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Trials question about graphs (1 Viewer)
- Thread starter BlueGas
- Start date
kawaiipotato
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- Joined
- Apr 28, 2015
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- HSC
- 2015
Notice that there are asynptotes at x = 3 and y = 2. For vertical asymptotes (x=3) you need to find restrictions on the DOMAIN of the function. In a fraction the denominator cannot equal zero.
And from the graph,
x =/= 3
x - 3 =/= 0
So the denominator has to be x-3 which rules out C and D.
Then for horizontal asymptotes we assess what happens when limit x approaches +/- infinity. When x approaches +/- infinity, the graph tends to y = 2.
This indicates that it must be "+2" since 1/(anything) cannot equal zero.
Which means 1/(anything) + 2 cannot equal 2
Which is our horizontal asymptote
And from the graph,
x =/= 3
x - 3 =/= 0
So the denominator has to be x-3 which rules out C and D.
Then for horizontal asymptotes we assess what happens when limit x approaches +/- infinity. When x approaches +/- infinity, the graph tends to y = 2.
This indicates that it must be "+2" since 1/(anything) cannot equal zero.
Which means 1/(anything) + 2 cannot equal 2
Which is our horizontal asymptote