The equation of the horizontal tangent to the graph of the function $y=e^x+e^{-x}$ is

The slope of the horizontal tangent is 0, so $0=\mathrm{d}y/\mathrm{d}x=e^x-e^{-x}\Rightarrow e^{2x}=1\Rightarrow x=0$ If $x=0$ then $y=$2. Hence the equation of horizontal tangent at (0, 2) is $y=2$.