Question

The gradient function of a curve is (x-3) . It is given P (2,1) lies on a curve.a) Find the gradient of the tangent at point Pb) Find the equation of the curve

Answer 1

Given:

The gradient of the curve is x-3.

`(dy)/(dx)=x-3\ldots\ldots\ldots(1)`

a) To find the gradient of tangent at point P,

`\begin{gathered} (dy)/(dx)\text{ at point P(2,1)} \\ (dy)/(dx)=2-3=-1 \end{gathered}`

So, the gradient of function at point P is -1.

b) To find the equation of the curve integrate the equation (1)

`\begin{gathered} (dy)/(dx)=x-3 \\ d(y)=(x-3)dx \\ \text{Take the integration,} \\ \int dy=\int (x-3)dx \\ y=\int xdx-3\int dx \\ y=(x^2)/(2)-3x+C \end{gathered}`

For point P,

`\begin{gathered} (x,y)=(2,1) \\ y=(x^2)/(2)-3x+C \\ 1=(2^2)/(2)-3(2)+C \\ 1=-4+C \\ C=5 \end{gathered}`

Answer: the equation of curve is,

`y=(x^2)/(2)-3x+5`