# Why is this an incorrect sketch for the curve $(2x^2 - x +5 ) /( x^2 -1 )$

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?

https://i.stack.imgur.com/rZuKs.png

• 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)? – Michael Burr May 21 '17 at 21:55
• 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. – user440261 May 21 '17 at 21:57
• Plot your function for $7 <x <15$ and 0 <y <2$. – hamam_Abdallah May 21 '17 at 21:59 • Looking at the google plot, it looks like the minimum is just really, really flat. – Michael Burr May 21 '17 at 22:01 • Even if I zoom in, I don't see anything there.. – user440261 May 21 '17 at 22:01 ## 1 Answer 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.

• Could you link me to the plot? Google doesn't show the turning point.. – user440261 May 21 '17 at 22:01
• – marty cohen May 21 '17 at 22:03
• Oh, I think I see it. Thanks. – user440261 May 21 '17 at 22:07