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How do you find the vertical, horizontal or slant asymptotes for #9/(1+2e^-x)#?

Answer 1

Sophie Adams

Horizontal asymptotes: #y=0 larr and y=9 rarr#. See the illustrative graph.

#x in (-oo, oo) and y in (0, 9)#

y-intercept ( x = 0 ) ; 3

y to 0#, as # x to -oo and y to 9#, as #x to oo#.

graph{y(1+2e^(-x))-9=0 [-35.48, 35.47, -17.66, 17.81]}

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Answer 2

Anna Allen

To find the vertical asymptote, set the denominator equal to zero and solve for ( x ). There is no vertical asymptote for the given function.

To find the horizontal asymptote, examine the behavior of the function as ( x ) approaches positive or negative infinity. As ( x ) approaches positive or negative infinity, ( e^{-x} ) approaches zero, so the function approaches ( \frac{9}{1+0} = 9 ). Therefore, the horizontal asymptote is ( y = 9 ).

To find the slant asymptote, divide the numerator by the denominator using polynomial division. Since the degree of the numerator is less than the degree of the denominator, there is no slant asymptote for the given function.

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Answer from HIX Tutor

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.