Question

the point on the curve y=12x-x^2 where the slope of the tangent is zero

Answer 1

Answer:

Let the point be P(x,y)

y=12x−x

2

dx

dy

=12−2x

(

dx

dy

)

(x

1

,y

1

)

=12−2x

1

since slope of tangent is zero

so (

dx

dy

)

x

1

,y

1

=0

12−2x

1

=0

2x

1

=12

x

1

=6

Also curve passing through tangent

y

1

=12x

1

−x

1

2

y

1

=12×6−36

y

1

=72−36

y

1

=36

The points are (6,36).

Step-by-step explanation:

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