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]
September 11th 2010, 11:13 AM
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 ' ' 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.
September 11th 2010, 11:21 AM
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]
September 11th 2010, 11:27 AM
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.
September 11th 2010, 11:45 AM
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. :)