Question

Write the equation of a tangent to the graphs of the following curves at the indicated points
$y=x^2{e}^{-x}$ at the point x = 1.

Answer 1

$y=x^2{e}^{-x}$ at $x=1$

To find : - equation of a tangent

First, put $x=1$ in $y=x^2{e}^{-x}$ and get,

$y_0(1{)}^2{e}^{-1}=e^{-1}$

Now, $y^1=x^2\left(e^{-x}\right)(-1)+\left(e^{-x}\right)\left(2x\right)$

Put $x=1$ in $y^{\prime }$

$y^{\prime }=slope=m=\left(1\right)\left(e^{-1}\right(-1)+(e^{-1}\left)\right(2)$

$=-e^{-1}+2e^{-1}$

$=e^{-1}$

Use point slope equation : -

$(y-y_0)=m(x-x_0)$

$(y-e^{-1})=e^{-1}(x-1)$

$y-\frac{1}{e}=\frac{x}{e}-\frac{1}{e}$

$y=\frac{1}{e}-\frac{1}{e}+\frac{x}{e}=\frac{x}{e}$

$y=\frac{x}{e}$.