Thursday

October 27, 2016
Posted by **Jessica** on Tuesday, August 4, 2009 at 3:28pm.

(a) State the y-intercept.

(b) State the vertical asymptote(s).

(c) State the horizontal asymptote.

- Algebra -
**MathMate**, Tuesday, August 4, 2009 at 4:53pmFirst 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. - Algebra -
**Jessica**, Tuesday, August 4, 2009 at 5:13pmYes 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)? - Algebra -
**MathMate**, Tuesday, August 4, 2009 at 6:13pmThe 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. - Algebra -
**Jessica**, Tuesday, August 4, 2009 at 7:46pmI got 0 for the y-intercept why -3??? and

The vertical asymptotes are x = -2,

x = 2 ???? - Algebra -
**MathMate**, Tuesday, August 4, 2009 at 8:20pmYou 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. - Algebra -
**MathMate**, Tuesday, August 4, 2009 at 8:21pm0 for the y-intercept is also correct, since there is no constant term in the function expression.