Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Ok, well, I have a function f(x)=2x^2/(x+2).

1* Df = R \ {-2}

2* x-axis intersection is: A(0,0) y-axis intersection is: B(0,0)

3* f(-x)=(2(-x)^2)/(-x+2) -> it's not even neither odd.

4* asymptotes.

horizontal = none vertical = 2 obliques = 2

5* extreme values The first derivative is: f'(x)=(2x(x+4))/(x+2)^2.

we can calculate it to f'(x)=(2x^2+8x)/(x+2)^2.

Then we do f'(x)=0

We find x1/2.

Then we find y1/2.

And find the points M1(x1,y1), M2(x2,y2)

M1(0,0) M2(-4,-16)

Now the second derivative of f'(x), f''(x) = 16/(x+2).

f''(0) > 0 - min (concave) f''(-4)< 0 - max (convex)

6* We don't have inflection points in this case, but if we have, we calculate it the following way? please check.

f''(x)=0 x=0?

We find the value of x to equal 0, then we use the value to check the function to find y, and find the point.

To find interval of concave, we basically check for f''(x)<0, and the interval is x(-infinity, value)

To find interval of convex, we check the f''(x)>0, the interval is x(infinity,value).

Now, once I have done all that, how do I write the graph? I am really confused.

I have made a small table if that helps somehow...

x -infinity -4 -2 0 +infinity
y -16 0
y' increasing 0(max) min increasing y'' convex convex convex concave concave

I also made a calculation for the oblique, y=2x-4

x -1 0 1 2 y -6 -4 -2 0

Any help writing the graph???

share|cite|improve this question
up vote 3 down vote accepted

It's a little hard to read all the details of your post, unformatted, but it's great to have included all the "leg-work" you put into this. I'll proceed with a description and a few corrections, plus a few graphing tips.

Note: there seems to be the one oblique asymptote. And there is a vertical asymptote at $x = -2$ (indeed, that's the equation of the vertical asymptote: the vertical line perpendicular to the x-axis and intersecting it at $x = -2$.

The range of $f(x)$ is $(-\infty, -16] \cup [0, \infty)$. And the function increases on $(-\infty, -4)$, reaches a local maximum at $(-4, -16),$ decreases from $x\in (-4, -2)$. As $x \to -2^-,\; f(x) \to -\infty$. And as $x \to -2^+,\; f(x) \to +\infty$. Then, $f(x)$ decreases on $(-2, 0)$, has a local minimum at $(0, 0)$, and increases on $(0, +\infty)$. So there are two local extrema.

The graph below will help to show some of the behavior of the function, and give you some intuition about how to proceed graphing the function given the asymptotic information you have (draw those lines first, as they "cradle" or provide bounds for the two parts of the curve). Place the points where $f'(x) = 0$ on the graph (along with plotting a few points evaluated at strategic values close to the local maximums and minimums, and to give guide posts as you'll want your curve/function to intersect those, and again, they'll help layout the "shape" of the curve.. Then, use the increasing, decreasing, concave/convex information you have to fill in the curve.

$\quad f(x) = \dfrac{2x^2}{x+2}$

enter image description here

And here's a link to Wolfram Alpha's two plots: one of which is a close-up of the curve surrounding the local minimum $(0,0)$.

Can you see why the asymptotes exist where they do? Can you confirm where the curve is concave and convex? Do you see the local maximum and minimum? Do you see how the asymptotes provide a framework in which to situate/sketch your curve, or put differently how they separate the two sections of the curve?

share|cite|improve this answer
Thanks for the help to flash. – Babak S. Mar 7 '13 at 6:01
My pleasure @Babak :-) I'll keep my eyes out, too, for other future potential questions from flash. – amWhy Mar 7 '13 at 6:04
The bottom one I believe is convex and the upper one is concave. The local min/max are the points where the functions change their direction, where they are very very close to the line. As for the horizontal/vertical asymptote I don't understand it. (The graph part, I understand the part to calculate horizontal/vertical asymptote, but I don't see them in the graph) – user2041143 Mar 7 '13 at 10:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.