Question

How do you find the equation of the tangent line to the graph `y=x^2e^x-2xe^x+2e^x` through point (1,e)?

Answer 1

Slope=`e`

Explanation:

You differentiate of course.

The derivative of a function at a point `x` (or `t` for parametric equations) is defined to be the slope of the graph of said equation at the given point.

Now, the given function is `y=x^2e^x-2xe^x+2e^x`

So since we just found out that a derivative finds us the slope of the equation at said point, we'll differentiate `y`, hence (using product rule)
`dy/dx= e^xdx^2/dx+x^2{de^x}/dx-2x{de^x}/dx-2e^xdx/dx+2{de^x}/dx`

(The constant `2` is left out as per the elementary equations in calculus).

Thus, the equation is simplified to

`dy/dx=cancel{2xe^x}+x^2e^x-cancel{2xe^2}-cancel{2e^x}+cancel{2e^x}\impliesdy/dx=x^2e^x`

The slope of the function at `(1,e)` is thus found by substituting for `x` from the co-ordinates given `(x,y)`.

Answer 2

`y=ex`, See the tangent-inclusive Socratic graph.

Explanation:

`y=e^x(x^2-2x+2)`

The slope of the tangent is

`y'=e^x(x^2-2x+2)+(x^2-2x+2)(e^x)'`

`=e^x x^2=e`, at `x = 1`

The point of contact of the tangent is `P( 1, e )`

The equation to the tangent at P is

`y-e=e(x-1`, giving

`y=ex`

graph{(e^x(x^2-2x+2)-y)(y-e^1 x)((x-1)^2+(y-2.73)^2-.0007)=0 [-1, 2, 1.5, 4]}