Question

find the equations of the tangent and the normal to the curve y=x³+2x²-4x+5 at origin​

Answer 1

Answer:

y=x³−2x+4

∴dx/dy=3x²−2

∴( dx/dy ) (1.3) =3(1)−2=3−2=1

∴ slope of tangent =1

∴ slope of normal =−1

∴ We get equation of Normal using

u−y 1 =m(x−x 1)

∴y−3=−1(x−1)

∴y−3=−x+1

∴x+y−3−1=0

∴ Required equation of Normal is x+y−4=0