`y = x^4 + 2e^x, (0, 2)` Find equation of the tangent line and normal to the curve at the given point.
Given curve is `y=x^4+2e^x`
` (dy)/(dx)=4x^3+2e^x `
the slope of the tangent is equal to the derivative of the equation of the curve at that point.
So the slope of the tangent at (0,2) is 2
The equation of tangent is y-2=2(x-0)
that is y=2x+2
The slope of the normal will be -1/2.
So the equation of the nornal will be y-2 = (-1/2)(x-0)
that is y = (-1/2)x + 2