Determine the zeroes, and the domain. Write the equation for each asymptote. Then graph the function and estimate the range.
g(x)=$\displaystyle x/x^2-4$ h(x)=$\displaystyle x^2-4/x$
Thank you
First factor the numerator and denominator and see if anything cancels out.
$\displaystyle g(x) = \frac{x}{(x + 2)(x - 2)}$
No cancellations.
So to find vertical asymptotes find out where the denominator is equal to 0. This gives x = 2 and x = -2 as vertical asymptotes.
This also gives the domain as all real numbers except x = 2, and -2.
As far as the zeros are concerned, solve
$\displaystyle g(x) = \frac{x}{(x + 2)(x - 2)} = 0$
This has a solution of x = 0, so there is your zero.
Is there a horizontal asymptote? For that we need to see what the behavior of g(x) is for very large x. I think it is easy to see that as x goes to either plus or minus infinity that g(x) goes to 0. So there is a horizontal asymptote at y = 0.
We do not have a slant asymptote because the degree of the numerator is not one more than the degree of the denominator.
I think that about covers it. I'll leave you to graph it yourself.
-Dan
Hello,
some remarks about the function h:
1. $\displaystyle h(x) = \frac{x^2-4}{x} = x-\frac4x = \frac1{g(x)}~,~x\ \in \ \mathbb{R} \setminus \{0\}$
Therefore: The zeros of g indicates the vertical asymptotes of h.
The vertical asymptotes of g pass through the zeros of h.
2. h has a slanted asymptote y = x and a vertical asymptote at x = 0
3. h has 2 zeros: x = -2, x = 2
4. The graph of h is drawn in red, the asymptotes in brown.
The blue graph with it's green asymptotes is the graph of g.