Graph the following showing all working

4X^3/((X-4)^2(X+2))

The main problem i'm having with this one is finding the stationary point and the assymptotes. any help appreciated.

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- Aug 13th 2007, 11:34 PMmathsBHere's another (harder) one
Graph the following showing all working

4X^3/((X-4)^2(X+2))

The main problem i'm having with this one is finding the stationary point and the assymptotes. any help appreciated. - Aug 14th 2007, 05:14 AMearboth
Hello,

. (the**black**curve)

1. Domain: . Thus the vertical asymptotes are: . (the blue lines)

2. Horizontal asymptote:

. Thus the horizontal asymtote is: . (the red line)

3. Zeros:

f(x) = 0 if 4x³ = 0 that means there is one zero at x = 0.

4. Stationary points. Calculate the 1st drivative of f. Use quotient rule:

=

f'(x) = 0 if the numerator equals zero. Thus you get 2 stationary points at x = 0 or x = -4. Minimum at x = -4: f(-4) = 2

5. Points of inflection:

Calculate the 2nd drivative of f. (You'll need at least half an hour :D ...)

f''(x) = 0 if the numerator equals zero. First you get x = 0 that means at x = 0 is**no**stationary point but a point of inflection with a horizontal slope. (Such a point is called in German a terrace point).

The 2nd solution you'll get if

(I used a computer to get this solution)

6. Graph (see attachment): Be careful the axes are scaled differently! - Aug 14th 2007, 09:50 PMmathsB
Thanx, Gr8 help :):)