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.

*Mathematical errors? Message me.*