy+y(2xdx)+4y^3dy=2dx}}}
{{{(x^2+4y^3)dy=(2-2xy)dx}}}
{{{dy/dx=(2-2xy)/(x^2+4y^3)}}}
When {{{x=1}}}, {{{y=0}}},
{{{m=dy/dx=(2-0)/(1+0)=2}}}
So then the tangent line is,
{{{y-0=2(x-1)}}}
{{{y=2x-2}}}
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When {{{x=1}}}, {{{y=-1}}},
{{{m=dy/dx=(2-2(1)(-1))/(1-4)=-4/3}}}
So then the tangent line is,
{{{y+1=(-4/3)(x-1)}}}
{{{y=-(4/3)x+4/3-3/3}}}
{{{y=-(4/3)x+1/3}}}
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*[illustration dv8.JPG].