pre calculus
posted by Muffy on .
Use the following conditions to create a sketch of the function by hand.
a. vertical asymptote at x=3
b. xintercept 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.

Think 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 xaxis 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 180degree rotational symmetry around the origin!