Originally Posted by

**NeF** Given $\displaystyle h(x)=x^3-3x^2-9x-5 $

$\displaystyle =(x+1)^2(x-5) $

1 Calculate the co-ords of the $\displaystyle x$ and $\displaystyle y$ intercepts of the curve of $\displaystyle h$

2 Find the co-ords of the turning point(s) of the curve

3 Use the info above to draw a graph

4 Use the graph to determine the co-ords of the turning points of the curve of

$\displaystyle y=x^3-3x^2-9x+5$

Some other question:

$\displaystyle p(x)=x^3-5x^2-8x+12$ at $\displaystyle (-1:41)$

My answers:

1 $\displaystyle x=-1 x=5 $

$\displaystyle y=-5 $

2 $\displaystyle \frac{dy}{dx}= 3x^2-6x-9 $

$\displaystyle 3(x^2-2x-3) = 0 $

$\displaystyle (x-3)(x-1) = 0 $

$\displaystyle x=3 $ $\displaystyle x=1 $

$\displaystyle f(3) = (3)^3 -(3)^2 -9(3) -5 = -32 $

$\displaystyle f(-1) = (-1)^3 -3(-1)^2-9(-1)-5 = 0 $

$\displaystyle (3;-32) (1;0) $

3 I got a wierd one, want to see what u guys get.

Thanks guys xD