e question in Calculus?

Question

Find the point on the curve y=e^-3x at which the tangent passes through the origin.

The question is confusing because the tangent passes through the origin, meaning the point we are looking for is the origin? But they say the answer is -1/3 and e. Please helppp. :/

I do believe the answer is -1/3 because it comes with the sheet my teacher gave me. Hehe.

Answer 1: I dont get why the slope between (0,0) and (x,y) is y/x. And how would I solve for x if there is nothing to make it equal to. (y'= y/x = ?)

Answer 1

Let (x, y) be the coordinates of the tangent point. The slope between (0,0) and (x,y) is y/x. Also the slope equals to y'. Therefore, we have

y' = e^(-3x)(-3) = -3y = y/x

Solve for x,
x = -1/3

Answer 2

If you are talking about the tangent, you need the derivative of the function. That in turn will be a function, which will give the slope of the tangent line at any point on the curve. Any straight line can be represented by an equation such as y = ax + b. But, if that line is to pass through the origin, then b must be zero, so the tangent line must be of the form y = ax, where a is the derivative at the point in question. You should be able to work it out from here.

Answer 3

If you graph the function you will see that there is no point on the curve where the tangent goes trough the origin.when x-> inf you will get a horizontal tangent that passes ( in the limit case trough the origin.

It is certainly NOT 1/3 and not e.
(the line orthogonal on the tangent oin the point 1/3 passes through the 0 )

Answer 4

Using the first answer

x = - 1/3
y = e^(- 3 x -1/3) = e^1 = e = 2.7.......

Point on the curve >> (- 1/3, e)

Answer 5

y’=-3exp(-3x); and tangent y=k·x, i.e. if u=0, then v=0; at tangent point both functions must be equal and their derivatives must be also equal; thus exp(-3x)=kx, y’ =k =-3exp(-3x), thence exp(-3x)= -3exp(-3x)·x, hence 1=-3x and x=-1/3 and y=exp(-3·(-1/3))=e, while k=y’ = -3exp(-3·(-1/3)) = -3e;