Question

Find the question to tangent and normal to the curve y=x³+3x-2 at the point (1,2)
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Answer 1

Answer:

The equation of the given curve is y=x

3

−3x

2

−9x+7 .

dx

dy

=3x

2

−6x−9

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.

∴3x

2

−6x−9⇒x

2

−2x−3=0

⇒(x−3)(x+1)=0

⇒x=3orx=−1

When x=3,y=(3)

3

−3(3)

2

−9(3)+7=27−27−27+7=−20.

When x=1,y=(1)

3

−3(1)

2

−9(1)+7=1−3−9+7=−4.

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