Question

Find the tangent y x 3 2x 4 where the curves y=x^3-3x+4

Answer 1

Tangent:

y

=

2

x

−

11

Normal:

y

=

−

1

2

x

−

6

Explanation:

A tangent line would have a gradient equal to that of the curve at x=2, while a normal line would have a gradient which is the negative reciprocal of that of the curve at x=2

d

y

d

x

=

3

x

2

−

4

x

−

2

When x=2,

d

y

d

x

=

3

(

2

)

2

−

4

(

2

)

−

2

=

2

Hence, tangent at x=2:

y

=

(

d

y

d

x

)

x

+

c

=

2

x

+

c

Solve for c by setting

x

=

2

,

y

=

(

2

)

3

−

2

(

2

)

2

−

2

(

2

)

−

3

=

−

7

Thus,

c

=

−

7

−

2

(

2

)

=

−

11

The equation of the tangent line is thus

y

=

2

x

−

11

graph{(y-x^3+2x^2+2x+3)(y-2x+11)=0 [1, 3, -7.5, -6.5]}

Similarly, for the normal line,

y

=

−

1

2

x

+

c

c

=

−

7

−

(

−

1

2

(

2

)

)

=

−

6

Thus the equation of the normal line is

y

=

−

1

2

x

−

6

graph{(y-x^3+2x^2+2x+3)(y+1/2x+6)=0 [1, 3, -7.5, -6.5]}

I hope this will help you

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