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anonymous

  • one year ago

Help anyone???? find the asymptotes, if any, of the graph of the rational function f(x) = x/x^2 + 1

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  1. campbell_st
    • one year ago
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    if the curve is \[y = \frac{x}{x^2 + 1}\] well as for the vertical asymptotes: there are no real numbers that make the denominator zero. so the curve is continuous you can find the 1st and 2nd derivatives and as a result can find a maximum value, a minimum value and point of inflection. the only thing of not is that as x approaches infinity the curve approaches zero from above. similar for negative infinity, the curve approaches zero from below. To get a good understanding graph the curve using https://www.desmos.com/calculator if the curve is \[y = \frac{x}{x^2} + 1\] then there will be a vertical and horizontal asymptote... and to see them, just graph the curves

  2. anonymous
    • one year ago
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    if the graph is (x/x^2) + 1 ==> (x+1)/x You have a horizontal asymptote at y = 1 because your top degree, of numerator is 1 and degree of denominator is 1, too thus, if your top and bottom degrees are the same, Horizontal asymptote is the ratio of the co-efficients in this case 1 over 1 for the V.asymptote, equal the denominator to zero x=0 thus V.A at x = 0

  3. anonymous
    • one year ago
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    Thanks y'all. Sorry it took so long to reply, but it helped. I got it right!

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