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

$\frac{dy}{dx}=kx-6$

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

$\frac{dy}{dx}=3x-6$

therefore

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