# Graphing simple fractional function

I am trying to graph $$f(x) = \frac{x^2-4}{x^2+4}$$

It seems pretty simple to me but I can't finish it correctly.

I know that there is a horizontal asymptote at $1$ beacuse the degrees on the variable at the same and $\frac{x}{x}$ is 1.

I know that it is a negative function until zero, and then positive because the only critical number of the derivative $$\frac{16x}{(x^2+4)^2}$$ is going to be zero since the denominator can't be zero and the only root of the top is $0$.

This then tells me that there is no local max but only a minimum which is at $0$ which gives me $-1$.

Trying to find concavity $$\frac{16(x^2+4)^2 - 16x (4x(x^2+4))}{(x^2+4)^4}= \frac {-64x^5 + 16x^4 + 256x^3 + 128x^2 + 256}{(x^2+4)^4}$$ Which I have no idea how to really work with I can't see to get anything workable out of that.

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Have you considered the zeros of $f(x)$? – Mark Bennet Apr 6 '12 at 21:40
Or that $f(x)=f(-x)$? – Mark Bennet Apr 6 '12 at 21:41
Your second derivative has an incorrect sign (you should have $x^2+4$ in the second summand, not $x^2-4$); factor out $x^2+4$ in the numerator and cancel it with one of the factors in the denominator before simplifyin. Then simplify some more. – Arturo Magidin Apr 6 '12 at 22:02
The GPA without the knowledge it supposedly represents is worthless. – Arturo Magidin Apr 6 '12 at 22:03
Colleges care about GPA; but unless you plan to be a career student, the GPA will not matter later. A good GPA without the knowledge it supposedly represent may fool someone into hiring you, but won't keep them from firing you if you don't exhibit the knowledge that is needed. – Arturo Magidin Apr 6 '12 at 22:08

Your second derivative looks wrong. You should get:

$$\frac{64-48x^{2}}{(x^2+4)^{2}}$$

If you would mind showing your work on taking the derivative, that would be helpful. Did you use quotient rule or rewrite and use product rule?

From that, you can test a few points to see where the graph is concave up or concave down.

The graph should look something like that (to give you an idea for test interval points if you aren't aware of what to note when doing the second derivative test). The picture was taken from Wolfram.

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I am almost certain my derivative is correct since that is what wolfram gives me. Just to keep it realistic I had wolfram multiply out everything like I was entering in the quotient rule by hand. That way it wouldn't reduce it so much. It matched what I did on my own. – user138246 Apr 6 '12 at 21:53
Are you sure you typed it in correctly? Check wolframalpha.com/input/…. – Joe Apr 6 '12 at 21:55
The second derivative is $16\frac{4-3x^2}{(4+x^2)^3}$. Don't really care about Alpha's opinion on the matter. – André Nicolas Apr 6 '12 at 21:55

Correct second derivative: \begin{align*} f(x) &= \frac{x^2-4}{x^2+4}\\ f'(x) &= \frac{(x^2+4)(x^2-4)' - (x^2-4)(x^2+4)'}{(x^2+4)^2}\\ &= \frac{(x^2+4)(2x) - (x^2-4)(2x)}{(x^2+4)^2}\\ &= \frac{2x(x^2+4-x^2+4)}{(x^2+4)^2}\\ &= \frac{16x}{(x^2+4)^2}.\\ f''(x) &= \frac{(x^2+4)^2(16x)' - 16x\Bigl( (x^2+4)^2\Bigr)'}{\Bigl((x^2+4)^2\Bigr)^2}\\ &= \frac{16(x^2+4)^2 - 16x(2(x^2+4)(x^2+4)')}{(x^2+4)^4}\\ &= \frac{16(x^2+4)^2 - 16x(2)(x^2+4)(2x)}{(x^2+4)^4}\\ &= \frac{16(x^2+4)\Bigl( x^2+4 - 4x^2\Bigr)}{(x^2+4)^4}\\ &= \frac{16(4-3x^2)}{(x^2+4)^3}. \end{align*} The denominator is never zero; the numerator is zero exactly when $4-3x^2=0$. This is a parabola that opens down, so it will be positive between the roots and negative before and after both roots.

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This particular function and its derivatives can benefit from using several perspectives: $$f(x)=\frac{x^2-4}{x^2+4}=\frac{(x+2)(x-2)}{x^2+4}=1-8(x^2+4)^{-1}$$ $$f\,'(x)=16x(x^2+4)^{-2}$$ $$f\,''(x)=16(x^2+4)^{-3}(4-3x^2)=-16\frac{3x^2-4}{x^2+4}=-48\frac{\left(x+\frac2{\sqrt3}\right)\left(x-\frac2{\sqrt3}\right)}{x^2+4}$$ Since $x^2+4$ is always positive, the factored forms are workable -- they give you all the critical points and allow you to draw number lines below. $$\matrix{ x & -\infty & -2 & -\tfrac2{\sqrt3} & 0 & \tfrac2{\sqrt3} & 2 & \infty \\ f &+&0&-&-&-&0&+\\ f' &-&-&-&0&+&+&+\\ f'' &-&-&0&+&0&-&-\\ }$$ The columns for $\pm\infty$ are meant to represent the signs for $x$ sufficiently large, i.e. past the last finite critical point of $f$ or one of its derivatives; for example, the $-48$ above tells us $f\,''$ should be negative for $x$ larger than all of its roots.

From this number line, you can infer the graphical behavior and make a rough sketch.

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