Posted by Jessica on Tuesday, August 4, 2009 at 3:28pm.
First you would factorize the denominator to visualize what's going on.
f(x) = x/(x²-4) = x/((x+2)(x-2)
A vertical asymptote is a value of x whereby as the graph nears this value, the y-value approaches positive or negative infinity.
At what value(s) does the given graph approach infinity?
A horizontal asymptote is the y-value approaches as x approaches ± ∞.
What value does f(x)=x/(x²-4) take when x → ± ∞?
You will need the l'HÃ´pital's rule since both the numerator and denominator become infinity when evaluating f(∞).
Using l'HÃ´pital's rule, we calculate the derivative at both the numerator and denominator to evaluate the resulting expression, thus
Lim x→&infin 1/(2x) = 0
So the horizontal asymptotes are at y=0.
You can view a sketch of the graph at:
http://i263.photobucket.com/albums/ii157/mathmate/Jessica-1.png
This kind of problem takes a lot of practice. I hope you will work out a few more similar problems to give yourself the facility to tackle the graph sketching questions.
Yes I will need the practice without a doubt! Now you say the
The horizontal asymptote is y = 0
What about the
the y-intercept and the
vertical asymptote(s)?
The y-intercept to a function f(x) is the constant term, i.e. the term that does not contain any variables.
For example, in
f(x) = x²+4x-3
the y-intercept is -3.
In your particular case, there is no constant term, so the y-intercept is zero, as illustrated in the figure.
Inspect the denominator of the function:
f(x) = x/(x²-4) = x/((x+2)(x-2))
and figure out what values of x would make the denominator zero. These are the values where the vertical asymptotes are located. You can confirm you answer from the figure, or you can post again for confirmation.
I got 0 for the y-intercept why -3??? and
The vertical asymptotes are x = -2,
x = 2 ????
You have correctly identified the vertical asymptotes, namely x=2 and x=-2.
You may have taken the -3 out of context, which was:
For example, in
f(x) = x²+4x-3
the y-intercept is -3.
0 for the y-intercept is also correct, since there is no constant term in the function expression.