how do i find the co - ordinates to sketch the curve y = x^3 - 3x - 1 ?

Answer 1

First of all, you differentiate the equation to obtain the stationary points. So :

f'(x) = 3(x^2) -3 = 0
3[ (x^2) -1] =0
(x^2) -1 = 0
(x-1)(x+1) = 0
so x= -1 , 1 are the x-coordinates of the max and the min, but which is which?
So you do a 2nd derivative, and substitute the answers above. If the answer is negative, then the x you substituted belongs to the max, and if positive, it belong to the min.

f"(x) = 6x
x= 1, 6x = 6 which is the x-coordinate of the min.
the y-coordinate is: y = (1^3) -3(1) -1 = 1-3-1 = -3
so the min has the coordinates (1 , -3)
Now we know that x = -1 is for the max., and the y-coordinate for this point is y= ( -1^3) -3(-1) -1 = -1 +3 -1 = 1
so the coordinates of the max is ( -1 , 1).

the y-intercept is easy to find by substituting x=0
so y= 0 -3*0 -1 = -1

Now make the shape of a capital n (N) which has a max point
( -1,1), and a min ( 1, -3), and a y-intercept of -1

Answer 2

Hi Tikki,
This is how its done: First of all find the d/dx to find the min and max of the function. Then find the second derivative in order to work out the saddle points. Min and max are +1 and -1 respectively and the saddle is at x=0. This will give you an idea as to how the graph should look. Remember to plug in these values in order to get representative points on the graph. Also, take note of Amber's suggestion which is the most practical way of going about it....
As well as that Desolator has only gone to the trouble of actually explaining the full calculation as laid out by me, so pay heed to him, well done Deolator. And Tom, such a question asks for a plot, why in the name of god would you want to go to the trouble of using the Newton-Raphson mathod. If in an exam, time would crucify u.....I would admit though that the laymans ways of looking for the min and max is acceptable here and should always be employed whenever the second derivative gives x=0...

Answer 3

Do what Harry_H says and then test a point on either side of x=-1 and x = 1. You will find there is a local maximum at x=-1 and a local minimum at x=1. The curve passes through the origin but there are two more x-intercepts. To find these you can use Newton's approximation formula:

x_2 = x_1 - f(x_1)/f'(x_1)
x_3 = x_2 - f(x_2)/f'(x_2)
... etc
when the difference is small between the x_n, then you are close to the root. So, to find the negative root, set x_1 = -3 and for the positive root, set x_1 = 3 and it should converge to the x-intercept. It appears this graph is symmetrical about the y-axis, therefore you may only need to find one root and then you'll know the other.

The curve will appear similar to the sine wave in shape.

Answer 4

y = x^3 ? 3x + a million y' = 3x^2 ? 3 At factor (2, 3) y' = 3(2)^2 ? 3 = 9 Tangent line has slope 9, regular has slope ?a million/9 Tangent line: y ? 3 = 9 (x ? 2) y = 9x ? 18 + 3 y = 9x ? 15 regular: y ? 3 = ?a million/9 (x ? 2) y = ?a million/9 x + 2/9 + 3 y = ?a million/9 x + 29/9

Answer 5

try a graphin calculator, type the formula in and then go to table

if not:

do

b/2a = x and plug that in to the equation, thats your midpoint sort of and then do a couple x's above and below that number (-1.5) and solve them for y in the equation too

hope this helps!

Answer 6

if you want to see the graph try the following link (first picture)

http://qwpey.com/mathindex/gl/tra.php