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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: mozart on January 12, 2008, 10:54:28 pm

Title: reasonable question, solveable
Post by: mozart on January 12, 2008, 10:54:28 pm: 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
Title: Re: reasonable question, solveable
Post by: cara.mel on January 13, 2008, 09:34:29 am: 1) Let the speed of the boat be v, time it takes to go upstream = t
Also, speed = distance / time

Going upstream: $v-5 = \frac{10}{t}$

Going downstream: $v+5 = \frac{10}{6-t}$ (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 $3+\frac{3}{x-1}$
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
Title: Re: reasonable question, solveable
Post by: Mao on January 13, 2008, 02:15:07 pm: 2 b)
factorising gives us $\frac{1}{x\cdot (x-2)}$
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= $-\infty$
y approaches $\infty$ as x approaches 0 from negative
y= $-\infty$ immediately to the right of x=0
maximum turning point at (1,-1), then y approaches $-\infty$ as x approaches 2
y= $\infty$ immediately to the right of x=2
y approaches 0 as x approaches $\infty$

(http://obsolete-chaos.wikispaces.com/space/showimage/FSN+MATHS+GRAPH.BMP)

:D