# `y = x^4 + 2e^x, (0, 2)` Find equation of the tangent line and normal to the curve at the given point.

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### 1 Answer

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