How do you find the vertical, horizontal and slant asymptotes of: #f(x) = (x - 3) / (x + 2)#?

Question

Sadie Alexander · Accepted Answer

vertical asymptote x = -2
horizontal asymptote y = 1

The denominator of f(x) cannot be zero as this would make f(x) undefined.Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve: x +2 = 0 #rArrx=-2" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by x #(x/x-3/x)/(x/x+2/x)=(1-3/x)/(1+2/x)# as #xto+-oo,f(x)to(1-0)/(1+0)# #rArry=1" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1 ) Hence there are no slant asymptotes. graph{(x-3)/(x+2) [-20, 20, -10, 10]}

Sadie Alexander · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for x. In this case, the denominator $x + 2$ equals zero when $x = -2$. So, the vertical asymptote is $x = -2$.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since both have the same degree (1), divide the leading coefficients. In this case, the leading coefficient of the numerator is 1, and the leading coefficient of the denominator is also 1. Therefore, the horizontal asymptote is $y = \frac{1}{1} = 1$.

To find the slant asymptote (if it exists), perform polynomial long division between the numerator and the denominator. After dividing, the quotient represents the equation of the slant asymptote. However, in this case, the degree of the numerator is less than the degree of the denominator, so there is no slant asymptote.