reasonable question, solveable

[mozart]

1) The speed of a current in a river is 5km/h. A boat traveled 10km upstream and 10km down stream for a total of 6 hours. ehat is the speed of hte boat in still water, to the nearest tenth of a km/h



2)Graph each function: Indicate the equation of any vertical and/ or horizontal asymptotes

a)3x/x-1

b)1/x^2-2x

[cara.mel]

1) Let the speed of the boat be v, time it takes to go upstream = t
Also, speed = distance / time

Going upstream:

Going downstream: (the 6-t is there because the two times add up to 6, so the second time is = 6h - first time)

And then you have your 2 simultaneous equations and can go from there, which unless someone else suggests something better, I'd say ask your calculator , especially given the invitation of 'to the nearest tenth of a km/h)
=> v=6.9km/hr

2 - for both of these I am assuming they are 1 fraction. I also don't know how to make graphs on the computer but you can ask your calculator that 0=)

a)long divide it etc to get
From there you can see it is the hyperbola graph dilated 3 and translated 1 to the right and 3 upwards
=> asymptotes are y=3 and x=1

b)You'd draw x^2-2x and then reciprocate it
Vertical asymptotes are when x^2-2x = 0 => at x=0 and x=2
Also it has a horizontal asymptote y=0 as because as x^2-2x gets bigger and bigger either way, 1/x^2-2x approaches 0

[Mao]

2 b)
factorising gives us
hence the vertical asymptotes are x=0 and x=2
turning point is midway between x=0 and x=2, at point (1,-1)

since the graph is positive, from left to right:
y=0 @ x=
y approaches as x approaches 0 from negative
y= immediately to the right of x=0
maximum turning point at (1,-1), then y approaches as x approaches 2
y= immediately to the right of x=2
y approaches 0 as x approaches