# Thread: Stuck with cal graph

1. ## Stuck with cal graph

Given $h(x)=x^3-3x^2-9x-5$
$=(x+1)^2(x-5)$
1 Calculate the co-ords of the $x$ and $y$ intercepts of the curve of $h$
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
$y=x^3-3x^2-9x+5$

Some other question:
$p(x)=x^3-5x^2-8x+12$ at $(-1:41)$

1 $x=-1 x=5$
$y=-5$

2 $\frac{dy}{dx}= 3x^2-6x-9$
$3(x^2-2x-3) = 0$
$(x-3)(x-1) = 0$
$x=3$ $x=1$
$f(3) = (3)^3 -(3)^2 -9(3) -5 = -32$
$f(-1) = (-1)^3 -3(-1)^2-9(-1)-5 = 0$

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

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

Thanks guys xD

2. Originally Posted by NeF
Given $h(x)=x^3-3x^2-9x-5$
$=(x+1)^2(x-5)$
1 Calculate the co-ords of the $x$ and $y$ intercepts of the curve of $h$
The y-intercept is when $x=0$ thus,
$(0+1)^2(0-5)=1^2(-5)=-5$
The x-intercept is when $y=0$ thus,
$(x+1)^2(x-5)^2=0$ thus,
$x=-1,5$

2 Find the co-ords of the turning point(s) of the curve
Find derivative, (product rule)
$y'=[(x+1)^2]'(x-5)+(x+1)^2(x-5)'$
$y'=2(x+1)(x-5)+(x+1)^2$
$y'=(x+1)[2(x-5)+(x+1)]$
$y'=(x+1)(3x-9)$
$y'=3(x+1)(x-3)$
Set $y'=0$ thus,
$3(x+1)(x-3)=0$ thus,
$x=-1,3$
Now,
$f(-1)=0,f(3)=-32$
Thus, the turning points are on,
$(-1,0),(3,-32)$
3 Use the info above to draw a graph
You should be able to do that now.
4 Use the graph to determine the co-ords of the turning points of the curve of
$y=x^3-3x^2-9x+5$
This curve is the same as the other curve in the sense that it is shifted up by 10 units.

3. Originally Posted by NeF
Given $h(x)=x^3-3x^2-9x-5$
$=(x+1)^2(x-5)$
1 Calculate the co-ords of the $x$ and $y$ intercepts of the curve of $h$
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
$y=x^3-3x^2-9x+5$

Some other question:
$p(x)=x^3-5x^2-8x+12$ at $(-1:41)$

1 $x=-1 x=5$
$y=-5$

2 $\frac{dy}{dx}= 3x^2-6x-9$
$3(x^2-2x-3) = 0$
$(x-3)(x-1) = 0$
$x=3$ $x=1$
$f(3) = (3)^3 -(3)^2 -9(3) -5 = -32$
$f(-1) = (-1)^3 -3(-1)^2-9(-1)-5 = 0$

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

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

Thanks guys xD
A technical point... You might need this depending on how picky your Math professor is. The x and y intercepts are actually points, so really the x intercepts are (-1, 0) and (5, 0) and the y intercept is (0, -5). Normally professors don't care, but you do run into the occasional one that does.

-Dan

4. Hectic thanks guys. One more question though, I post my questions in pre-calculus but it always ends up in the uni calculus section.