document.write( "Question 1001042: Basic question. f(x) equals y because plugging in some value x into the function will give you what the corresponding y-value will be. And the y-value of a LINEAR FUNCTION will always be the same thing as its horizontal asymptote?\r
\n" ); document.write( "\n" ); document.write( "I know the function will output differently for a rational function, and then the rules change. but for something like lim x -> ∞ where the function is f(x) = 5+e^(-x^2). By plugging in ∞ into the function this will output the horizontal asymptote because the horizontal asymptote is the same as y-value. In this case y = 5 and that is the horizontal asymptote.\r
\n" ); document.write( "\n" ); document.write( "not sure if this is right or not.\r
\n" ); document.write( "\n" ); document.write( "Please help!
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Algebra.Com's Answer #618337 by ikleyn(52793)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "Below is the plot of the functions  y1 = \"5+%2B+e%5E%28-x%5E2%29\"  (in red)  and  y2 = \"e%5E%28-x%5E2%29\"  (in green)  with the horizontal asymptote  y = 5  (in blue).\r
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\n" ); document.write( "\n" ); document.write( "Figure. Plots y1 = \"5+%2B+e%5E%28-x%5E2%29\" and y2 = \"e%5E%28-x%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "The function  y2 = \"e%5E%28-x%5E2%29\"  tends to zero at  x ---> +/-\"infinity\".\r
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\n" ); document.write( "\n" ); document.write( "Therefore,  the function  y1 = \"5+%2B+e%5E%28-x%5E2%29\"  tends to  5 at  x ---> +/-\"infinity\".\r
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\n" ); document.write( "\n" ); document.write( "y = 5  is the horizontal asymptote for the function  y1.\r
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\n" ); document.write( "\n" ); document.write( "Notice that the function  y1  is not a rational function,  as well as the function  y2.\r
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