Question

Find the equation of tangent to the curve y = x2 +3x+2 at (0,3).
​

Answer 1

Answer:

y=x

2

−3x+2 [ Given ]

y=x [ Given ]

Differentiating both sides w.r.t. x, we get

⇒

dx

dy

=1

Let (x

1

,y

1

) be the required point.

It is given that point lies on the curve.

∴ y

1

=x

1

2

−3x

1

+2

⇒ y=x

2

−3x+2

Differentiating both sides w.r.t. x, we get

∴

dx

dy

=2x−3

Slope of the tangent =(

dx

dy

)

(x

1

,y

1

)

=2x

1

−3

The tangent is perpendicular to the line.

∴ Slope of the tangent =

slopeoftheline

−1

=

1

−1

=−1

Now,

2x

1

−3=−1

⇒ 2x

1

=2

⇒ x

1

=1

⇒ y

1

=x

1

2

−3x

1

+2

⇒ y

1

=(1)

2

−3(1)+2

⇒ y

1

=1−3+2

∴ y

1

=0

∴ (x

1

,y

1

)=(1,0)

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