Question

find the equation of tangent and normal to the curve. y=
$2 {x}^{3} - {x}^{2} + 2$
at (1/2 ,2) ​
please it is so urgent

Answer 1

Step-by-step explanation:

y=2x

3

−x

2

+3

Differentiate both sides w.r.t. x,

⇒

dx

dy

=6x

2

−2x

⇒ Slop of tangent =(

dx

dy

)

(1,4)

=6(1)

2

−2(1)=6−2=4

⇒ Slop of normal =

Slopoftangent

−1

=

4

−1

⇒ (x

1

,y

1

)=(1,4) [ Given ]

Equation of normal is,

y−y

1

=m(x−x

1

)

⇒ y−4=

4

−1

(x−1)

⇒ 4y−16=−x+1

⇒ x+4y=17