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'Hôpital's rule since both the numerator and denominator become infinity when evaluating f(∞).
Using l'Hôpital'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.
Let f(x) = (x)/(x^2-4)
(a) State the y-intercept.
(b) State the vertical asymptote(s).
(c) State the horizontal asymptote.
6 answers
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 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.
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 ????
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.
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.