# Thread: graph sketching qn

1. ## graph sketching qn

I shifted this qn to the uni section, apparently its too challenging for pre-uni lvl. . Hope you guys can enlighten me.

The function g is defined by $g(x)=\dfrac {1}{(x-5)(x+3)}$, 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:

2. 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}{(x-5)(x%2B3)}

[And this is a pre-university question]

3. my graphs from graphmatica are correct.
the black one is y=(x-5)(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 x-axis with asymptotes at x=3 and 5.

How do i get the L-shapes? How do I get the range of the function g?

[Pre-uni do not have calculus, hence I post it here]

4. Pre-uni 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 ' $\cap$' 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.

5. 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 pre-uni]

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

7. 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 x-axis, we sub x= 6 and x=-4. Since the respective y values are positive. The Ls are above the x-axis.

Is that the right way of reasoning?

8. Yes, you can think of this like that