Does A Protractor Help?

If $y_1 = e^x$ and $y_2 = e^{-x}$ intersect then $y_1 = y_2$ or

$\begin{aligned} e^x & = e^{-x} \\ e^{2x} & = 1 \\ 2x & = \ln 1 = 0 \\ \implies x & = 0 \\ \implies y_1 & = y_2 = e^0 = 1 \end{aligned}$ .

The two curves intersect at $(0,1)$ .

The gradients of the two curves are:

$\begin{cases} \dfrac{dy_1}{dx} \big|_0 & = e^x \big|_0 & = 1 \\ \dfrac{dy_2}{dx} \big|_0 & = -e^{-x} \big|_0 & = -1 \end{cases}$

therefore, the two curves are perpendicular at the intersection.

That is:

$\boxed{\text{They intersect at (0,1) and created a right angle at the point of intersection.}}$