# How To Graph Y = Xe^-x?

There are a lots of solutions.!

i) Differentiate to find the gradient function of the tangent

dy/dx = e^x + xe^2 ........(1)

ii) Substitute x = 0 into (1) to find the slope of the tangent

slope of tangent m = e^(0) + (0)e^(0)

= 1

slope of normal = - 1

Equation of normal

y - 0 = -1(x - 0)

y = - x

I highly suggest you that try this online graphing calculator, through which you can easily do all the graphing calculations.

thanked the writer.
The graph would just look like the e^x graph slightly altered. The shape would be the same, but because you multiply by 'X', all the y-values before 1 will be below the original e^x graph (because you multiply by a value < 1), all the y-values after 1 will be higher (multiplying by values > 1). So, it would just be an e^x shaped graph, a bit lower before 1 and higher after 1.
thanked the writer.
Anonymous commented
To see the graph there are lots of graphing sites. But, you should be able to figure out the graph fairly easy, as long as you know how an e^x graph looks. The way you would do it analytically is, you know as e^x gets large, e^x grows much faster than x, and thus it controls the shape of the function. But as x is small e^x is changing slowly, and x becomes more prominent. (Just look at x=100 vs x=0). On negative side, values increase linear-like to 1, then approach 0 as x becomes large (x=-100)