Sketching

[clıppy]

Hi guys,

Would just like to confirm something.

If anybody has the time or motivation, would they be kind enough to please post up full working out and the sketch of the graph for y=-2sec(2x-pi/3)+1

It is quite urgent and I would be forever grateful.


Thanks.

[Special At Specialist]

You could do the transformations method, but I find that very confusing and annoying.

I'll show you my way which works for every single graph. It may be long and confusing at first, but once you get the hang of it it is actually a lot faster than you think.
First find the asymptotes:
y = -2sec(2x - pi/3) + 1
y = -2 / cos(2x - pi/3) + 1
Asymptotes at cos(2x - pi/3) = 0
2x - pi/3 = pi/2, 3pi/2, 5pi/2, 7pi/2 etc.
2x = 5pi/6, 11pi/6, 17pi/6, 23pi/6 etc.
x = 5pi/12, 11pi/12, 17pi/12, 23pi/12 etc.
So draw vertical dotted lines at those values of x.

Now you want to find intercepts. We'll start with y-intercept, since that is the easiest.
y int, let x = 0
y = -2 / cos(- pi/3) + 1
y = -2 / (1/2) + 1
y = -4 + 1
y = -3
(0, -3)

x int, let y = 0
0 = -2 / cos(2x - pi/3) + 1
-2 / cos(2x - pi/3) = -1
cos(2x - pi/3) / -2 = -1
cos(2x - pi/3) = 2
since -1 ≤ cos(u) ≤ 1 then there is no solution.
No x-intercepts.

Now to find the general shape of the graph, find what it looks like as it approaches an asymptote. Ignore the 1, since that is negligible. Also, pretend the -2 is -1 to simplify things.
y = -1 / cos(2x - pi/3)
As x --> 5pi/12 + (from the positive side)
y --> -1 / cos(5pi/6 - pi/3 + ?)
y --> -1 / cos(pi/2 + ?)
y --> -1 / (very small negative number)
y --> + infinity

Try as x --> 5pi/12 - (from the negative side)
y --> -1 / cos(5pi/6 - pi/3 + ?)
y --> -1 / (pi/2 - ?)
y --> -1 / (very small positive number)
y --> - infinity

As x --> 11pi/12 + (from the positive side)
y --> -1 / cos(11pi/6 - pi/3 + ?)
y --> -1 / cos(3pi/2 + ?)
y --> -1 / - cos(pi/2 + ?)
y --> 1 / cos(pi/2 + ?)
y --> 1 / (very small negative number)
y --> - infinity

As x --> 11pi/12 - (from the negative side)
y --> -1 / cos(11pi/6 - pi/3 - ?)
y --> 1 / cos(pi/2 + ?)
y --> 1 / (very small positive number)
y --> + infinity

We can predict that the pattern will repeat. At one asymptote it is negative on the left, positive on the right. At the other it is positive on the left and negative on the right. And it will keep repeating like that.
We can also predict that there will be a turning point between every intercept and that that turning point will be before it crosses an axis. To find the turning point:

y = -2 / cos(2x - pi/3) + 1
y = -2(cos(2x - pi/3))^(-1) + 1
Max/min at dy/dx = 0
d/dx (cos(2x - pi/3)) = -2sin(2x - pi/3)
d/du (-2(u)^(-1)) = -2*-1*u^(-2) = 2u^(-2)
Therefore dy/dx = -4sin(2x - pi/3) / (cos(2x - pi/3))^2 = 0
For that to equal 0, the numerator must equal 0.
sin(2x - pi/3) = 0
2x - pi/3 = 0, pi, 2pi, 3pi, 4pi etc.
x - pi/6 = 0, pi/2, pi, 3pi/2, 2pi etc.
x = 2pi/3, 7pi/6, 5pi/3, 13pi/6 etc.

When x equals those numbers:
y(2pi/3) = -2 / cos(4pi/3 - pi/3) + 1
y(2pi/3) = -2 / cos(pi) + 1
y(2pi/3) = -2 / -1 + 1
y(2pi/3) = 3
(2pi/3, 3) is a turning point.

y(7pi/6) = -2 / cos(7pi/3 - pi/3) + 1
y(7pi/6) = -2 / cos(2pi) + 1
y(7pi/6) = -2 + 1
y(7pi/6) = -1
(7pi/6, -1)

This pattern will repeat. The y value will go to +3 then to -1 then +3 again and so on. So label those points. Then just graph it and you're done.

This is what the graph should look like:
http://img845.imageshack.us/img845/6341/mathsgraph.png