How do you find the vertical, horizontal or slant asymptotes for #(2x)/(x^2+16)#?

Question

Sadie Ackerman · Accepted Answer

#"horizontal asymptote at " y=0#

#f(x)=(2x)/(x^2+16)#

The denominator of f(x) cannot be zero as this would make f(x) 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+16=0rArrx^2=-16#

#"this has no real solutions hence there are no"# #"vertical 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)=((2x)/x^2)/(x^2/x^2+16/x^2)=(2/x)/(1+16/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{(2x)/(x^2+16) [-10, 10, -5, 5]}

Serenity Jones · Answer

undefined To find the vertical asymptotes of the function $ f(x) = \frac{2x}{x^2 + 16} $, we need to identify the values of $ x $ for which the denominator becomes zero. Setting the denominator equal to zero, we get:

$$ x^2 + 16 = 0 $$

Solving this equation, we find that there are no real solutions, which means there are no vertical asymptotes.

To find the horizontal asymptote, we examine the behavior of the function as $ x $ approaches positive or negative infinity. As $ x $ becomes large, the terms involving $ x $ in the numerator and the denominator dominate the function. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $.

To determine if there are any slant asymptotes, we perform polynomial long division or use partial fraction decomposition to rewrite the function. After simplifying, if the resulting expression has a linear term that cannot be cancelled out, there will be a slant asymptote. In this case, performing polynomial long division:

$$ \frac{2x}{x^2 + 16} = \frac{2x}{(x - 0)(x^2 + 16)} = \frac{2x}{x(x^2 + 16)} $$

Thus, there is no slant asymptote.