Find the equation of the tangent line and normal line to the curve at a given point y=x^4+2e^x

The slope of the line is given by the derivative

m = dy/dx =  4x^3 + 2e^x

using the point slope form of a straight line 
y - y1 = m(x - x1)     where m is the above and (x1,y1) is  the point of contact of the tangent  we have

y - y1 = (4x^3 + 2e^x)(x - x1)   which is equation of the tangent 

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