**Question 2:**Given the equation of a curve is:

*y*=

*x*

^{2}(

*x*– 3) + 1

(a) Find the gradient of the curve at the point where

*x*= –1.

(b) Find the coordinates of the turning points.

*Solution*:**(a)**

$\begin{array}{l}y={x}^{2}\left(x-3\right)+1\\ y={x}^{3}-3{x}^{2}+1\\ \frac{dy}{dx}=3{x}^{2}-6x\\ \text{When}x=-1\\ \frac{dy}{dx}=3{\left(-1\right)}^{2}-6\left(-1\right)\\ \text{}=9\\ \text{Gradientofthecurveis9}\text{.}\end{array}$

**(b)**

At turning points, $\frac{dy}{dx}=0$

3

*x*

^{2}– 6

*x*= 0

*x*

^{2}– 2

*x*= 0

*x*(

*x*– 2) = 0

*x*= 0, 2

*y*=

*x*

^{2}(

*x*– 3) + 1

When

*x*= 0, y = 1

When

*x*= 2,

*y*= 2

^{2}(2 – 3) + 1

*y*= 4 (–1) + 1 = –3

Therefore, coordinates of the turning points are

**(0, 1) and (2, –3)**.

**Question 3:**It is given the equation of the curve is

*y*= 2

*x*(1 –

*x*)

^{4}and the curve pass through

*T*(2, 4).

Find

(a) the gradient of the curve at point

*T*.

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

*T*.

*Solution*:**(a)**

$\begin{array}{l}y=2x{\left(1-x\right)}^{4}\\ \frac{dy}{dx}=2x\times 4{\left(1-x\right)}^{3}\left(-1\right)+{\left(1-x\right)}^{4}\times 2\\ \text{}=-8x{\left(1-x\right)}^{3}+2{\left(1-x\right)}^{4}\\ \\ \text{At}T\left(2,4\right),x=2.\\ \frac{dy}{dx}=-16\left(-1\right)+2\left(1\right)\\ \text{}=16+2\\ \text{}=18\end{array}$

**(b)**

$\begin{array}{l}\text{Equationofnormal:}\\ y-{y}_{1}=-\frac{1}{\frac{dy}{dx}}\left(x-{x}_{1}\right)\\ y-4=-\frac{1}{18}\left(x-2\right)\\ 18y-72=-x+2\\ x+18y=74\end{array}$