Question

find the equations of tangents and normals to the curve at the point on it .y=x²+2ex+2 at (0,4)​

Answer 1

Answer:

Step-by-step explanation:

y=x²+2e^x+2

dy/dx=2x+2e^x

dy/dx=m of the tangent

The equation of the tangent

y-y1=m(x-x1)

y-4=(2x+2e^x)(x-0)=y-4=2x²+2xe^x

The slope of normal=-1/(dy/dx)

m=-1/(2x+2e^x)

The equation

y-4=-1/(2x+2e^x)(x-0)

y-4(2x+2e^x)=-x

y-4(2x+2e^x)+x=0

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Answer 2

Step-by-step explanation:

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