# Differentiation (Paper 2), Question 2 & 3

Question 2:
Given the equation of a curve is:
y = x2 (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)

(b)
At turning points, $\frac{dy}{dx}=0$
3x2 – 6x = 0
x2 – 2x = 0
x (x – 2) = 0
x = 0, 2

y = x2 (x – 3) + 1
When x = 0, y = 1
When x = 2,
y = 22 (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 = 2x (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)

(b)