**Question 1**:

The curve

*y*=*x*^{3}– 6*x*^{2}+ 9*x*+ 3 passes through the point*P*(2, 5) and has two turning points,*A*(3, 3) and*B.*Find

**(a)**the gradient of the curve at

*P*.

**(b)**the equation of the normal to the curve at

*P*.

**(c)**the coordinates of

*B*and determine whether

*B*is a maximum or the minimum point.

*Solution:***(a)**

*y*=

*x*

^{3}– 6

*x*

^{2}+ 9

*x*+ 3

d

*y*/d*x*= 3*x*^{2}– 12*x*+ 9At point

*P*(2, 5),d

*y*/d*x*= 3(2)^{2}– 12(2) + 9 = –3**Gradient of the curve at point**

*P*= –3.**(b)**

Gradient of normal at point

*P*= 1/3Equation of the normal at

*P*(2, 5):*y*–

*y*

_{1 }=

*m*(

*x*–

*x*

_{1})

*y*– 5 = 1/3 (

*x*– 2)

3

*y*– 15 =*x*– 2**3**

*y*=*x*+ 13**(c)**

At turning point, d

*y*/d*x*= 0.3

*x*^{2}– 12*x*+ 9 = 0*x*

^{2}– 4

*x*+ 3 = 0

(

*x*– 1)(*x*– 3) = 0*x*– 1 = 0 or

*x*– 3 = 0

*x*= 1

*x*= 3 (Point

*A*)

Thus at point

*B*:*x*= 1

*y*= (1)

^{3}– 6(1)

^{2}+ 9(1) + 3 = 7

Thus, coordinates of

$$\begin{array}{l}\text{when}x=1,\text{}\\ \frac{{d}^{2}y}{d{x}^{2}}=6x-12\\ \frac{{d}^{2}y}{d{x}^{2}}=6\left(1\right)-12\\ \frac{{d}^{2}y}{d{x}^{2}}=-60\\ \text{Since}\frac{{d}^{2}y}{d{x}^{2}}0,\text{}B\text{isamaximumpoint}\text{.}\end{array}$$
*B*= (1, 7)