# Thread: Finding T.P using differentiation

1. ## Finding T.P using differentiation

Hi
I am having trouble finding the turning point and POI for this function:
$\displaystyle \frac{3x}{(x+8)^2}$

i know that the asymptote is x=-8 and y=0.

x an y intercepts are 0.

this is what i get for the first derivative: $\displaystyle \frac{-3x^2+96x+192}{(x+8)^4}$
This is where i am stuck.

P.S

2. Originally Posted by Paymemoney
Hi
I am having trouble finding the turning point and POI for this function:
$\displaystyle \frac{3x}{(x+8)^2}$

i know that the asymptote is x=-8 and y=0.

x an y intercepts are 0.

this is what i get for the first derivative: $\displaystyle \frac{-3x^2+96x+192}{(x+8)^4}$
This is where i am stuck.

P.S
Well to find the turning points, just solve for $\displaystyle \frac{dy}{dx} = 0$

and for the points of inflection just find the 2nd derivative of position i.e.
$\displaystyle \frac{d2y}{dx^{2}}$ and let that = 0 and solve for x. Then sub back in value(s) of x and find y

These formula's should be in your class notes... and i would hazard a guess that this is pre-uni calculus...?

3. i have found the turning point from the derivative $\displaystyle \frac{-3x+24}{(x+8)^3}$. values $\displaystyle (8, \frac{3}{32})$
However i cannot seen to get a Point of Inflection value.