How do I find if the graph lies above or below an oblique asymptote?
plot a point!!
how though?
? pick an x find the y plot the point see if it above or below the line
you got a specific example?
but how would i know if its above or below the line?
look!
yeah, (x^2+1)/x
but what if i am required to sketch it?
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
ohh... okay. So we are comparing the vertical distances of the two points, right?
hold the phone your function is \(\frac{x^2+1}{x}=x+\frac{1}{x}\) right?
yes
so the "oblique asymptote is \(y=x\)
yes
then it should be pretty clear without plotting a point that for \(x>0\) that \(x+\frac{1}{x}>x\) right?
and similarly for \(x<0\) you would have \(x+\frac{1}{x}<x\)
ok
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
i mean of course above the oblique asymptote
got it!! thank you so much :)
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