Find the gradient of the curve y=e^-3x at the point where it crosses the y-axis.?

y = e⁻³ˣ
dy/dx = -3e⁻³ˣ

Gradient of the curve where it crosses the y-axis
= dy/dx at (x = 0)
= -3e⁰
= --3

That's when x = 0.

So first we need the first derivative of that function, which I presume is:

y = e^(-3x)

We'll need the chain rule to do this:

y = e^u and u = -3x
dy/du = e^u and du/dx = -3

dy/dx = dy/du * du/dx
dy/dx = e^u * (-3)
dy/dx = -3e^u

Put the expression in terms of "x" back in:

dy/dx = -3e^(-3x)

Now that we have this, solve for dy/dx when x = 0:

dy/dx = -3e^(-3 * 0)
dy/dx = -3e^0
dy/dx = -3(1)
dy/dx = -3

The only way y = 0 is for x = infinity.  Since the slope m(x) of the curve y is given by the first derivative of y

m(x) = dy/dx = -3 e^(-3x)  then as x --> infinity m(x) --> 0 .  so the curve is parallel to the x axis