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)?

2 Answers
Feb 12, 2017

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).

Feb 12, 2017

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]}