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September 3, 2014

September 3, 2014

Posted by **Muffy** on Monday, October 26, 2009 at 11:09pm.

a. vertical asymptote at x=3

b. x-intercept of 1 and 5

c. f(2)=f(4)=f(6)=3

d. increasing on [0,3) and [5,infinity)

e. decreasing on (3,5]

f. f(x) is an odd function

I don't know how to graph this. It ends up looking like a triangle with a line down it.

I am unfamiliar with this type of graph since I have never graphed with only a vertical asymptote and not a horizontal.

- pre calculus -
**jim**, Tuesday, October 27, 2009 at 8:38amThink of it as two curves to draw.

One starts at x just greater than 0, somewhere in negative y, passes through (1,0) and (2,3) and off to infinity at x=3.

The other comes down from infinity at x=3, passing through (4,3), and just touches the x-axis at a turning point (5,0) before heading off to infinity again through (6,3).

What I can't see from the given description is how far into negative y it goes near x=0. It definitely must "start" negative, since it increases on [0,3) and passes through (1,0), but we don't know.

And just when you thought it was safe to put down your pencil :-) -- it's an odd function, so you have to draw it all again upside down on negative x, since it has 180-degree rotational symmetry around the origin!

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