Question 142854: g(x)=5x/x^2+1
Can you help me find the verticle and horizontal asymptote I know y = none for the horizontal is this correct? Then what is the verticle?
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! First, let's make sure we are dealing with the correct equation:
g(x)=5x/(x^2+1) is this correct? If not, this answer will be incorrect.
Vertical asymptotes correspond to the zeros of the denominator of a rational function. Can the denominator in this example ever equal 0 when using real numbers? No, x^2 is always positive, thus the denominator will always be >0.
No Vertical Asymptote.
For horizontal asymptotes, we look at the degree of the highest x term in both the numerator and denominator. There are 3 cases:
"# If the degree (the largest exponent) of the denominator is bigger than the degree of the numerator, the horizontal asymptote is the x-axis (y = 0).
# If the degree of the numerator is bigger than the denominator, there is no horizontal asymptote.
# If the degrees of the numerator and denominator are the same, the horizontal asymptote equals the leading coefficient (the coefficient of the largest exponent) of the numerator divided by the leading coefficient of the denominator"
The degree of the numerator for our problem here is 1, the denominator's degree is 2. Thus, the first case applies (and is reflected in the graph) that y=0 is a horizontal asymptote.
|
|
|
