Question 862194
To find the tangent, find the value of the derivative at that point.
{{{y=(2+x)e^(-x)}}}
{{{dy/dx=(2+x)(-e^(-x))+(e^(-x))(1)=e^(-x)(1-2-x)=-e^(-x)(x+1)}}}
So when {{{x=0}}}
{{{dy/dx=-e^(-0)(0+1)=-1}}}
{{{y=-x+b}}}
Use the point to solve for {{{b}}}.
{{{2=-0+b}}}
{{{b=2}}}
{{{highlight(y=-x+2)}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(0,2,0.2),graph(300,300,-5,5,-5,5,(2+x)e^(-x),-x+2))}}}