How do you find the vertical, horizontal or slant asymptotes for #(4x)/(x^2-25)#?

Question

Theodore Amidon · Accepted Answer

#"vertical asymptotes at " x=+-5#

#"horizontal asymptote at " y=0# The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes. solve : #x^2-25=0rArrx^2=25rArrx=+-5# #rArrx=-5" and " x=5" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by the highest power of x, that is #x^2# #f(x)=((4x)/x^2)/(x^2/x^2-25/x^2)=(4/x)/(1-25/x^2)# as #xto+-oo,f(x)to0/(1-0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1,denominator-degree 2 ) Hence there are no slant asymptotes. graph{(4x)/(x^2-25) [-10, 10, -5, 5]}

Christian Antunez · Answer

undefined To find the vertical, horizontal, or slant asymptotes for the function $ \frac{4x}{x^2 - 25} $, follow these steps:

Vertical asymptotes:
1. Vertical asymptotes occur where the denominator of the rational function equals zero but the numerator doesn't.
2. Set the denominator $ x^2 - 25 $ equal to zero and solve for $ x $.
3. Any roots found in step 2 represent vertical asymptotes.

Horizontal asymptotes:
1. Horizontal asymptotes occur when the degree of the numerator is equal to or less than the degree of the denominator.
2. Compare the degrees of the numerator and the denominator.
3. If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $ y = 0 $.
4. If the degrees are equal, divide the leading coefficients of the numerator and the denominator to find the horizontal asymptote.

Slant asymptotes:
1. Slant asymptotes occur when the degree of the numerator is one greater than the degree of the denominator.
2. Perform polynomial long division to divide the numerator by the denominator.
3. The quotient obtained represents the equation of the slant asymptote.

Applying these steps to $ \frac{4x}{x^2 - 25} $:
1. Vertical asymptotes: Set $ x^2 - 25 = 0 $ and solve for $ x $. The solutions are $ x = 5 $ and $ x = -5 $.
2. Horizontal asymptotes: The degree of the numerator is 1 and the degree of the denominator is 2, so there is no horizontal asymptote.
3. Slant asymptotes: Since the degree of the numerator is not one greater than the degree of the denominator, there are no slant asymptotes.

In summary, the function $ \frac{4x}{x^2 - 25} $ has vertical asymptotes at $ x = 5 $ and $ x = -5 $, and no horizontal or slant asymptotes.