# Thread: Here's another (harder) one

1. ## Here'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.

2. Originally Posted by mathsB
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.
Hello,

$f(x)=\frac{4x^3}{(x-4)^2 \cdot (x+2)}$. (the black curve)

1. Domain: $D = \mathbb{R} \setminus \{-2, 4\}$. Thus the vertical asymptotes are: $\boxed{x = 4~ or~ x = -2}$ . (the blue lines)

2. Horizontal asymptote:

$\lim_{|x| \rightarrow \infty}f(x) = \lim_{|x| \rightarrow \infty}\left(\frac{4x^3}{(x-4)^2 \cdot (x+2)} \right) = \lim_{|x| \rightarrow \infty}\left(\frac{4x^3}{x^3-6x^2+32}\right) = 4$. Thus the horizontal asymtote is: $\boxed{y = 4}$ . (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:

$f'(x) = \frac{(x-4)^2 \cdot (x+2) \cdot 12x^2 - 4x^3\cdot((x-4)^2+(x+2)\cdot 2 \cdot (x-4))}{\left( (x-4)^2 \cdot (x+2)\right)^2}$ = $\frac{-24x^2(x+4)}{(x-4)^3 \cdot (x+2)^2}$

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 ...)

$f''(x) = \frac{48x(x^3+8x^2+16x+32)}{(x-4)^4 \cdot (x+2)^3}$

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
$x^3+8x^2+16x+32 = 0 ~\Longrightarrow~x\approx -6.26079...$ (I used a computer to get this solution)

6. Graph (see attachment): Be careful the axes are scaled differently!

3. Thanx, Gr8 help