Math Help - Sketching Graphs

1. Sketching Graphs

So I'm supposed to sketch f(x)= (-2x)/(x^2-4)
I've found that domain is all real numbers except +/-2
Intercepts are (0,0)
Asymptotes - y=0 is the line of the horizontal one

I'm not sure how to get the vertical asymptote???

Then I found the derivative of f(x), which is f'(x)= 2(x^2+4) / (x^2-4)^2

Now I need to find the critical points.. but I'm really confused at this part because x^2 can't equal -4 right? But critical points are also points that don't exist.. so do I use +/-2 as critical points??

2. Originally Posted by Jiyongie
So I'm supposed to sketch f(x)= (-2x)/(x^2-4)
I've found that domain is all real numbers except +/-2
Intercepts are (0,0)
Asymptotes - y=0 is the line of the horizontal one

I'm not sure how to get the vertical asymptote???

Then I found the derivative of f(x), which is f'(x)= 2(x^2+4) / (x^2-4)^2

Now I need to find the critical points.. but I'm really confused at this part because x^2 can't equal -4 right? But critical points are also points that don't exist.. so do I use +/-2 as critical points??
$x^2+4$ has no real roots, which means that the function is monotonous.

how did you get y=0 asymptote?

the function is discontinuous at x=2,-2.

3. If $y=f(x)$ is your curve then the vertical asymptotes are found by solving $\frac1y=0$, and the horizontal asymptotes by solving $\frac1x=0$.

In this case, $\frac1y=\frac{x^2-4}{-2x}=\frac{(x-2)(x+2)}{-2x}$, so vertical asymptotes are $x-2=0$ and $x+2=0$.

For the horizontal asymptote, write $y=\frac{-2x}{x^2-4}$ in the form $y=\frac{-2/x}{1-4/x^2}$ obtained by dividing numerator and denominator by $x^2$.

Now when you put $1/x=0$ (strictly speaking, you let $1/x$ tend to zero) you obtain $y=\frac{-2\times0}{1-4\times0^2}=0$, i.e. $y=0$ is the horizontal asymptote.

Another useful tip: try to find the sign of $f(x)$ -- it helps to sketch the curve when you know which parts lie above the $x$-axis and which parts below.

Since $f(x)=\frac{-2x}{(x-2)(x+2)}$, $f(x)$ changes sign at $x=0$, $x=2$ and at $x=-2$. You will discover (with the aid of a sign line for instance) that $f(x)$ is positive when $x<-2$ and when $0, and it is negative for $-2 and for $x>2$.

Incidentally the curve passes through the origin and is symmetrical about it (since $f(-x)=-f(x)$ we are dealing with the graph of an odd function which always has this property.)

So what does the graph look like?

1. In the region $x<-2$: the curve is above the $x$-axis; as $x$ gets more negative the curve approaches the asymptote $y=0$ from above; as $x$ gets closer to $-2$ the curve approaches the vertical asymptote $x=-2$ and it must be going up to it.

2. In the region $-2: going to the right, the curve lies below the $x$-axis, then passes through the origin, then lies above the x-axis; near the asymptote $x=-2$ the curve must be going down; near the asymptote $x=2$ the curve must go up.

3. In the region $x>2$: the curve lies below the $x$-axis and goes down to its asymptote $x=2$; as $x$ increases the curve approaches the asymptote $y=0$ from below.

The curve has no stationary points, as you found out. And although it may be monotonous the function is not monotonic.