Answer
The equation of the tangent line: $$(l):y=2x+2$$
The equation of the normal line: $$(m):y=-\frac{1}{2}x+2$$
Work Step by Step
$$y=f(x)=x^4+2e^x$$
1) Find the derivative of $y$$$y'=\frac{d}{dx}(x^4)+2\frac{d}{dx}(e^x)$$
$$y'=4x^3+2e^x$$
2) Find $y'(0)$ $$y'(0)=4\times0^3+2e^0$$$$y'(0)=0+2\times1=2$$
3) $y'(0)$ is the slope of the tangent line $(l)$ to the curve $f(x)$ at point $(0,2)$
Knowing the slope of the tangent line $(l)$ $y'(0)$ and 1 point of (l) $(0,2)$, the equation of the tangent line $(l)$ would be: $$(l):(y-2)=y'(0)(x-0)$$
$$(l):y-2=2x$$
$$(l):y=2x+2$$
4) The normal line $(m)$ at $(0,2)$ is the line perpendicular to the tangent line $(l)$ at $(0,2)$.
So, the slope $a$ of $(m)$ is: $$a=\frac{-1}{y'(0)}=\frac{-1}{2}$$ (the product of the slopes of 2 perpendicular lines equals $-1$)
Knowing the slope of $(m)$ and 1 point of $(m)$ $(0,2)$, the equation of the normal line $(m)$ would be $$(m):(y-2)=a(x-0)$$
$$(m):y-2=-\frac{1}{2}x$$
$$(m):y=-\frac{1}{2}x+2$$
