Question

The gradient function of a curve is dy/dx=kx-6 where k is a constant. It is given that the curve has a turning point at (2,1)Finda) the value of kb) the equation of the curve

Answer 1

we have that

`(dy)/(dx)=kx-6`

At (2,1) the curve has a turning point ----> that means ---> the derivative is equal to zero

so

`0=kx-6`

substitute the x-coordinate of point (2,1)

x=2

`\begin{gathered} 0=k(2)-6 \\ 2k=6 \\ k=3 \end{gathered}`

Part a)

the valuie of k=3

Part b

the equation of the curve

we have

`(dy)/(dx)=3x-6`

therefore

`y=\int ^{}_{}(3x-6)dx`
`y=(3x^2)/(2)-6x+C`