To sketch the curve
f(x) = 2x+5
First, we find the incercept
When x = 0; f(x) = 5/2 (y-intercept)
When f(x) = 0; 0 = 2x+5
=> 0 = 2x+5
=> x = -5/2
5/2 is the y-intercept.
-5/2 is the x-intercept.
Secondly, we find the turning point
f(x) = (2x+5)(x+2)^-1
Find the derivative
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.
, there are no turning points for this curve.
Next, we find the asymptote
s 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
. | 5/2 . = asymptote lines
. o o = worked out incercepts
. | 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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