Let f(x) = (x)/(x^2-4)
(a) State the y-intercept.
(b) State the vertical asymptote(s).
(c) State the horizontal asymptote.
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.
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.
To find the y-intercept of the function f(x), we need to substitute x = 0 into the function and evaluate it.
(a) Y-intercept:
To find the y-intercept, substitute x = 0 into the function f(x) = (x)/(x^2-4):
f(0) = (0)/((0)^2-4)
= 0/(-4)
= 0
Therefore, the y-intercept of the function f(x) is 0.
To find the vertical asymptotes of the function f(x), we need to determine the values of x that make the denominator equal to zero, because dividing by zero is undefined.
(b) Vertical asymptotes:
Set the denominator of f(x) equal to zero and solve for x:
x^2 - 4 = 0
(x + 2)(x - 2) = 0
Setting each factor equal to zero, we have:
x + 2 = 0, or x - 2 = 0
x = -2 or x = 2
So, the vertical asymptotes of the function f(x) are x = -2 and x = 2.
To find the horizontal asymptote of the function f(x), we need to look at the behavior of the function as x approaches positive or negative infinity.
(c) Horizontal asymptote:
To determine the horizontal asymptote, we examine the degrees of the numerator and denominator polynomials.
The degree of the numerator is 1 (since it is just x), and the degree of the denominator is 2 (x^2).
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote can be found by dividing the coefficient of the highest degree term in the numerator by the coefficient of the highest degree term in the denominator.
In this case, the coefficient of the highest degree term in the numerator is 1 and the coefficient of the highest degree term in the denominator is also 1. Therefore, the horizontal asymptote is y = 1.
So, the vertical asymptotes of the function f(x) are x = -2 and x = 2, and the horizontal asymptote is y = 1.