# Thread: Co-ordinates and a point C

1. ## Co-ordinates and a point C

A question on my practice exam paper says:

The curve C has equation y = x³ - 3x² - 9x + 2. Find the coordinates of the stationary points of C and determine the nature of these points.

I worked part out as:

dy/dx = 3x² - 6x - 9 = 0
(/3) = x² - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3, x = -1

When x=3:
(3)³ - 3(3)² - 9(3) + 2
= 27 - 27 - 27 + 2
= -25

When x=-1:
(-1)³ - 3(-1)² - 9(-1) + 2
= -1 - 3 + 9 + 2
= 7

Coordinates are (3,-25),(-1,7).

Is this right so far and what do I need to do to complete the rest?

2. Originally Posted by db5vry
A question on my practice exam paper says:

The curve C has equation y = 3x³ - 3x² - 9x + 2. Find the coordinates of the stationary points of C and determine the nature of these points.

I worked part out as:

dy/dx = 3x² - 6x - 9 = 0
(/3) = x² - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3, x = -1

When x=3:
(3)³ - 3(3)² - 9(3) + 2
= 27 - 27 - 27 + 2
= -25

When x=-1:
(-1)³ - 3(-1)² - 9(-1) + 2
= -1 - 3 + 9 + 2
= 7

Coordinates are (3,-25),(-1,7).

Is this right so far and what do I need to do to complete the rest?
use the 2nd derivative test to determine if the points are maximums or minimums.

3. Originally Posted by skeeter
use the 2nd derivative test to determine if the points are maximums or minimums.
So if I had 3x² - 6x - 9 I could use:

d²y / dx² - which would be 6x - 6

then substituting in the values, I would get:

for x=3 - 18-6=12, so d²y / dx² < 0, meaning it is a maximum point

and for x=-1, -6-6=-12, so d²y / dx² >0, meaning it is a minimum point?

4. you typed them in backwards ...

for x=3, 18-6=12, so d²y / dx² < 0, meaning it is a maximum point

y'' > 0, not less

and for x=-1, -6-6=-12, so d²y / dx² > 0, meaning it is a minimum point?

y'' < 0, not greater