
1 Attachment(s)
graph sketching qn
I shifted this qn to the uni section, apparently its too challenging for preuni lvl. (Giggle). Hope you guys can enlighten me.
The function g is defined by $\displaystyle g(x)=\dfrac {1}{(x5)(x+3)}$, x not equal to 3 and 5.
i) Find the coordinates of the turning point of the curve y=g(x).
ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.
PS: Using calculus, I found the turning point to be (1, 1/16).
Extra stuff:Attachment 18889

Yes, the turning point is correct.
Your graph however is not... what is the actual graph...
Put x = 2. Then, y = 1/15, which is lower than the turning point. This means that the curve goes down from the turning point.
And this being an 'inverted' square function (ie like y = 1/x^2), you have some sort of 'volcano' shape.
See here:
http://www.wolframalpha.com/input/?i=+y%3D+\frac{1}{(x5)(x%2B3)}
[And this is a preuniversity question]

my graphs from graphmatica are correct.
the black one is y=(x5)(x+3)
the pink one is the reciprocal.
I can determine that the turning point is a maximum point by finding the dy/dx of the points before and after x=1.
since x not equal to 3 and 5, i learnt to sketch the 'n' shape below the xaxis with asymptotes at x=3 and 5.
How do i get the Lshapes? How do I get the range of the function g?
[Preuni do not have calculus, hence I post it here]

Preuni does have... I don't know whether there is a section for that though.
You know that x cannot be equal to 3 or 5, because otherwise, you get y = infinity, hence the ever going up of the 'L' shapes and the ever going down '$\displaystyle \cap$' shape.
As x goes to ve infinity, y tends to zero.
As x tends to +ve infinity, y tends to zero again.
In your diagram, maybe it's because I didn't zoom in... and didn't see the other graph. Sorry.

is there a horizontal asymptote at y=0? if so, why?
How do you even know that there are 2 "L"s at those particular positions?
[there isn't a calculus section at preuni]

Yes, because the curve never meets the xaxis. You can try putting y = 0, and solve for x, but you will end up with imaginary roots, or no solution if you prefer.

Since there is a horizontal asymptote at y=0 and vertical asymptotes at x=3 and 5, there are 2 L shapes.
To find out if the Ls are above of below the xaxis, we sub x= 6 and x=4. Since the respective y values are positive. The Ls are above the xaxis. :)
Is that the right way of reasoning?

Yes, you can think of this like that (Happy)