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another graph the relation:
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\[y=\frac{ x(x+2) }{ x+3 }\]
intercepts: x= 0 y=0 y=0 x=>x^2+2=0 --> x^2= -2 then?
@waterineyes , @ganeshie8 @Tushara
vertical asymptote where the denominator is 0, i.e. at \(x=3\)
oops i meant \(x=-3\)
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y=0 0 = x(x + 2)
\(x(x+2)=0\implies x=0\) or \(x=-2\)
slant asymptote as well
since the numerator degree is 1 greater than the denominator so i divide N/D right?
yes divide, get a degree one polynomial (aka a line) and a remainder. the line is the slant asymptote, ignore the remainder
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okay :) thank you!
yw
@satellite73 the S.A is x-1?
\(y=x-1\) yes you have it
okay :)
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