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

graph{(x^2+2)(y-2)-5=0 [-20, 20, -10, 10]}

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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).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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