Question

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the given point.

y = x^2 e^-x, (1,1/e)

Answer 1

Equation of tangent will be
`y=(x)/(e)`

Explanation:

We have given the function
`y=x^2e^(-x)`

We have to find the equation of tangent at the point
`(1,(1)/(e))`

Equation of tangent is equal to
`(dy)/(dx)`

So
`(dy)/(dx)=x^2(d)/(dx)e^(-x)+e^(-x)(d)/(dx)2x=-x^2e^(-x)+2e^(-x)`

Now we have given point
`(1,(1)/(e))`

So putting these points in the equation of tangent

`(dy)/(dx)=-1^2e^(-1)+2e^(-1)`

`(dy)/(dx)=(1)/(e)`

Now equation of tangent passing through
`(1,(1)/(e))`

`y-(1)/(e)=(1)/(e)(x-1)`

`y=(x)/(e)-(1)/(e)+(1)/(e)=(x)/(e)`