Question

how do i solve this problem?

find an equation for a line that is tangent to the graph of y=e^x and goes through the origin.

Answer 1

let the point of contact be (a,e^a)
so the lope of the tangent is (e^a - 0)/(a-0) = e^a/a

also y' = e^x
at (a,e^a) , y' = e^a

so e^a/a = e^a
then a = 1
and the slope of the tangent is e^1 = e
and the eqution of the tangent, using y = mx+b, is
y = ex + b
but the y=intercept is zero, it goes through the origin
our equation of the tangent is y = ex