# Find the horizontal asymptotes of the function's graph

y = 200 / (1+10e^(0.5x))

If you let x --> minus infinty the exponential tends to zero and the function approaches 200. If x --> infinity then the exponential becomes infinite and therefore the limit is zero. So, the asymptotes are y = 200 and y = 0.

i have no clue

## To find the horizontal asymptotes of the function y = 200 / (1+10e^(0.5x)), we can observe the behavior of the function as x approaches negative infinity and positive infinity.

1. As x approaches negative infinity:

When x approaches negative infinity, the exponential term e^(0.5x) tends to 0 since any negative exponent approaches zero. This means that the denominator of the function approaches (1 + 10 * 0) = 1. So, as x approaches negative infinity, the function simplifies to y = 200 / 1 = 200.

Therefore, as x approaches negative infinity, the function approaches y = 200. This gives us one of the horizontal asymptotes.

2. As x approaches positive infinity:

When x approaches positive infinity, the exponential term e^(0.5x) increases without bound since any positive exponent becomes larger and larger. This means that the denominator of the function becomes larger and larger, approaching infinity. So, as x approaches positive infinity, the function simplifies to y = 200 / infinity = 0.

Therefore, as x approaches positive infinity, the function approaches y = 0. This gives us the second horizontal asymptote.

Hence, the horizontal asymptotes of the given function are y = 200 and y = 0.