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Mathematics 10 Online
OpenStudy (anonymous):

How do I find if the graph lies above or below an oblique asymptote?

OpenStudy (anonymous):

plot a point!!

OpenStudy (anonymous):

how though?

OpenStudy (anonymous):

? pick an x find the y plot the point see if it above or below the line

OpenStudy (anonymous):

you got a specific example?

OpenStudy (anonymous):

but how would i know if its above or below the line?

OpenStudy (anonymous):

look!

OpenStudy (anonymous):

yeah, (x^2+1)/x

OpenStudy (anonymous):

but what if i am required to sketch it?

OpenStudy (anonymous):

suppose you find \((2,3)\) on the graph of your function and the line is \(y=2x\) then on the line you would have \((2,4)\) which is evidently above \((2,3)\) so the curve would be below the line

OpenStudy (anonymous):

ohh... okay. So we are comparing the vertical distances of the two points, right?

OpenStudy (anonymous):

hold the phone your function is \(\frac{x^2+1}{x}=x+\frac{1}{x}\) right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so the "oblique asymptote is \(y=x\)

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

then it should be pretty clear without plotting a point that for \(x>0\) that \(x+\frac{1}{x}>x\) right?

OpenStudy (anonymous):

and similarly for \(x<0\) you would have \(x+\frac{1}{x}<x\)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so to the right of the vertical asymptote at \(x=0\) the curve lies above the asymptote, and to the left of \(x=0\) it lies below the asymptote

OpenStudy (anonymous):

i mean of course above the oblique asymptote

OpenStudy (anonymous):

got it!! thank you so much :)

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