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^2e^-x , (2,4/e^2)

Answer 1

The equation of the tangent line for the function
`f\left(x\right)=x^(2) e^(- x)` at
`\left(2,\:(4)/(e^2)\right)` is
`y=(4)/(e^(2))`

Explanation:

To find the tangent line to
`f\left(x\right)=x^(2) e^(- x)` at
`x_0=2`

Firstly, find the slope of the tangent line, which is the derivative of the function, evaluated at the point:
`m=f^(\prime)\left(2\right)`

`\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\\\\f=x^2,\:g=e^(-x)`

`(d)/(dx)\left(x^2e^(-x)\right)=(d)/(dx)\left(x^2\right)e^(-x)+(d)/(dx)\left(e^(-x)\right)x^2\\\\=2e^(-x)x-e^(-x)x^2`

Next, evaluate the derivative at the given point to find the slope.

`\mathrm{Plug\:}x=2\mathrm{\:into\:the\:equation\:}2e^(-x)x-e^(-x)x^2\\\\2e^(-2)\cdot \:2-e^(-2)\cdot \:2^2=0`

`m=f^(\prime)\left(2\right)=0`

Finally, the equation of the tangent line is
`y-y_0=m(x-x_0)`

Plugging the found values, we get that

`y-\left((4)/(e^(2))\right)=0\left(x-\left(2\right)\right)\\\\y=(4)/(e^(2))`

The equation of the tangent line for the function
`f\left(x\right)=x^(2) e^(- x)` at
`\left(2,\:(4)/(e^2)\right)` is
`y=(4)/(e^(2))`