1.

Find the equation of tangent and normal to the Curve y = x^3- 2x^2+4 at x=2​

Answer»

Answer:

18

Step-by-step EXPLANATION:

x+4y=18

At x=2, y=2

3

−2.2

2

+4=4

So the point is, (2,4)

Now y=x

3

−2x

2

+4

dx

dy

=3x

2

−4x

∴(

dx

dy

)

(2,4)

=3.2

2

−4.2=4=m (say)

∴ Slope of normal at the given point is =−

m

1

=−

4

1

THEREFORE, equation of normal is, (y−4)=−

4

1

(x−2)

⇒x+4y=18


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