How do you find vertical, horizontal and oblique asymptotes for #sqrt(x^2-3x) - x#?

Question

Sadie Ackerman · Accepted Answer

Horizontal asymptote :# - -3/2#. The graph has a gap, when # 0< x< 3/2 #.

#y=sqrt((x-3/2)^2-9/4)-x#, giving # |x-3/2|>=3/2#.. So, x is not in #(0, 3/2)#. Moreover, y has the form # x((1-3/x)^0.5-1)# #=x(1-3/2(1/x)-9/8(1/x^2)+ ... -1)# #=-3/2-9/8(1/x)# + higher powers of (1/x)) #to -3/2# , as x to +-oo#. #to +-oo#, as # x to +-oo#. So, #y = -3/2# is the asymptote.

Christian Antunez · Answer

undefined To find the vertical asymptotes of the function, we need to identify any values of $ x $ that make the denominator of the function equal to zero. In this case, the denominator is $ \sqrt{x^2 - 3x} - x $. Setting the denominator equal to zero and solving for $ x $ will give us the values where vertical asymptotes occur.

To find horizontal asymptotes, we analyze the behavior of the function as $ x $ approaches positive or negative infinity. If the function approaches a constant value as $ x $ goes to infinity or negative infinity, then that constant value is the horizontal asymptote.

For oblique (slant) asymptotes, we need to divide the function $ \sqrt{x^2 - 3x} - x $ by $ x $ and then simplify. If the resulting quotient approaches a non-zero constant as $ x $ goes to positive or negative infinity, then the equation of the line is the oblique asymptote.

Remember that for rational functions, oblique asymptotes exist only when the degree of the numerator is one more than the degree of the denominator.