# Thread: graphing with asymptotes help

1. ## graphing with asymptotes help

hey guys!

I just want to make sure what i'm doing is right:

to find horizontal asymptote, you look at the degree of the numerator and denominator. If they are the same degree, its the numerators coefficient divided by the denominators coefficient

to find vertical asymptote, find when denominator = 0 (factor)

What i DONT get is in the attached image, the numerator is -4x^3, but in the graph, it shows 4x^3 as if it was not negative.

the second thing i don't understand is before and after the asymptotes, how do you know (the part that curves down at -3 and then comes back up at 3) that it is underneath the asymptote line and not above it?

2. Originally Posted by snypeshow
hey guys!

I just want to make sure what i'm doing is right:

to find horizontal asymptote, you look at the degree of the numerator and denominator. If they are the same degree, its the numerators coefficient divided by the denominators coefficient

to find vertical asymptote, find when denominator = 0 (factor)

(A) What i DONT get is in the attached image, the numerator is -4x^3, but in the graph, it shows 4x^3 as if it was not negative.

(B) the second thing i don't understand is before and after the asymptotes, how do you know (the part that curves down at -3 and then comes back up at 3) that it is underneath the asymptote line and not above it?
to (A): Re-write the equation of the function: $f(x)=\dfrac{-4x^3}{(x-3)(x+3)^2}$
Now examine the signs of the values of f(x) for
$-3
or
$0\leq x <3$

By the way $h(x)=4x^3$ has no vertical asymptote so the given graph looks quite different to h.

to (B): Use long division to re-write the equation of the function:

$f(x)=-4-\dfrac{12(x^2-3x-9)}{(x-3)(x+3)^2}$

Now examine the sign of the remainder:

If x < -3 then the numerator of the remainder is positiv and the denominator is negative. That means you subtract a fraction from -4 so that the graph runs below the asymptote.

If x > 4.85 the remainder is positive so that the graph runs above the asymptote. The peak is at P(9, -3.75). The given drawing is definitely wrong.