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Mathematics 8 Online
OpenStudy (anonymous):

another graph the relation:

OpenStudy (anonymous):

\[y=\frac{ x(x+2) }{ x+3 }\]

OpenStudy (anonymous):

intercepts: x= 0 y=0 y=0 x=>x^2+2=0 --> x^2= -2 then?

OpenStudy (anonymous):

@waterineyes , @ganeshie8 @Tushara

OpenStudy (anonymous):

vertical asymptote where the denominator is 0, i.e. at \(x=3\)

OpenStudy (anonymous):

oops i meant \(x=-3\)

OpenStudy (cwrw238):

y=0 0 = x(x + 2)

OpenStudy (anonymous):

\(x(x+2)=0\implies x=0\) or \(x=-2\)

OpenStudy (anonymous):

slant asymptote as well

OpenStudy (anonymous):

since the numerator degree is 1 greater than the denominator so i divide N/D right?

OpenStudy (anonymous):

yes divide, get a degree one polynomial (aka a line) and a remainder. the line is the slant asymptote, ignore the remainder

OpenStudy (anonymous):

okay :) thank you!

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

@satellite73 the S.A is x-1?

OpenStudy (anonymous):

\(y=x-1\) yes you have it

OpenStudy (anonymous):

okay :)

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