Let f be the function given by f(x)=2x/sqrt(x^2+x+1)

b)Write an equation for each horizontal asymptote of the graph f.

Note: f'(x)=(x+2)/(x^2 +x+1)^.5

2. Originally Posted by Vigo
Let f be the function given by f(x)=2x/sqrt(x^2+x+1)

The fraction is underfined when the denominator is zero. Also, when the radical is negative. Thus, $x^2+x+1=0$ or $x^2+x+1<0$. Thus, $x^2+x+1>0$ is the domain. But $x^2+x+1$ is always positive, because the leading coefficient is positive and its discrimanat is negative. Thus, this function is defined for all real numbers.
Originally Posted by Vigo
b)Write an equation for each horizontal asymptote of the graph f.
The horizontal tangents is when the derivative approaches infinity. Since is derivative is,
$f'(x)=\frac{x+2}{\sqrt{x^2+x+1}}$ that happens when the denominator is zero. Thus, $\sqrt{x^2+x+1}=0$ thus, $x^2+x+1$ but that never happens because we said that $x^2+x+1$ is always a positive number. Thus, it has no horizontal tangents.

3. Originally Posted by ThePerfectHacker
The horizontal tangents is when the derivative approaches infinity. Since is derivative is,
$f'(x)=\frac{x+2}{\sqrt{x^2+x+1}}$ that happens when the denominator is zero. Thus, $\sqrt{x^2+x+1}=0$ thus, $x^2+x+1$ but that never happens because we said that $x^2+x+1$ is always a positive number. Thus, it has no horizontal tangents.
No a horizontal tangent is when the derivative approaches zero.

$
\frac{d}{dx}\frac{2x}{\sqrt{x^2+x+1}}=\frac{x+2}{( x^2 + x + 1)^{3/2}}$

There is a zero of this at $x=-2$, but that is not an asymtote.
Also the derivative goes to zero as $x \to \pm \infty$, so if
the limits of $f(x)$ as $x \to \pm \infty$ are finite these
will give horizontal asymtotes.

Now:

$
f(x)=\frac{2x}{\sqrt{x^2+x+1}}=\frac{2\ \mbox{sign}(x)}{\sqrt{1+1/x+1/x^2}}
$

So as $x \to \infty\ \ f(x) \to 2$, and as $x \to -\infty\ \ f(x) \to -2$, giving horizontal asymtotes at $y=\pm 2$

RonL

4. I see how you got the equation for the horizontal asympototes but when I graph the function, y=-2 does not quite fit the horizontal asymptote on the bottom. Why is that so?

5. Originally Posted by Vigo
I see how you got the equation for the horizontal asympototes but when I graph the function, y=-2 does not quite fit the horizontal asymptote on the bottom. Why is that so?
Probably because you have not extended the plot far enough to the left.
Here is a little table which shows what is going on:

$
\begin{array}{cc}x&f(x)\\-1&-2\\-2&-2.31\\-3&-2.27\\-4&-2.22\\-5&-2.18\\-10&-2.10\\-100&-2.01 \end{array}
$

RonL