How do you identify all horizontal and slant asymptote for #f(x)=2+5/(x^2+2)#?

Question

Lincoln Anderson · Accepted Answer

y = 2.

#y> 2# As # x to +-oo, y to 2. So, y = 2 is the horizontal asymptote. y-intercept (x = 0 ) is the maximum y, 4.5 Thus, #y in (2,4.5]#. graph{(x^2+2)(y-2)-5=0 [-20, 20, -10, 10]}

Madelyn Anderson · Answer

undefined To identify horizontal asymptotes, we need to look at the behavior of the function as x approaches positive or negative infinity. For rational functions like f(x) = 2 + 5/(x^2 + 2), horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. In this case, the degree of the numerator is 0, and the degree of the denominator is 2. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.

To find slant asymptotes, we can use polynomial long division or synthetic division to divide the numerator by the denominator. When the division yields a quotient that is a linear function, it represents a slant asymptote. In this case, when we divide 5 by (x^2 + 2), we find that the quotient approaches 0 as x approaches positive or negative infinity. Therefore, there are no slant asymptotes for the function f(x) = 2 + 5/(x^2 + 2).