How do you find any asymptotes of #h(x)=(x-5)/(x^2+2x-4)#?

Question

Evelyn Agard · Accepted Answer

By (a) seeing if the denominator is zero for any value(s) of #x#, and by examining the behaviour for large positive and negative values of #x#.

The denominator is zero where #x=-1+-sqrt 5# so there #h(x)# has vertical asympototes at #x=-1+sqrt 5# and #x=-1=sqrt 5#. For large #x# positive or negative #h(x)# approximates to #1/x# (#h(x)to 0# as #x to +- oo#), as you can see by ignoring the #-5# and the #2x-4# for large #x# magnitudes. So the #x# axis is another asymptote.

Emily Ammerman · Answer

undefined To find the asymptotes of the function h(x) = (x-5)/(x^2+2x-4), we need to consider the behavior of the function as x approaches positive or negative infinity.

First, we check for vertical asymptotes by finding the values of x that make the denominator equal to zero. Solving x^2+2x-4 = 0, we find that x = -1 ± √5. Therefore, there are two vertical asymptotes at x = -1 + √5 and x = -1 - √5.

Next, we check for horizontal asymptotes by examining the degrees of the numerator and denominator. Since the degree of the numerator is 1 and the degree of the denominator is 2, we have a slant asymptote instead of a horizontal asymptote. To find the slant asymptote, we perform long division or synthetic division to divide the numerator by the denominator.

Performing the division, we get h(x) = 1 - (3x + 11)/(x^2 + 2x - 4). The quotient is 1 and the remainder is -(3x + 11)/(x^2 + 2x - 4). As x approaches positive or negative infinity, the remainder approaches zero. Therefore, the slant asymptote is y = 1.

In summary, the function h(x) = (x-5)/(x^2+2x-4) has two vertical asymptotes at x = -1 + √5 and x = -1 - √5, and a slant asymptote at y = 1.