Sketching Asymptotes (idea)

To sketch the curve:

f(x) = 2x+5
       ----
        x+2

First, we find the incercept/s:

    When x = 0; f(x) = 5/2 (y-intercept)

 When f(x) = 0; 0 = 2x+5
                    ----
                     x+2

             => 0 = 2x+5
             => x = -5/2

5/2 is the y-intercept.
-5/2 is the x-intercept.

Secondly, we find the turning points:

f(x) = (2x+5)(x+2)^-1

Find the derivative of f(x)

f'(x) = 2(x+2)^-1+(2x+5).1(x+2)^-2.1
	2/(x+2) + (2x+5)/(x+2)^2

Then, we let f'(x) = 0

 0 = (2(x+2)-(2x+5))/(x+2)^2
 x = Undefined.

Therefore, there are no turning points for this curve.

Next, we find the asymptotes of the curve. By looking at the formula for the curve, we work out what the values of x approaching certain points would be.

For this curve, it would look something like this:

x -> -2 (from below) ; f(x) -> -inf
x -> -2 (from above) ; f(x) -> -inf
x -> -inf ; f(x) -> +2
x -> inf ; f(x) -> +2

From this data, we can now draw our graph:

        1  y
        .  |
        .  |
        .  | 5/2                   . = asymptote lines
        .  o                       o = worked out incercepts
...........|.............2
        .  |
        .  |
-------o---+------------ x
    -5/2.  |
        .  |
        .  |     It's difficult to show asymptotes in ASCII, therefore I'll just explain how
        .  |     the graph goes. The first line comes from the top of line 1 (reaching it at 
        .  |     y=infinity), curves through the y-incercept and goes along line 2 (reaching
        .  |     it at x=infinity).
                 The second asymptote follows a similar journey, but from line 2, through the
                 y-intercept, reaching line 1 at y = -infinity.



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