The asymptotes are $x=\pm1$ and $y=2$. The curve intersects the horizontal asymptote at $x=7$. It intersects with the y-axis at $y=-5$. It has two turning points at $x \approx 0.1$ and $13.9$. Here is the sketch I made. However, plotting the graph with a graph plotter does not show the turning point on the right-most branch. Is this because I haven't 'zoomed enough', or is it something else?


  • $\begingroup$ You state that a turning point is at $x=13.9$, but you drew the minimum at $x=7$? Also, what range are you looking at for the plotter? Is this an online plotter (if so, can you provide a link)? $\endgroup$ – Michael Burr May 21 '17 at 21:55
  • $\begingroup$ Sorry, mistake on my part. I'll fix that. And I just googled the equation of my curve. The range can be extended by moving your mouse to the sides. $\endgroup$ – user440261 May 21 '17 at 21:57
  • $\begingroup$ Plot your function for $7 <x <15$ and 0 <y <2$. $\endgroup$ – hamam_Abdallah May 21 '17 at 21:59
  • 1
    $\begingroup$ Looking at the google plot, it looks like the minimum is just really, really flat. $\endgroup$ – Michael Burr May 21 '17 at 22:01
  • $\begingroup$ Even if I zoom in, I don't see anything there.. $\endgroup$ – user440261 May 21 '17 at 22:01

Putting this into Wolfy, it looks like your sketch is correct.

That min on the right is at $7+4\sqrt{3} \approx 13.928 $, with a value of $2\sqrt{3}-\frac32 \approx 1.9641 $ so you may not have gone out far enough.

  • $\begingroup$ Could you link me to the plot? Google doesn't show the turning point.. $\endgroup$ – user440261 May 21 '17 at 22:01
  • $\begingroup$ Go to wolframalpha.com/input/?i=(2x%5E2+-+x+%2B5+)+%2F(+x%5E2+-1+) $\endgroup$ – marty cohen May 21 '17 at 22:03
  • $\begingroup$ Oh, I think I see it. Thanks. $\endgroup$ – user440261 May 21 '17 at 22:07

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