What are the asymptotes and removable discontinuities, if any, of #f(x)=(3x^2+x-5)/(x^2+1) #?

Question

What are the asymptotes and removable discontinuities, if any, of #f(x)=(3x^2+x-5)/(x^2+1)  #?

Eva Adamson · Accepted Answer

#"horizontal asymptote at " y=3#

The values that x cannot be are obtained by equating the denominator to zero and solving the problem; if the numerator is non-zero for these values, they are vertical asymptotes. The denominator of f(x) cannot be zero since this would render f(x) undefined.

#"solve " x^2+1=0rArrx^2=-1#

There are no vertical asymptotes because there are no true solutions to this.

#"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)=((3x^2)/x^2+x/x^2-5/x^2)/(x^2/x^2+1/x^2)=(3+1/x-5/x^2)/(1+1/x^2)#

as #xto+-oo,f(x)to(3+0-0)/(1+0)#

#rArry=3" is the asymptote"#

There are no removable discontinuities (holes) in this graph{(3x^2+x-5)/(x^2+1) [-10, 10, -5, 5]} because there are no duplicate factors on the numerator or denominator.

Eva Adamson · Answer

undefined The function $ f(x) = \frac{3x^2 + x - 5}{x^2 + 1} $ has no asymptotes but has one removable discontinuity.

To find the vertical asymptotes, we look for values of $ x $ that make the denominator $ x^2 + 1 $ equal to zero. Since $ x^2 + 1 = 0 $ has no real solutions (the minimum value of $ x^2 $ is 0 and adding 1 to it results in a positive value), there are no vertical asymptotes.

To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Both have the highest degree of 2. So, we divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote. In this case, it's $ \frac{3}{1} = 3 $. Therefore, the horizontal asymptote is $ y = 3 $.

To find any removable discontinuities, we look for values of $ x $ that make both the numerator and denominator equal to zero. Setting the numerator $ 3x^2 + x - 5 = 0 $ equal to zero, we find no real solutions. Thus, there are no removable discontinuities.

In summary, the function $ f(x) = \frac{3x^2 + x - 5}{x^2 + 1} $ has no vertical asymptotes, a horizontal asymptote at $ y = 3 $, and no removable discontinuities.